Free math worksheets

- Read the bar values in order. - Look for a change that repeats. - Check the pattern using more than one pair of bars.

1. The bar values are \(10, 15, 20, 25,\) and \(30\). 2. Each value is \(5\) greater than the value before it.

The number of math puzzles solved increases by \(5\) each day.

- Read the bar values in order. - Find the change that repeats from day to day. - Apply the same change to Thursday’s value.

1. The values increase by \(4\) each day: \(12, 16, 20, 24\). 2. The next value is \(24+4=28\).

Friday’s bar should have a value of \(28\).

- Compare the last digit of each number in both sequences. - Subtract one pair of numbers in the same position. - Compare the hundreds places in the two sequences.

1. Every number in both sequences has \(7\) in the ones place. 2. Subtract numbers in the same position: for example, \(207-7=200\) and \(217-17=200\). 3. Each number in Sequence B is \(200\) greater than the number in the same position in Sequence A.

a) Every number has \(7\) in the ones place. b) Each number in Sequence B is \(200\) greater than the number in the same position in Sequence A.

- Recall which digits can appear in the ones place of an even number. - Find the first even number after \(777\). - Count forward by twos.

1. The first even number greater than \(777\) is \(778\). 2. Add \(2\) repeatedly to generate the remaining even numbers. 3. The last even number less than \(793\) is \(792\).

\(778, 780, 782, 784, 786, 788, 790, 792\)

- Recall which digits can appear in the ones place of an odd number. - Find the odd number that comes after \(85\). - Count forward by twos and pay attention to the change from \(99\) to \(101\).

1. Start with the odd number \(85\). 2. Add \(2\) each time to find the next odd number. 3. Continue through the change from \(99\) to \(101\) until you reach \(105\).

\(85, 87, 89, 91, 93, 95, 97, 99, 101, 103, 105\)

- Find the difference between consecutive terms. - Determine how much must be added to \(225\) to reach \(250\). - Check whether that difference stays the same. - Continue using the same change until you reach \(375\).

1. The difference between the first two terms is \(250-225=25\), and \(275-250=25\). 2. The rule is to add \(25\) each time. 3. The missing terms are \(275+25=300\), \(300+25=325\), and \(325+25=350\). 4. The next term is \(350+25=375\), which matches the given endpoint.

The missing numbers are \(300, 325,\) and \(350\). Rule: add \(25\) each time.

- List several positive multiples of \(5\) and \(10\). - Look at the ones digit of each multiple. - Compare the two lists to find the values they share.

1. The positive multiples of \(5\) are \(5, 10, 15, 20, 25, 30, \ldots\). Their ones digits alternate between \(5\) and \(0\). 2. Every positive multiple of \(10\) has a ones digit of \(0\). 3. Every multiple of \(10\) is also a multiple of \(5\). Through \(100\), the common multiples are \(10, 20, 30, 40, 50, 60, 70, 80, 90\), and \(100\).

a) \(0\) and \(5\) b) \(0\) c) \(10, 20, 30, 40, 50, 60, 70, 80, 90, 100\)

- Compare the first factors in each pair. - How can the known product help you find the other product? - What happens to a product when one factor is doubled or halved and the other factor stays the same?

1. \(3 \times 4 = 12\). Doubling the first factor from \(3\) to \(6\) doubles the product, so \(6 \times 4 = 24\). 2. \(10 \times 7 = 70\). Halving the first factor from \(10\) to \(5\) halves the product, so \(5 \times 7 = 35\). 3. \(4 \times 8 = 32\). Doubling the first factor from \(4\) to \(8\) doubles the product, so \(8 \times 8 = 64\).

a) \(3 \times 4 = 12\) and \(6 \times 4 = 24\); the first factor and the product are doubled. b) \(10 \times 7 = 70\) and \(5 \times 7 = 35\); the first factor and the product are halved. c) \(4 \times 8 = 32\) and \(8 \times 8 = 64\); the first factor and the product are doubled.

- List the 3s products and check which also appear in the 6s facts. - To keep the product the same when the factor \(3\) becomes \(6\), what must happen to the other factor? - How are \(3\) and \(6\) related?

1. Common products in the 3s and 6s facts include \(6, 12, 18, 24\), and \(30\). 2. For product \(6\), \(2 \times 3 = 6\) and \(1 \times 6 = 6\). 3. For product \(12\), \(4 \times 3 = 12\) and \(2 \times 6 = 12\). 4. For product \(18\), \(6 \times 3 = 18\) and \(3 \times 6 = 18\). 5. Other correct pairs are possible.

One possible set is: \(2 \times 3 = 6\) and \(1 \times 6 = 6\) \(4 \times 3 = 12\) and \(2 \times 6 = 12\) \(6 \times 3 = 18\) and \(3 \times 6 = 18\)

- Find how much each term decreases. - Think about the number that is \(2\) less than \(700\). - Check that you wrote exactly eight new terms.

1. Each term is \(2\) less than the term before it. 2. Starting from \(708\), subtract \(2\) eight times. 3. The sequence passes from \(700\) to \(698\).

\(706, 704, 702, 700, 698, 696, 694, 692\)

- What happens when you take exactly one group of a quantity? - What happens when you take zero groups of a quantity? - In each product with a factor of \(1\), which number remains unchanged?

1. The products are: a) \(1 \times 9 = 9\) and \(0 \times 9 = 0\); b) \(14 \times 1 = 14\) and \(14 \times 0 = 0\); c) \(1 \times 1 = 1\) and \(0 \times 1 = 0\); d) \(0 \times 25 = 0\) and \(1 \times 25 = 25\). 2. Multiplying by \(1\) leaves the other factor unchanged, while multiplying by \(0\) gives a product of \(0\).

a) \(9\) and \(0\) b) \(14\) and \(0\) c) \(1\) and \(0\) d) \(0\) and \(25\) Pattern: A number multiplied by \(1\) stays the same, and a number multiplied by \(0\) has a product of \(0\).

- Find the amount you count by in each part. - Do not include the starting or ending number. - Keep adding the same amount until the next number would reach or pass the end.

1. Count by \(100\) from \(100\) through \(900\). 2. Count by \(10\) from \(470\) through \(530\). 3. Count by \(50\) from \(750\) through \(900\).

a) \(100, 200, 300, 400, 500, 600, 700, 800, 900\) b) \(470, 480, 490, 500, 510, 520, 530\) c) \(750, 800, 850, 900\)

- Compare the first two terms in each sequence. - Decide whether the terms increase or decrease. - Find the amount of the repeated change. - In b), pay attention to the tens and hundreds transition.

1. Sequence a) increases by \(10\), giving \(700, 710, 720, 730, 740,\) and \(750\). 2. Sequence b) decreases by \(2\), giving \(306, 304, 302, 300, 298,\) and \(296\). 3. Sequence c) increases by \(50\), giving \(600, 650, 700, 750,\) and \(800\).

a) \(670, 680, 690, 700, 710, 720, 730, 740, 750\) b) \(312, 310, 308, 306, 304, 302, 300, 298, 296\) c) \(450, 500, 550, 600, 650, 700, 750, 800\)

- Find the change between two adjacent known terms. - Check whether the same change works throughout the sequence. - Verify each missing term using the next known term.

1. Sequence a) increases by \(25\). The missing terms are \(150+25=175\) and \(225+25=250\). 2. Sequence b) decreases by \(20\). The missing terms are \(820-20=800\) and \(780-20=760\).

a) \(175\) and \(250\) b) \(800\) and \(760\)

- Find the change from the first term to the second term. - Check whether the same change works from the second term to the third. - State the repeated rule in words. - Count the terms carefully so that each sequence has six terms.

1. In a), each term is \(15\) greater than the previous term. Continue with \(60, 75,\) and \(90\). 2. In b), each term is \(30\) less than the previous term. Continue with \(820, 790,\) and \(760\).

a) Rule: add \(15\). Sequence: \(15, 30, 45, 60, 75, 90\) b) Rule: subtract \(30\). Sequence: \(910, 880, 850, 820, 790, 760\)

- Compare each expression with the one above it. - Focus on the place value that changed. - Use the previous sum to update the next one.

1. \(300 + 400 = 700\). 2. The first addend increases by \(50\), so \(350 + 400 = 750\). 3. The second addend increases by \(20\), so \(350 + 420 = 770\). 4. The first addend increases by \(8\), so \(358 + 420 = 778\). 5. The second addend increases by \(1\), so \(358 + 421 = 779\).

\(300 + 400 = 700\) \(350 + 400 = 750\) \(350 + 420 = 770\) \(358 + 420 = 778\) \(358 + 421 = 779\)

- Compare each expression with the previous one. - Identify whether the first number or the number being subtracted changed and by how much. - Adjust the previous difference instead of starting over.

1. \(800 - 300 = 500\). 2. The first number increases by \(70\), so \(870 - 300 = 570\). 3. The number being subtracted increases by \(40\), so \(870 - 340 = 530\). 4. The first number increases by \(9\), so \(879 - 340 = 539\). 5. The number being subtracted increases by \(5\), so \(879 - 345 = 534\).

\(800 - 300 = 500\) \(870 - 300 = 570\) \(870 - 340 = 530\) \(879 - 340 = 539\) \(879 - 345 = 534\)

- Calculate all three differences. - Compare the numbers being subtracted. - Relate the change in the number being subtracted to the change in the difference.

1. \(1000 - 20 = 980\). 2. \(1000 - 40 = 960\). 3. \(1000 - 60 = 940\). 4. The number being subtracted increases by \(20\) each time, so the difference decreases by \(20\) each time.

a) \(980\) b) \(960\) c) \(940\) The differences decrease by \(20\) each time.

- Increase the first factor by \(1\) in each new expression. - Compare expressions with the same first factor across the two sequences. - Relate \(80\) to \(40\).

1. Sequence A continues with \(6 \times 40 = 240\), \(7 \times 40 = 280\), and \(8 \times 40 = 320\). The products increase by \(40\). 2. Sequence B continues with \(6 \times 80 = 480\), \(7 \times 80 = 560\), and \(8 \times 80 = 640\). The products increase by \(80\). 3. Each product in Sequence B is twice the matching product in Sequence A because \(80\) is twice \(40\).

Sequence A: \(6 \times 40 = 240\), \(7 \times 40 = 280\), \(8 \times 40 = 320\) Sequence B: \(6 \times 80 = 480\), \(7 \times 80 = 560\), \(8 \times 80 = 640\) Each product in Sequence B is twice the matching product in Sequence A.

- Notice what changes from the first expression to the second. - Use the first two expressions to combine both changes in the third. - Keep track of whether you are adding tens or ones.

1. a) \(450 + 30 = 480\) and \(450 + 6 = 456\). Since \(36 = 30 + 6\), \(450 + 36 = 486\). 2. b) \(720 + 5 = 725\) and \(720 + 40 = 760\). Since \(45 = 40 + 5\), \(720 + 45 = 765\).

a) \(480\), \(456\), \(486\) b) \(725\), \(760\), \(765\)

- Subtract in steps: remove the tens first, then the ones. - Watch what happens to the tens place when you subtract the ones. - The third expression combines the two separate subtractions.

1. a) \(960 - 40 = 920\) and \(960 - 8 = 952\). Since \(48 = 40 + 8\), \(960 - 48 = 912\). 2. b) \(390 - 70 = 320\) and \(390 - 2 = 388\). Since \(72 = 70 + 2\), \(390 - 72 = 318\).

a) \(920\), \(952\), \(912\) b) \(320\), \(388\), \(318\)

- Counting backward means each term is smaller than the term before it. - Subtract the stated amount each time.

1. In a), subtract \(5\) repeatedly after \(710\): \(705, 700, 695, 690, 685, 680\). 2. In b), subtract \(50\) repeatedly after \(450\): \(400, 350, 300, 250, 200\).

a) \(720, 715, 710, 705, 700, 695, 690, 685, 680\) b) \(550, 500, 450, 400, 350, 300, 250, 200\)

- Compare each first number with the one before it. - The second number stays the same. How should the answer change? - Find how much greater \(400\) is than \(370\).

1. The first number increases by \(10\) each time. 2. The second number stays the same, so the answer also increases by \(10\) each time. 3. From \(370\) to \(400\) is an increase of \(30\). 4. Increase the known answer by \(30\): \(490+30=520\).

a) The first number increases by \(10\) each time. b) The answer increases by \(10\) each time. c) \(400+120=520\)

- Test each change with an equation such as \(100 - 20 = 80\). - What happens when the amount you start with increases? - What happens when the amount you subtract increases?

1. Increasing the first number by \(65\) increases the distance between the two numbers by \(65\), so the difference increases by \(65\). 2. Increasing the number being subtracted by \(65\) means \(65\) more is subtracted, so the difference decreases by \(65\).

a) The difference increases by \(65\). b) The difference decreases by \(65\).

- Subtract the skateboard cost from the amount saved. - For part b), decide how starting with more money changes the remainder. - For part c), decide how spending more changes the remainder. - Use your answer from part a) instead of starting over.

1. Subtract the cost from the amount saved: \(\$350 - \$120 = \$230\). 2. If Anna started with \(\$100\) more, the remainder would also be \(\$100\) more: \(\$230 + \$100 = \$330\). 3. If the skateboard cost \(\$20\) more, the remainder would be \(\$20\) less: \(\$230 - \$20 = \$210\).

a) \(\$230\) b) \(\$330\) c) \(\$210\)

- Compare the two starting amounts. - Both bakers sell the same number of rolls. - Use the difference in their starting amounts to adjust the first answer.

1. The first baker has \(150 - 65 = 85\) rolls left. 2. The second baker starts with \(20\) fewer rolls but sells the same number, so the second remainder is also \(20\) less. 3. The second baker has \(85 - 20 = 65\) rolls left.

a) \(85\) rolls remain. b) \(65\) rolls remain.

- Identify the number being subtracted. - Think about what happens when more is subtracted. - Test the pattern with a simple equation such as \(10 - 2 = 8\).

1. When the number being subtracted increases, more is subtracted, so the difference decreases. 2. The difference decreases by the same amount, \(60\). 3. The new difference is \(350 - 60 = 290\).

The difference decreases by \(60\), so its new value is \(290\).

- Calculate the three differences first. - Compare consecutive results as the first number increases. - In part c), first find the new number being subtracted after increasing \(100\) by \(50\). - Decide whether subtracting more makes the difference greater or smaller.

1. The differences are \(400 - 100 = 300\), \(450 - 100 = 350\), and \(500 - 100 = 400\). 2. Each time the first number increases by \(50\), the difference also increases by \(50\). 3. Increasing the number being subtracted in the original equation gives \(400 - 150 = 250\). 4. Compared with \(300\), the difference decreases by \(50\).

a) \(300\), \(350\), \(400\) b) The difference increases by \(50\). c) The difference decreases by \(50\), from \(300\) to \(250\).

- Subtract the amount saved from the goal. - For part b), decide how saving more changes the remaining amount. - For part c), decide how a higher goal changes the remaining amount.

1. The amount still needed is \(\$450 - \$280 = \$170\). 2. Saving \(\$30\) more reduces the amount still needed by \(\$30\): \(\$170 - \$30 = \$140\). 3. Increasing the cost by \(\$30\) increases the amount still needed by \(\$30\): \(\$170 + \$30 = \$200\).

a) \(\$170\) b) \(\$140\) c) \(\$200\)

- Find how many cards Luke has left. - Find how many cards Maya has left. - Compare the two starting amounts and the two amounts given away. - What happens to a difference when both numbers increase by the same amount?

1. Luke has \(360 - 140 = 220\) cards left. 2. Maya has \(390 - 170 = 220\) cards left. 3. They have the same number of cards left. 4. Maya started with \(30\) more cards but also gave away \(30\) more cards, so the difference stays the same.

They each have \(220\) cards left.

- Compare the first numbers in the two equations. - What happens to a difference when the number being subtracted increases? - Find the difference between \(180\) and \(140\).

1. Both equations have the same first number, \(500\). 2. Ben subtracts \(180\), which is \(40\) more than \(140\). 3. Subtracting more produces a smaller difference, so Anya's difference is greater. 4. The two differences differ by \(180 - 140 = 40\).

Anya gets the greater difference. Her result is \(40\) greater than Ben's.

- Think about what happens when you start with more but subtract the same amount. - Think about what happens when you subtract more from the same amount. - Compare each new result with \(350\).

1. Increasing the first number gives \(620 - 230 = 390\), so the difference increases by \(40\). 2. Increasing the number being subtracted gives \(580 - 270 = 310\), so the difference decreases by \(40\).

a) The difference increases by \(40\), to \(390\). b) The difference decreases by \(40\), to \(310\).

- Find all four sums. - Compare the results. - Look for how a change in one addend is balanced by a change in the other.

1. The sums are \(260 + 380 = 640\), \(450 + 190 = 640\), \(170 + 470 = 640\), and \(540 + 100 = 640\). 2. Each pair of addends has a total of \(640\). When one addend changes, the other changes by the opposite amount, preserving the sum.

a) \(640\) b) \(640\) c) \(640\) d) \(640\) All four sums are equal.

- Think about what happens when you subtract less. - Test the pattern with \(10 - 5 = 5\). - Decide whether the difference changes in the same or opposite direction as the number being subtracted.

1. Decreasing the number being subtracted means less is subtracted. 2. Therefore, the difference increases by \(40\). 3. The new difference is \(380 + 40 = 420\).

The new difference is \(420\).

- Try two whole numbers that are next to each other. - Is one even and the other odd? - What kind of number do you get when you add an even number and an odd number?

1. Two whole numbers that are next to each other have different types: one is even and the other is odd. 2. An even number plus an odd number is odd. For example, \(4+5=9\). 3. Therefore, the statement is false.

False. One of two consecutive whole numbers is even and the other is odd, so their sum is odd. For example, \(4+5=9\).

- Find the value between two labeled ticks. - Count how many equal intervals divide that distance. - Find the value of one small interval. - Count forward from the nearest labeled value to each marker.

1. The distance from \(0\) to \(200\) is divided into \(10\) equal intervals, so each small interval represents \(20\). 2. Point \(A\) is \(6\) intervals to the right of \(0\), so \(A=120\). 3. Point \(B\) is \(7\) intervals to the right of \(200\), so \(B=340\). 4. Point \(C\) is \(5\) intervals to the right of \(400\), so \(C=500\). 5. Point \(D\) is \(8\) intervals to the right of \(600\), so \(D=760\). 6. Point \(E\) is \(5\) intervals to the right of \(800\), so \(E=900\).

\(A=120\), \(B=340\), \(C=500\), \(D=760\), \(E=900\)

- Use neighboring pairs in the bottom row to build the middle row. - Find the top only after both middle-row values are known. - Compare the top with the special sum requested in part b).

1. The middle row is \(15 + 22 = 37\) and \(22 + 13 = 35\). 2. The top is \(37 + 35 = 72\). 3. The bottom-row sum is \(15 + 22 + 13 = 50\). 4. Adding the middle number again gives \(50 + 22 = 72\), which equals the top brick.

a) The middle row is \(37\), \(35\), and the top is \(72\). b) \(15 + 22 + 13 + 22 = 72\). The result equals the number in the top brick.

- Each brick is the sum of the two bricks below it. - Start with the bottom row and work upward. - Add neighboring bottom bricks to find each brick above.

1. Find the second-row bricks: \(12 + 15 = 27\) and \(15 + 20 = 35\). 2. Find the top: \(27 + 35 = 62\).

Second row: \(27\), \(35\) Top: \(62\)

- Use what you know about adding hundreds. - The number-wall rule stays the same for larger numbers. - Check that the two middle bricks add to the top.

1. Find the second-row bricks: \(120 + 250 = 370\) and \(250 + 380 = 630\). 2. Find the top: \(370 + 630 = 1000\).

Second row: \(370\), \(630\) Top: \(1000\)

- Begin with the bottom row. - Add each neighboring pair to find the brick above it.

1. The second row is \(4 + 2 = 6\), \(2 + 5 = 7\), and \(5 + 3 = 8\). 2. The third row is \(6 + 7 = 13\) and \(7 + 8 = 15\). 3. The top brick is \(13 + 15 = 28\).

Second row: \(6\), \(7\), \(8\) Third row: \(13\), \(15\) Top: \(28\)

- Work upward one complete row at a time. - Use each pair of adjacent values exactly once for the brick above them. - Check that the final value remains within the stated range.

1. The second row is \(100 + 50 = 150\), \(50 + 120 = 170\), and \(120 + 130 = 250\). 2. The third row is \(150 + 170 = 320\) and \(170 + 250 = 420\). 3. The top brick is \(320 + 420 = 740\).

Second row: \(150\), \(170\), \(250\) Third row: \(320\), \(420\) Top: \(740\)

- Read the rows and dots per row from the array. - Do not invent or count a second array; use the product you found in part a). - Think of twice as many rows as two copies of the original array.

1. The array has \(4\) rows of \(6\) dots, so \(4 \times 6 = 24\). 2. The second array has twice as many equal rows, so its total is twice \(24\): \(2 \times 24 = 48\). 3. Each original row still contains \(6\) dots. Doubling the number of equal rows makes two copies of the original \(24\)-dot array, so the total doubles.

a) \(4 \times 6 = 24\) b) \(2 \times 24 = 48\), so the second array has \(48\) dots. c) Doubling the rows makes two copies of the same \(24\)-dot array, so the total doubles.

- Halve the number of rows first. - Halve the product of the large array to check.

1. The large array represents \(6 \times 8 = 48\). 2. Half of \(6\) rows is \(3\) rows. 3. The small array therefore represents \(3 \times 8 = 24\), which is half of \(48\).

The small array represents \(3 \times 8 = 24\). The number of rows is halved from \(6\) to \(3\), so the product is halved from \(48\) to \(24\).

- Compare each bar’s value with the value before it. - Add \(4\) to predict each next value. - Check whether the final bar fits after correcting the error.

1. The pattern begins \(8, 12, 16\), increasing by \(4\) each time. 2. The fourth value should be \(16+4=20\), but the chart shows \(16\). 3. The fifth value, \(24\), fits the intended pattern after the corrected fourth value.

Bar \(4\) is incorrect. Its value should be \(20\).

- Compare the numbers being subtracted in the three expressions. - Use the first difference to predict the next one. - Think about what happens when the number being subtracted increases by \(100\).

1. For a), \(500 - 142 = 358\). 2. In b), the number being subtracted is \(100\) greater than in a), so the difference is \(100\) less: \(358 - 100 = 258\). 3. In c), the number being subtracted increases by another \(100\), so the difference decreases by another \(100\): \(258 - 100 = 158\).

a) \(358\) b) \(258\) c) \(158\) The differences decrease by \(100\) each time.

- Decide which digits are in the tens places and which are in the ones places. - After choosing the tens digits, place the remaining two digits in the ones places in both possible orders. - Compare consecutive sums to find the change.

1. For a), the ones digits are \(6\) and \(7\): \(46 + 57 = 103\) and \(47 + 56 = 103\). 2. For b), the ones digits are \(5\) and \(7\): \(45 + 67 = 112\) and \(47 + 65 = 112\). 3. For c), the ones digits are \(5\) and \(6\): \(45 + 76 = 121\) and \(46 + 75 = 121\). 4. The sums increase by \(9\): \(112 - 103 = 9\) and \(121 - 112 = 9\).

a) \(46 + 57 = 103\) and \(47 + 56 = 103\) b) \(45 + 67 = 112\) and \(47 + 65 = 112\) c) \(45 + 76 = 121\) and \(46 + 75 = 121\) d) The sums are \(103\), \(112\), and \(121\), increasing by \(9\) each time.

- Calculate all four sums first. - Compare the tens digits and ones digits in equations with equal sums. - Consider how a digit's value changes when it moves from the ones place to the tens place.

1. The sums are \(13 + 46 = 59\), \(16 + 43 = 59\), \(31 + 64 = 95\), and \(34 + 61 = 95\). 2. In the first two equations, the same digits occupy the tens places and the same digits occupy the ones places. Only the ones digits switch addends, so the total does not change. 3. In the last two equations, the larger digits occupy the tens places. A digit in the tens place has ten times its value in the ones place, making the total greater.

a) \(59\), \(59\), \(95\), \(95\) b) The same two digits are in the tens places and the same two digits are in the ones places; only their order between addends changes. c) The larger digits are in the tens places in the last two equations.

- Observe how both digits change from one line to the next. - Compare the difference between the two digits in each original number. - For part c), look for other two-digit numbers whose tens digit is \(2\) greater than the ones digit.

1. Increasing both digits by \(1\) gives \(74 - 47 = 27\) and \(85 - 58 = 27\). 2. The difference stays \(27\) because the tens digit is always \(3\) greater than the ones digit. 3. \(42 - 24 = 18\). A digit difference of \(2\) gives the same result, so examples include \(53 - 35 = 18\) and \(64 - 46 = 18\).

a) \(74 - 47 = 27\) and \(85 - 58 = 27\) b) Every difference is \(27\). c) The new difference is \(18\). Examples: \(53 - 35 = 18\) and \(64 - 46 = 18\).

- Compare each digit difference with the matching multiple of \(9\). - Notice how the result changes when the digit difference increases by \(1\). - For part c), find two digits that differ by \(5\).

1. The subtraction result is \(9\) times the difference between the digits. 2. For \(61 - 16\), the digits differ by \(5\), so \(5 \times 9 = 45\). 3. For \(81 - 18\), the digits differ by \(7\), so the prediction is \(7 \times 9 = 63\). Direct subtraction confirms \(81 - 18 = 63\). 4. To obtain \(45\), the digits must differ by \(5\). One example is \(72 - 27 = 45\).

a) \(45\) b) \(63\); indeed, \(81 - 18 = 63\). c) One example is \(72 - 27 = 45\).

- Start by finding the multiplication facts. - Compare \(5\times5\) with \(4\times6\) by finding both products. - Do the same for \(4\times4\) and \(3\times5\).

1. \(4\times4=16\), \(5\times5=25\), and \(6\times6=36\). 2. \(5\times4=20\), \(5\times6=30\), \(4\times6=24\), and \(6\times4=24\). 3. \(5\times5=25\) is \(1\) greater than \(4\times6=24\). 4. The same relationship happens because \(4\times4=16\) is \(1\) greater than \(3\times5=15\).

a) \(16,25,36\) b) \(5\times4=20\), \(5\times6=30\), \(4\times6=24\), \(6\times4=24\) c) In both comparisons, the product that uses the same number twice is \(1\) greater.

- Start with the product in the middle column. - How many groups of \(10\) make that product? - Once the product is known, find the missing factor in the other fact.

1. In the first row, \(1 \times 10 = 10\), so the missing factor is \(1\). 2. In the second row, \(4 \times 5 = 20\), so the missing factor is \(4\). 3. In the third row, \(6 \times 5 = 30\). The matching 10s fact is \(3 \times 10 = 30\). 4. In the fourth row, \(8 \times 5 = 40\), so the missing factor is \(8\).

Row 1: \(1 \times 10\) Row 2: \(4 \times 5\) Row 3: product \(30\); \(3 \times 10\) Row 4: \(8 \times 5\)

- Find the left side first. - Then ask which number multiplied by \(2\) gives that product. - Compare each first factor on the left with its missing factor.

1. The left-side products are \(4, 8, 12, 16\), and \(20\). 2. The matching 2s facts are \(2 \times 2 = 4\), \(4 \times 2 = 8\), \(6 \times 2 = 12\), \(8 \times 2 = 16\), and \(10 \times 2 = 20\). 3. The missing factor is always twice the factor multiplied by \(4\) on the left.

a) \(2\) b) \(4\) c) \(6\) d) \(8\) e) \(10\) Pattern: Each missing factor is twice the first factor on the left.

- Find the first two facts and compare their products. - How are \(5\) and \(10\) related? - Test the relationship with another factor.

1. \(2 \times 5 = 10\) and \(2 \times 10 = 20\). 2. The product \(20\) is twice the product \(10\). 3. Because \(10\) is twice \(5\), multiplying the same number by \(10\) gives twice the product of multiplying it by \(5\). 4. For example, \(3 \times 5 = 15\) and \(3 \times 10 = 30\); also, \(4 \times 5 = 20\) and \(4 \times 10 = 40\).

The product doubles. Possible examples: \(3 \times 5 = 15\) and \(3 \times 10 = 30\) \(4 \times 5 = 20\) and \(4 \times 10 = 40\)

- Evaluate the left side of each equation. - Which 8s fact has the same product? - Compare the first factors on the two sides.

1. The left-side products are \(16, 24, 32\), and \(40\). 2. Dividing each product by \(8\) gives the missing factors \(2, 3, 4\), and \(5\). 3. The factor paired with \(8\) is half the factor paired with \(4\). Equivalently, when one factor doubles from \(4\) to \(8\), the other factor must be halved to keep the product equal.

a) \(4 \times 4 = 2 \times 8\) b) \(6 \times 4 = 3 \times 8\) c) \(8 \times 4 = 4 \times 8\) d) \(10 \times 4 = 5 \times 8\) Rule: Doubling one factor and halving the other keeps the product the same.

- Evaluate each pair of facts. - Compare the product in the 3s facts with the product in the 6s facts. - How are \(3\) and \(6\) related?

1. For factor \(4\), \(4 \times 3 = 12\). Doubling \(12\) gives \(24\), and \(4 \times 6 = 24\). 2. For factor \(7\), \(7 \times 3 = 21\). Doubling \(21\) gives \(42\), and \(7 \times 6 = 42\). 3. Lucas is correct because \(6\) is twice \(3\). With the other factor unchanged, doubling one factor doubles the product.

Yes. \(4 \times 3 = 12\) and \(4 \times 6 = 24\); \(7 \times 3 = 21\) and \(7 \times 6 = 42\). Because \(6\) is twice \(3\), the matching products double.

- Compare the numbers you divide by in each pair. - Then compare the two answers. - How are doubling one number and halving the answer connected here? - Use multiplication to check your answers.

1. a) \(16\div2=8\) and \(16\div4=4\); b) \(20\div2=10\) and \(20\div4=5\); c) \(40\div4=10\) and \(40\div8=5\). 2. In each pair, the number being divided stays the same while the number you divide by doubles. 3. The answer is cut in half each time.

a) \(8\) and \(4\) b) \(10\) and \(5\) c) \(10\) and \(5\) The answer is cut in half when the number you divide by doubles in these exact division facts.

- What happens to the number being divided in each pair? - Find all six answers before comparing them. - What happens to the answer when the number being divided doubles?

1. a) \(8\div2=4\) and \(16\div2=8\); b) \(12\div3=4\) and \(24\div3=8\); c) \(15\div5=3\) and \(30\div5=6\). 2. In each pair, the number being divided doubles while the number you divide by stays the same. 3. The answer also doubles.

a) \(4\) and \(8\) b) \(4\) and \(8\) c) \(3\) and \(6\) The answer doubles.

- Find both answers in every pair. - Compare how both numbers in each division fact change. - What happens to the answer when both numbers are doubled?

1. a) \(6\div3=2\) and \(12\div6=2\); b) \(10\div2=5\) and \(20\div4=5\); c) \(14\div2=7\) and \(28\div4=7\). 2. In each pair, both numbers in the division fact are doubled. 3. The answer stays the same.

a) \(2\) and \(2\) b) \(5\) and \(5\) c) \(7\) and \(7\) The answer stays the same when both numbers are doubled in these exact division facts.

- Decide whether each sequence increases or decreases. - Find the constant change between consecutive terms. - Continue by adding or subtracting that same amount.

1. For a), add \(25\) repeatedly after \(825\): \(850, 875, 900, 925, 950\). 2. For b), subtract \(5\) repeatedly after \(95\): \(90, 85, 80, 75, 70\). 3. For c), add \(20\) each time. The missing terms are \(380, 420,\) and \(460\).

a) \(850, 875, 900, 925, 950\) b) \(90, 85, 80, 75, 70\) c) \(380, 420, 460\)

- Use the ones digit to find even and odd numbers. - A multiple of \(10\) has \(0\) in the ones place. - After listing the numbers, describe the pattern you see in the ones digits.

1. Even numbers have \(0, 2, 4, 6,\) or \(8\) in the ones place. Counting by twos gives \(896\) through \(910\). 2. Odd numbers have \(1, 3, 5, 7,\) or \(9\) in the ones place. Counting by twos gives \(397\) through \(411\). 3. Multiples of \(10\) have \(0\) in the ones place. The requested numbers are \(680\) through \(720\) in steps of \(10\).

a) \(896, 898, 900, 902, 904, 906, 908, 910\); ones digits repeat \(0,2,4,6,8\). b) \(397, 399, 401, 403, 405, 407, 409, 411\); ones digits repeat \(1,3,5,7,9\). c) \(680, 690, 700, 710, 720\); every ones digit is \(0\).

- Track the change in the first number from line to line. - Track the change in the number being subtracted. - Combine those two changes to predict the next difference.

1. \(542 - 199 = 542 - 200 + 1 = 343\). 2. \(552 - 198 = 552 - 200 + 2 = 354\). 3. \(562 - 197 = 562 - 200 + 3 = 365\). 4. The first number increases by \(10\) while the number being subtracted decreases by \(1\), so each difference increases by \(11\). 5. The next expressions are \(572 - 196 = 376\) and \(582 - 195 = 387\).

\(542-199=343\) \(552-198=354\) \(562-197=365\) \(572-196=376\) \(582-195=387\) The first number increases by \(10\), and the number being subtracted decreases by \(1\). Both changes make the answer larger, so it increases by \(10+1=11\) each time.

- Compare the first numbers in consecutive expressions. - Compare the numbers being subtracted in the same way. - Think about the distance between two numbers when both increase equally.

1. \(735 - 399 = 735 - 400 + 1 = 336\). 2. \(736 - 400 = 336\). 3. \(737 - 401 = 336\). 4. From one expression to the next, both the first number and the number being subtracted increase by \(1\). Increasing both numbers by the same amount keeps their difference constant.

\(735 - 399 = 336\) \(736 - 400 = 336\) \(737 - 401 = 336\) The difference stays the same because both numbers increase by \(1\) each time.

- Compare one line with the line just before it. - In addition, if a number goes up, the sum goes up by the same amount. - In subtraction, changing the first number and changing the number being subtracted affect the answer in different directions.

1. a) Start with \(300+400=700\). The first number increases by \(20\), so the answer increases by \(20\) to \(720\). The second number increases by \(50\), so the answer increases by \(50\) to \(770\). The first number increases by \(6\), so the answer increases by \(6\) to \(776\). The second number increases by \(1\), so the answer increases by \(1\) to \(777\). 2. b) Start with \(800-500=300\). The first number increases by \(70\), so the answer increases by \(70\) to \(370\). The number being subtracted increases by \(30\), so the answer decreases by \(30\) to \(340\). The first number increases by \(9\), so the answer increases by \(9\) to \(349\). The number being subtracted increases by \(4\), so the answer decreases by \(4\) to \(345\).

a) Start: \(700\) The first number increases by \(20\), so the answer increases by \(20\) to \(720\). The second number increases by \(50\), so the answer increases by \(50\) to \(770\). The first number increases by \(6\), so the answer increases by \(6\) to \(776\). The second number increases by \(1\), so the answer increases by \(1\) to \(777\). b) Start: \(300\) The first number increases by \(70\), so the answer increases by \(70\) to \(370\). The number being subtracted increases by \(30\), so the answer decreases by \(30\) to \(340\). The first number increases by \(9\), so the answer increases by \(9\) to \(349\). The number being subtracted increases by \(4\), so the answer decreases by \(4\) to \(345\).

- Compare the two expressions in each pair and find what changed. - Decide whether that change makes the answer go up or down. - Say how the answer changes before giving the second answer.

1. a) \(450+300=750\). The second number goes up by \(20\), so the answer goes up by \(20\) to \(770\). 2. b) \(680-200=480\). The number being subtracted goes up by \(40\), so the answer goes down by \(40\) to \(440\). 3. c) \(512+100=612\). The second number goes up by \(7\), so the answer goes up by \(7\) to \(619\). 4. d) \(975-400=575\). The number being subtracted goes up by \(6\), so the answer goes down by \(6\) to \(569\).

a) The first answer is \(750\). The second number increases by \(20\), so the answer increases by \(20\) to \(770\). b) The first answer is \(480\). The number being subtracted increases by \(40\), so the answer decreases by \(40\) to \(440\). c) The first answer is \(612\). The second number increases by \(7\), so the answer increases by \(7\) to \(619\). d) The first answer is \(575\). The number being subtracted increases by \(6\), so the answer decreases by \(6\) to \(569\).

- Track how each digit changes from one line to the next. - Compare the three calculated differences. - Continue the same digit pattern for part c).

1. \(959 - 595 = 364\), \(848 - 484 = 364\), and \(737 - 373 = 364\). 2. Every difference is \(364\). 3. Each digit in both numbers decreases by \(1\) from one line to the next. 4. The next expressions are \(626 - 262 = 364\) and \(515 - 151 = 364\).

a) All three answers are \(364\). b) The answers are all the same. c) \(626 - 262 = 364\) and \(515 - 151 = 364\)

- Determine how the first number changes from line to line. - Notice that the number being subtracted remains fixed. - Look for a shared property of \(858\), \(747\), and \(636\).

1. The first number decreases by \(111\) each time, while \(121\) stays fixed. 2. Continue with \(646 - 121 = 525\), \(535 - 121 = 414\), and \(424 - 121 = 303\). 3. The results are also palindromes and decrease by \(111\) each time.

a) \(646 - 121 = 525\), \(535 - 121 = 414\), and \(424 - 121 = 303\) b) The results are palindromes that decrease by \(111\) each time.

- Notice how the first factor changes. - Compare \(30\) with \(60\). - Find the difference between consecutive products in each sequence.

1. Sequence a) continues with \(5 \times 60 = 300\), \(4 \times 60 = 240\), and \(3 \times 60 = 180\). The products decrease by \(60\). 2. Sequence b) continues with \(5 \times 30 = 150\), \(4 \times 30 = 120\), and \(3 \times 30 = 90\). The products decrease by \(30\). 3. Each product in b) is half the matching product in a) because \(30\) is half of \(60\).

a) \(5 \times 60 = 300\), \(4 \times 60 = 240\), \(3 \times 60 = 180\) b) \(5 \times 30 = 150\), \(4 \times 30 = 120\), \(3 \times 30 = 90\) The products decrease by \(60\) in a) and by \(30\) in b). Each b) product is half the matching a) product.

- Find what stays the same in each pair and what changes. - Compare \(2\) with \(4\). - Calculate both products in each pair before describing the pattern.

1. Part a: \(4 \times 2 = 8\) and \(4 \times 4 = 16\). 2. Part b: \(6 \times 2 = 12\) and \(6 \times 4 = 24\). 3. Part c: \(9 \times 2 = 18\) and \(9 \times 4 = 36\). 4. In each pair, the second factor doubles from \(2\) to \(4\), and the product also doubles.

a) \(8\) and \(16\) b) \(12\) and \(24\) c) \(18\) and \(36\) When one factor is doubled and the other factor stays the same, the product doubles.

- How does the first factor change in each pattern? - Which set of multiplication facts is used in Pattern A? - Which set of multiplication facts is used in Pattern B? - Can you find the next product by adding or subtracting one equal group?

1. In Pattern A, \(5 \times 6 = 30\) and \(6 \times 6 = 36\). The first factor increases by \(1\), so the product increases by \(6\) each step. 2. In Pattern B, \(7 \times 8 = 56\) and \(6 \times 8 = 48\). The first factor decreases by \(1\), so the product decreases by \(8\) each step.

Pattern A: \(30, 36\); the product increases by \(6\) each step. Pattern B: \(56, 48\); the product decreases by \(8\) each step.

- Find the sum of the completed first row. - Every row, column, and diagonal must have that same sum. - Next choose a row or column with only one missing number. - Check the final column and both diagonals.

1. The first row gives the common sum: \(250 + 130 + 220 = 600\). 2. In the first column, the missing number is \(600 - 250 - 180 = 170\). 3. In the second column, the missing number is \(600 - 130 - 200 = 270\). 4. In the second row, the remaining number is \(600 - 170 - 200 = 230\). 5. In the third row, the remaining number is \(600 - 180 - 270 = 150\). 6. Check the third column: \(220 + 230 + 150 = 600\). 7. Check the diagonals: \(250 + 200 + 150 = 600\) and \(220 + 200 + 180 = 600\).

The completed square is: <table border="1" style="text-align:center;"> <tr><td>250</td><td>130</td><td>220</td></tr> <tr><td>170</td><td>200</td><td>230</td></tr> <tr><td>180</td><td>270</td><td>150</td></tr> </table>

- Look at how the two smaller factors change from one line to the next. - Split each second factor as evenly as possible. - For an even number, the two addends are equal. For an odd number, they differ by \(1\).

1. The second factor is split into two addends that are equal or differ by \(1\). 2. Since \(6 = 3 + 3\), \(7 \times 6 = 7 \times 3 + 7 \times 3\). 3. Since \(7 = 3 + 4\), \(7 \times 7 = 7 \times 3 + 7 \times 4\). 4. Since \(8 = 4 + 4\), \(7 \times 8 = 7 \times 4 + 7 \times 4\).

\(7 \times 6 = (7 \times 3) + (7 \times 3)\) \(7 \times 7 = (7 \times 3) + (7 \times 4)\) \(7 \times 8 = (7 \times 4) + (7 \times 4)\)

- Consider each change separately. - One change raises the sum and the other lowers it. - Combine the two changes to find the overall change.

1. Increasing the first number being added increases the sum by \(210\). 2. Decreasing the second number being added decreases the sum by \(130\). 3. The overall change is \(210 - 130 = 80\). 4. Therefore the sum increases by \(80\).

The sum increases by \(80\).

- The answer increased by \(500\) altogether. - The first number caused \(320\) of that increase. - What increase is left for the second number?

1. The answer increases by \(500\) altogether. 2. Increasing the first number by \(320\) makes the answer increase by \(320\). 3. The rest of the increase is \(500-320=180\), so the second number increased by \(180\).

The second number increased by \(180\).

- Choose two numbers that are easy to add. - Increase one by \(15\) and decrease the other by \(15\). - Compare the old answer with the new answer.

1. For example, start with \(100+100=200\). 2. Increase the first number by \(15\): \(100+15=115\). 3. Decrease the second number by \(15\): \(100-15=85\). 4. The new equation is \(115+85=200\). 5. The answer stays the same because one change adds \(15\) and the other subtracts \(15\). The two changes cancel each other.

The answer stays the same. For example, \(100+100=200\) and \(115+85=200\). Adding \(15\) to one number and subtracting \(15\) from the other cancel each other.

- Decide how the first change affects the answer. - Decide how the second change affects the answer. - Combine the two changes, then use the starting answer of \(450\).

1. Raising the first number by \(80\) raises the answer by \(80\). 2. Lowering the second number by \(30\) lowers the answer by \(30\). 3. Together, the answer changes by \(80-30=50\). 4. The new answer is \(450+50=500\).

a) The sum increases by \(80\). b) The sum decreases by \(30\). c) The new sum is \(500\) because the overall change is an increase of \(50\).

- Decide how changing one number being added affects the sum. - The same change is made to both numbers being added. How many times does it affect the sum? - Combine the two decreases before changing the original sum.

1. Decreasing the first number being added by \(40\) decreases the sum by \(40\). 2. Decreasing the second number being added by \(40\) decreases the sum by another \(40\). 3. The total decrease is \(40 + 40 = 80\). 4. The new sum is \(500 - 80 = 420\).

The sum decreases by \(80\), so the new sum is \(420\).

- Think about what happens when you start with more. - Think about what happens when you subtract less. - Test the changes with an equation such as \(200 - 100\).

1. Increasing the first number by \(50\) increases the difference by \(50\). 2. Decreasing the number being subtracted by \(50\) also increases the difference by \(50\), because less is subtracted. 3. The total increase is \(50 + 50 = 100\).

The difference increases by \(100\).

- Decide how increasing one addend affects the sum. - Then decide how decreasing the other addend by the same amount affects it. - You may test the pattern with smaller numbers.

1. Increasing one addend by \(30\) increases the sum by \(30\): \(480 + 30 = 510\). 2. Decreasing the other addend by \(30\) decreases the new sum by \(30\): \(510 - 30 = 480\). 3. The final sum equals the original sum because the two changes cancel each other.

a) The new sum is \(510\). b) The resulting sum is \(480\), the same as the original sum.

- The first change raises the answer by \(50\). - The second change lowers the answer by \(80\). - Combine those two changes before applying them to \(550\).

1. The first change increases the answer by \(50\). 2. The second change decreases the answer by \(80\). 3. The decrease is \(30\) larger than the increase, so the answer decreases by \(30\). 4. The new answer is \(550-30=520\).

The answer decreases by \(30\). New answer: \(550-30=520\)

- Think of the difference as the distance between two numbers on a number line. - If one number moves \(40\) units to the right, what must happen to the other number to keep the distance unchanged? - You can represent the new number being subtracted with \(n\).

1. The new first number is \(600+40=640\). 2. Let \(n\) be the new number being subtracted. It must satisfy \(640-n=350\). 3. The new number being subtracted is \(640-350=290\), so \(n=290\). 4. Since \(290-250=40\), the number being subtracted must also increase by \(40\). 5. Adding the same amount to both numbers keeps their difference unchanged.

He must increase \(250\) by \(40\), making it \(290\). Then \(640-290=350\).

- Compare the increase of \(80\) with the target increase of \(50\). - Decide whether the second number being added must increase or decrease. - Test your idea with an equation such as \(100 + 100 = 200\).

1. The answer needs to increase by \(50\). 2. Increasing the first number by \(80\) would increase the answer by \(80\). 3. That is \(30\) too much. 4. Therefore, the second number must decrease by \(30\).

The second number must decrease by \(30\).

- Find each new difference and compare it with \(440\). - Think about how changing the first number differs from changing the number being subtracted. - On a number line, what happens to the distance if both numbers move the same amount?

1. Increasing the first number gives \(710 - 230 = 480\), which is \(40\) greater than \(440\). 2. Increasing the number being subtracted gives \(670 - 270 = 400\), which is \(40\) less than \(440\). 3. Increasing both numbers gives \(770 - 330 = 440\). The difference stays the same because the distance between the two numbers does not change.

a) The new difference is \(480\), which is \(40\) greater. b) The new difference is \(400\), which is \(40\) less. c) The difference remains \(440\).

- What happens to the difference when the amount you start with decreases? - What happens when the amount you subtract increases? - Try to find each change without doing the whole subtraction again.

1. The original difference is \(780 - 150 = 630\). 2. Decreasing the first number gives \(730 - 150 = 580\). The difference decreases by \(50\). 3. Increasing the number being subtracted gives \(780 - 170 = 610\). The difference decreases by \(20\).

a) \(630\) b) The difference decreases by \(50\), to \(580\). c) The difference decreases by \(20\), to \(610\).

- Decide how each change affects the sum. - Apply the two changes one at a time. - You may also combine them into one overall change.

1. Increasing the first number by \(50\) increases the sum to \(420+50=470\). 2. Decreasing the second number by \(30\) decreases the sum to \(470-30=440\).

The new sum is \(440\).

- Test the situation with an equation such as \(400 - 100 = 300\). - After decreasing the first number, decide whether you must subtract more or less. - Remember that changing both numbers by the same amount keeps their difference unchanged.

1. Decreasing the first number by \(50\) would decrease the difference from \(300\) to \(250\). 2. To restore the difference to \(300\), the difference must increase by \(50\). 3. Decreasing the number being subtracted by \(50\) increases the difference by \(50\). 4. Therefore, both numbers must decrease by the same amount.

The number being subtracted must also decrease by \(50\).

- First find how much water remains. - Decide how using less water changes the remainder. - Decide how starting with more water changes the remainder. - Adjust the answer from part a) instead of starting over.

1. The amount remaining is \(650 - 380 = 270\), so \(270\,\text{L}\) remain. 2. Using \(20\,\text{L}\) less increases the remainder by \(20\,\text{L}\): \(270 + 20 = 290\). 3. Starting with \(20\,\text{L}\) more also increases the remainder by \(20\,\text{L}\): \(270 + 20 = 290\).

a) \(270\,\text{L}\) b) \(290\,\text{L}\) c) \(290\,\text{L}\)

- Compare the first numbers from one line to the next. - Compare the numbers being subtracted from one line to the next. - Describe the rule when both numbers in a subtraction increase by the same amount. - Think of the difference as the distance between the two numbers.

1. Both numbers increase by \(10\), so the next equation is \(570 - 240 = 330\). 2. In each step, the first number and the number being subtracted both increase by \(10\). 3. Adding the same amount to both numbers does not change the distance between them, so the difference remains unchanged.

a) \(570 - 240 = 330\) b) Both numbers increase by \(10\). c) The difference stays the same because both numbers change by the same amount.

- Decide how increasing the first number affects the difference. - Decide how decreasing the number being subtracted affects the difference. - Apply the two changes one at a time.

1. Increasing the first number by \(30\) increases the difference to \(150 + 30 = 180\). 2. Decreasing the number being subtracted by \(10\) means less is subtracted, so the difference increases by another \(10\). 3. The new difference is \(180 + 10 = 190\).

The new difference is \(190\). Increasing the first number by \(30\) raises the difference to \(180\), and decreasing the number being subtracted by \(10\) raises it by another \(10\).

- Find the sum of the completed first row. - Use that same sum for every row and column. - Start with a row or column that has only one empty cell. - Check all three columns at the end.

1. The completed first row has sum \(25 + 45 + 20 = 90\). 2. In the first column, the missing number is \(90 - 25 - 30 = 35\). 3. In the second column, the missing number is \(90 - 45 - 30 = 15\). 4. In the second row, the last number is \(90 - 35 - 30 = 25\). 5. In the third row, the last number is \(90 - 30 - 15 = 45\). 6. The third column checks: \(20 + 25 + 45 = 90\).

The completed square is: <table border="1" style="width:150px; text-align:center;"> <tr><td>25</td><td>45</td><td>20</td></tr> <tr><td>35</td><td>30</td><td>25</td></tr> <tr><td>30</td><td>15</td><td>45</td></tr> </table>

- Find the sum of the completed second row. - Use that same sum for every row and column. - Start where only one number is missing.

1. The completed second row has sum \(350 + 200 + 50 = 600\). 2. The top-right number is \(600 - 150 - 250 = 200\). 3. The bottom-left number is \(600 - 150 - 350 = 100\). 4. The bottom-right number is \(600 - 100 - 150 = 350\). 5. The third column checks: \(200 + 50 + 350 = 600\).

The completed square is: <table border="1" style="width:150px; text-align:center;"> <tr><td>150</td><td>250</td><td>200</td></tr> <tr><td>350</td><td>200</td><td>50</td></tr> <tr><td>100</td><td>150</td><td>350</td></tr> </table>

- Identify which number changes in each pair of expressions. - Decide whether that change makes the result greater or less. - Use the size of the change to compare the results. - Compare the expressions closely before calculating from scratch.

1. \(270 + 80 = 350\) and \(270 + 90 = 360\). 2. The second addend increases by \(10\), so the sum also increases by \(10\). 3. \(640 - 50 = 590\) and \(660 - 50 = 610\). 4. The first number in the second subtraction is \(20\) greater while the number being subtracted is unchanged, so the second difference is \(20\) greater.

a) The sums are \(350\) and \(360\). The second is \(10\) greater. b) The differences are \(590\) and \(610\). The second is \(20\) greater because its first number is \(20\) greater.

- Consider how changing one addend changes the sum. - Apply each change to the known sum of \(350\). - For part c), combine the changes to both addends.

1. Increasing one addend by \(40\) increases the sum by \(40\), to \(350 + 40 = 390\). 2. Decreasing one addend by \(30\) decreases the sum by \(30\), to \(350 - 30 = 320\). 3. Increasing both addends by \(20\) increases the sum by \(20 + 20 = 40\), so the new sum is \(390\).

a) The sum increases by \(40\), to \(390\). b) The sum decreases by \(30\), to \(320\). c) The sum increases by \(40\), to \(390\).

- Change each number being added as described and find the new sum. - Compare the increase in one number being added with the decrease in the other. - For part c), combine the two increases.

1. The changed numbers being added are \(340 + 20 = 360\) and \(160 - 20 = 140\). 2. Their sum is \(360 + 140 = 500\), so the claim works for this example. 3. The increase of \(20\) and decrease of \(20\) cancel, producing a overall change of \(0\). 4. If both numbers being added increase by \(20\), the sum increases by \(40\), from \(500\) to \(540\).

a) The new numbers being added are \(360\) and \(140\), and their sum is \(500\). b) The changes cancel because one number being added increases by the same amount that the other decreases. c) The sum would increase by \(40\), to \(540\).

- In part a, compare the number removed with the number added. - In part b, find the difference between the number added and the number removed. - Use that change to find the new total.

1. For part a, the baker removes and adds the same number of rolls. These changes cancel, so the total remains \(480\). 2. For part b, compare the number added with the number removed: \(80 - 70 = 10\). The total increases by \(10\). 3. Find the new total: \(480 + 10 = 490\).

a) There are still \(480\) rolls altogether. b) The total increases by \(10\), so there would be \(490\) rolls.

- Decide how each change affects the sum. - Treat the increase and decrease as opposite changes. - Find the overall change.

1. Increasing the first number increases the sum by \(120\). 2. Decreasing the second number decreases the sum by \(150\). 3. The decrease is \(30\) larger than the increase, so the sum decreases by \(30\).

The sum decreases by \(30\).

- Think of subtraction as the first number minus the number being subtracted. - Decide how decreasing the first number affects the difference. - Decide how decreasing the number being subtracted affects the difference. - Test the two changes in order with a simple example such as \(100 - 50\).

1. Decreasing the first number by \(40\) decreases the difference by \(40\). 2. Decreasing the number being subtracted by \(15\) increases the difference by \(15\), because less is subtracted. 3. The increase of \(15\) offsets part of the decrease of \(40\). The difference therefore decreases by \(40-15=25\).

The difference decreases by \(25\).

- Decide how increasing the first number affects the difference. - Then decide how increasing the number being subtracted affects it. - Think of the difference as the distance between two numbers.

1. Increasing the first number by \(20\) increases the difference to \(350 + 20 = 370\). 2. Increasing the number being subtracted by \(20\) then decreases the difference by \(20\). 3. The resulting equation is \(520 - 170 = 350\), so the difference returns to its original value.

a) The new difference is \(370\). b) The difference is \(350\), the same as the original difference.

- Consider each change separately. - Increasing the first number and increasing the number being subtracted affect the difference in opposite ways. - What happens to the distance between two numbers when both increase by the same amount?

1. Increasing the first number gives \(540 - 200 = 340\), so the difference increases by \(40\). 2. Increasing the number being subtracted gives \(500 - 240 = 260\), so the difference decreases by \(40\). 3. Increasing both numbers gives \(540 - 240 = 300\). The difference stays the same because both numbers change by the same amount.

a) \(340\) b) \(260\) c) \(300\)

- Picture the two numbers as points on a number line. - Decide how moving the greater number toward the lesser number changes the distance. - Then decide how moving the lesser number toward the greater number changes it. - In each step, decide whether the gap becomes larger or smaller.

1. Decreasing the greater number by \(20\) reduces the difference to \(150 - 20 = 130\). 2. Increasing the lesser number by \(10\) reduces the difference by another \(10\). 3. The new difference is \(130 - 10 = 120\).

The new difference is \(120\).

- Add adjacent bricks to fill the middle row. - Add the two middle-row bricks to find the top. - Compare the bottom middle numbers and note how much they increase. - Compare the top numbers and look for a relationship. - Find how many times the bottom middle number is used in the top.

1. In a), the middle row is \(10 + 15 = 25\) and \(15 + 12 = 27\), so the top brick is \(25 + 27 = 52\). 2. In b), the middle row is \(10 + 20 = 30\) and \(20 + 12 = 32\), so the top brick is \(30 + 32 = 62\). 3. In c), the middle row is \(10 + 25 = 35\) and \(25 + 12 = 37\), so the top brick is \(35 + 37 = 72\). 4. The top bricks increase by \(10\). The bottom middle number is used in both middle-row sums, so increasing it by \(5\) increases the top by \(2 \times 5 = 10\).

1. a) Middle row: \(25\), \(27\); top: \(52\) b) Middle row: \(30\), \(32\); top: \(62\) c) Middle row: \(35\), \(37\); top: \(72\) 2. Each increase of \(5\) in the bottom middle number increases the top by \(10\), because that number is used in both bricks in the middle row.

- Look for a group of three bricks where two values are known. - When the upper brick and one lower brick are known, subtract to find the other lower brick. - Enter each new value before moving to nearby bricks. - Use the top brick to check your completed wall.

1. Find the second brick in the bottom row: \(19 - 7 = 12\). 2. Find the fourth brick in the bottom row: \(22 - 7 = 15\). 3. Find the left brick in the next row: \(8 + 12 = 20\). 4. Find the two bricks in the third row: \(20 + 19 = 39\) and \(19 + 22 = 41\). 5. Check the top: \(39 + 41 = 80\).

Bottom row: \(8\), \(12\), \(7\), \(15\) Second row: \(20\), \(19\), \(22\) Third row: \(39\), \(41\) Top: \(80\)

- Start with the brick labeled \(12\) and the two bricks below it. - When an upper brick and one lower brick are known, subtract to find the other lower brick. - Work upward where possible, then use the top brick to work backward on the other side.

1. The middle brick labeled \(12\) is above the unknown bottom brick and \(4\), so the unknown is \(12 - 4 = 8\). 2. The left brick in the second row is \(5 + 8 = 13\). 3. The left brick in the third row is \(13 + 12 = 25\). 4. Use the top to find the right brick in the third row: \(53 - 25 = 28\). 5. Find the right brick in the second row: \(28 - 12 = 16\). 6. Find the rightmost bottom brick: \(16 - 4 = 12\).

Bottom-row blanks: \(8\), \(12\) Second-row blanks: \(13\), \(16\) Third-row blanks: \(25\), \(28\)

- Start where an upper brick and one brick below it are known. - Use subtraction to find the other lower brick. - Use addition when both lower bricks are known. - After completing both walls, compare the same positions in a) and b).

1. In a), the bottom-left brick is \(32 - 15 = 17\), the right brick in the middle row is \(15 + 24 = 39\), and the top is \(32 + 39 = 71\). 2. In b), the bottom-left brick is \(35 - 15 = 20\), the right brick in the middle row is \(15 + 24 = 39\), and the top is \(35 + 39 = 74\). 3. The bottom-left brick increases by \(3\), from \(17\) to \(20\), and the top also increases by \(3\), from \(71\) to \(74\).

a) Bottom row: \(17\), \(15\), \(24\); middle row: \(32\), \(39\); top: \(71\) b) Bottom row: \(20\), \(15\), \(24\); middle row: \(35\), \(39\); top: \(74\) The bottom-left brick and the top brick both increase by \(3\).

- Complete the original wall before changing any bottom values. - Compare how much each bottom brick changes with how much the top changes. - Track how often each bottom brick contributes to the top.

1. The middle row is \(6 + 8 = 14\) and \(8 + 12 = 20\), so the top is \(14 + 20 = 34\). 2. The new bottom row is \(8\), \(10\), \(14\). The middle row is \(18\), \(24\), and the top is \(18 + 24 = 42\). It is \(42 - 34 = 8\) greater. 3. If every bottom number increases by the same amount, the top increases by four times that amount: the outer numbers contribute once each, and the middle number contributes twice. An increase of \(5\) in each bottom number increases the top by \(4 \times 5 = 20\).

a) Middle row: \(14\), \(20\); top: \(34\) b) New bottom row: \(8\), \(10\), \(14\); middle row: \(18\), \(24\); top: \(42\). The top is \(8\) greater. c) The top would be \(20\) greater.

- Start with the middle brick in the second row and the \(9\) below it. - Subtract a known lower brick from the brick above to find the other lower brick. - Work across the wall, then use the top brick to work backward.

1. The middle brick in the second row is \(21\), and one brick below it is \(9\). The other bottom brick is \(21 - 9 = 12\). 2. The left brick in the second row is \(4 + 12 = 16\). 3. The left brick in the third row is \(16 + 21 = 37\). 4. Use the top to find the right brick in the third row: \(70 - 37 = 33\). 5. Find the right brick in the second row: \(33 - 21 = 12\). 6. Find the rightmost bottom brick: \(12 - 9 = 3\).

Second bottom brick: \(12\) Fourth bottom brick: \(3\)

- Look for groups of three bricks with two known values. - Add two lower bricks to find the brick above them. - Subtract a known lower brick from an upper brick to find the other lower brick. - Find the missing bottom bricks first.

1. Find the second bottom brick: \(13 - 5 = 8\). 2. Find the rightmost bottom brick: \(15 - 6 = 9\). 3. Find the middle brick in the second row: \(8 + 6 = 14\). 4. Find the third-row bricks: \(13 + 14 = 27\) and \(14 + 15 = 29\). 5. Find the top: \(27 + 29 = 56\).

\(56\)

- Each pair of neighboring bricks adds to the brick above. - When an upper brick and one lower brick are known, subtract to find the other lower brick. - Begin with a group of three bricks that has two known values.

1. Find the second bottom brick: \(13 - 5 = 8\). 2. Find the third bottom brick: \(18 - 6 = 12\). 3. The requested brick is above \(8\) and \(12\): \(8 + 12 = 20\). 4. Check the upper rows: \(13 + 20 = 33\), \(20 + 18 = 38\), and \(33 + 38 = 71\).

\(20\)

- Each brick equals the sum of the two bricks below it. - If an upper brick and one lower brick are known, subtract to find the other lower brick. - Start where two of the three connected values are known. - Use each new value to solve the next connected group.

1. Wall a): \(23 - 8 = 15\), \(12 + 15 = 27\), and \(27 + 23 = 50\). 2. Wall b): \(39 - 14 = 25\), \(71 - 39 = 32\), and \(32 - 14 = 18\). 3. Wall c): \(7 + 22 = 29\), \(70 - 29 = 41\), and \(41 - 22 = 19\).

a) Bottom middle: \(15\); second-row left: \(27\); top: \(50\) b) Bottom left: \(25\); second-row right: \(32\); bottom right: \(18\) c) Second-row left: \(29\); second-row right: \(41\); bottom right: \(19\)

- Find a group of three connected bricks with only one missing value. - Begin with the brick labeled \(230\) and the \(90\) below it. - Use each new value to solve neighboring bricks. - Add when moving upward and subtract when moving downward. - Check the completed wall from bottom to top.

1. Find the second bottom brick: \(230 - 90 = 140\). 2. Find the left brick in the second row: \(110 + 140 = 250\). 3. Find the right brick in the second row: \(490 - 230 = 260\). 4. Find the rightmost bottom brick: \(260 - 90 = 170\). 5. Find the left brick in the third row: \(250 + 230 = 480\). 6. Find the top: \(480 + 490 = 970\).

Bottom row: \(110\), \(140\), \(90\), \(170\) Second row: \(250\), \(230\), \(260\) Third row: \(480\), \(490\) Top: \(970\)

- Start with the right brick in the second row and the two bricks below it. - When an upper brick and one lower brick are known, subtract to find the other lower brick. - Then add upward.

1. Find the middle bottom brick: \(35 - 23 = 12\). 2. Find the left brick in the second row: \(15 + 12 = 27\). 3. Find the top: \(27 + 35 = 62\).

Left brick in the second row: \(27\) Middle bottom brick: \(12\) Top brick: \(62\)

- Look for connected groups with two known values. - Use addition to find an upper brick when both lower bricks are known. - Use subtraction to find a lower brick when the upper brick and the other lower brick are known. - Continue until you reach the top.

1. Find the right brick in the second row: \(14 + 7 = 21\). 2. Find the middle brick in the second row: \(45 - 21 = 24\). 3. Find the missing bottom brick: \(24 - 14 = 10\). 4. Find the left brick in the second row: \(10 + 10 = 20\). 5. Find the left brick in the third row: \(20 + 24 = 44\). 6. Find the top: \(44 + 45 = 89\).

\(89\)

- If an upper brick and one lower brick are known, subtract to find the other lower brick. - Look for a connected group with only one missing value. - Work upward or downward depending on the given information.

1. Find the left brick in the second row: \(63 - 35 = 28\). 2. Find the middle bottom brick: \(35 - 21 = 14\). 3. Find the left bottom brick: \(28 - 14 = 14\).

Bottom row: \(14\), \(14\), \(21\) Second row: \(28\), \(35\) Top: \(63\)

- Finish each wall separately before comparing them. - Track which two bricks in the next row use the changed middle value. - Compare the two top values after both walls are complete.

1. In Wall 1, the middle row is \(8 + 10 = 18\) and \(10 + 12 = 22\), so the top is \(18 + 22 = 40\). 2. In Wall 2, the middle row is \(8 + 11 = 19\) and \(11 + 12 = 23\), so the top is \(19 + 23 = 42\). 3. The bottom middle brick increases by \(1\), while the top increases by \(42 - 40 = 2\).

a) In Wall 1, the middle row is \(18\), \(22\), and the top is \(40\). In Wall 2, the middle row is \(19\), \(23\), and the top is \(42\). b) Increasing the bottom middle brick by \(1\) increases the top by \(2\).

- Use subtraction to find a missing lower brick from an upper brick and the other lower brick. - Ask what number plus \(14\) equals \(30\). - Once the lower row is complete, add upward.

1. Find the left bottom brick: \(30 - 14 = 16\). 2. Find the right bottom brick: \(26 - 14 = 12\). 3. Find the top: \(30 + 26 = 56\).

Bottom row: \(16\), \(14\), \(12\) Top: \(56\)

- Find a connected group with only one missing value. - Most bricks can be found by adding upward, but one step requires subtraction.

1. Find the missing bottom brick: \(150 - 80 = 70\). 2. Complete the second row: \(50 + 70 = 120\) and \(80 + 60 = 140\). 3. Complete the third row: \(120 + 150 = 270\) and \(150 + 140 = 290\). 4. Find the top: \(270 + 290 = 560\).

Bottom row: \(50\), \(70\), \(80\), \(60\) Second row: \(120\), \(150\), \(140\) Third row: \(270\), \(290\) Top: \(560\)

- Begin where an upper brick and one of its two lower bricks are both known. - Subtract to recover a missing lower brick; add when both lower bricks are known. - Check the finished wall by rebuilding it upward.

1. Find the second bottom brick: \(90 - 30 = 60\). 2. Find the left brick in the second row: \(40 + 60 = 100\). 3. Find the right brick in the second row: \(170 - 90 = 80\). 4. Find the rightmost bottom brick: \(80 - 30 = 50\). 5. Find the left brick in the third row: \(100 + 90 = 190\). 6. Find the top: \(190 + 170 = 360\).

Bottom row: \(40\), \(60\), \(30\), \(50\) Second row: \(100\), \(90\), \(80\) Third row: \(190\), \(170\) Top: \(360\)

- When an upper brick and one lower brick are known, subtract to find the other lower brick. - After completing the bottom row, compare the sums of each adjacent pair. - Look for repeated values in the rows above and connect them to the bottom-row pattern.

1. Find the first and third bricks in the bottom row: \(150-50=100\) in each case. 2. The middle brick in the second row is \(50+100=150\), so all three bricks in that row are \(150\). 3. Each brick in the third row is \(150+150=300\). 4. The top brick is \(300+300=600\). 5. The bottom row alternates \(100,50,100,50\). Because every adjacent pair sums to \(150\), the entire second row repeats \(150\); that repetition then creates two equal \(300\) bricks above.

Bottom row: \(100\), \(50\), \(100\), \(50\) Second row: \(150\), \(150\), \(150\) Third row: \(300\), \(300\) Top: \(600\) Pattern: the bottom row alternates \(100\) and \(50\), so every adjacent pair adds to \(150\). That makes every number in the second row the same.

- Use the given \(250\) and the known lower brick beside its blank neighbor first. - After the bottom row is complete, build each higher row from adjacent pairs. - Verify the top after all lower blanks are filled.

1. The missing bottom brick is \(250 - 150 = 100\). 2. The left brick in the second row is \(50 + 100 = 150\). 3. The right brick in the second row is \(150 + 80 = 230\). 4. The third row is \(150 + 250 = 400\) and \(250 + 230 = 480\). 5. The top brick is \(400 + 480 = 880\).

Bottom row: \(50\), \(100\), \(150\), \(80\) Second row: \(150\), \(250\), \(230\) Third row: \(400\), \(480\) Top: \(880\)

- Start with a group of three connected bricks that has only one blank. - Add two neighboring lower bricks to find the brick above them. - If the upper brick and one lower brick are known, subtract to find the other lower brick.

1. Find the two missing bottom bricks: \(15 - 5 = 10\) and \(15 - 7 = 8\). 2. The middle brick in the second row is \(10 + 8 = 18\). 3. The two bricks in the third row are \(15 + 18 = 33\) and \(18 + 15 = 33\). 4. The top brick is \(33 + 33 = 66\).

Bottom row: \(5\), \(10\), \(8\), \(7\) Second row: \(15\), \(18\), \(15\) Third row: \(33\), \(33\) Top: \(66\)

- Start with the known \(250\) and the \(100\) directly below it. - Use the top value to work backward after the left side is known. - Compare the four completed bottom values only after the wall is finished.

1. The second bottom brick is \(250 - 100 = 150\). 2. The left brick in the second row is \(100 + 150 = 250\), so the left brick in the third row is \(250 + 250 = 500\). 3. The right brick in the third row is \(1000 - 500 = 500\). 4. The right brick in the second row is \(500 - 250 = 250\). 5. The fourth bottom brick is \(250 - 100 = 150\). 6. The bottom row alternates between \(100\) and \(150\).

Bottom row: \(100\), \(150\), \(100\), \(150\) Second row: \(250\), \(250\), \(250\) Third row: \(500\), \(500\) Top: \(1000\) The bottom-row pattern is alternating \(100, 150, 100, 150\).

- First use subtraction to fill the two blanks in the bottom row. - Then work upward one brick at a time. - Each upper brick is the sum of the two bricks directly below it.

1. Complete the bottom row: \(7 - 4 = 3\) and \(7 - 5 = 2\), giving \(2, 3, 4, 2, 5\). 2. Complete the second row: \(2 + 3 = 5\) and \(4 + 2 = 6\), giving \(5, 7, 6, 7\). 3. Add upward for the third row: \(5 + 7 = 12\), \(7 + 6 = 13\), and \(6 + 7 = 13\). 4. The fourth row is \(12 + 13 = 25\) and \(13 + 13 = 26\). 5. The top brick is \(25 + 26 = 51\).

Bottom row: \(2\), \(3\), \(4\), \(2\), \(5\) Second row: \(5\), \(7\), \(6\), \(7\) Third row: \(12\), \(13\), \(13\) Fourth row: \(25\), \(26\) Top: \(51\)

- Track where each bottom brick is used in the row above. - A change in the middle bottom brick affects both bricks in the next row. - Compare one change that is used twice with two changes that are each used once.

1. The original middle row is \(15,15\), so the original top is \(30\). 2. The middle bottom brick is used in both middle-row bricks. Increasing it by \(5\) therefore increases the top by \(5+5=10\). 3. Each outer bottom brick is used on one side only. Increasing both outer bricks by \(5\) increases the top by \(5+5=10\). 4. Both changes increase the top by \(10\), though they change different bottom bricks.

a) Middle row: \(15,15\); top: \(30\) b) Top change: \(+10\) c) Top change: \(+10\) d) The middle brick is counted twice in the top, while the two outer bricks are each counted once, so both changes add \(10\) altogether.

- Start at a known second-row brick that has one known lower neighbor. - Use subtraction to fill the remaining bottom values before building upward. - Check each higher brick against its two lower neighbors.

1. Find the second bottom brick: \(220 - 120 = 100\). 2. Find the last bottom brick: \(200 - 120 = 80\). 3. The left brick in the second row is \(150 + 100 = 250\). 4. The third row is \(250 + 220 = 470\) and \(220 + 200 = 420\). 5. The top brick is \(470 + 420 = 890\).

Bottom row: \(150\), \(100\), \(120\), \(80\) Second row: \(250\), \(220\), \(200\) Third row: \(470\), \(420\) Top: \(890\)

- Focus first on the left middle brick and the known \(15\) below it. - Once the missing bottom value is known, finish the right side and then the top. - Verify that each upper brick equals its two lower neighbors added together.

1. The bottom-left value is \(32 - 15 = 17\). 2. The right brick in the second row is \(15 + 12 = 27\). 3. The top brick is \(32 + 27 = 59\).

The bottom-left brick is \(17\), the right brick in the second row is \(27\), and the top brick is \(59\).

- The known \(14\) and bottom-left \(8\) determine the adjacent bottom blank. - Once the bottom row is complete, move upward one row at a time. - Check each result against the two bricks below it.

1. Find the missing bottom brick: \(14 - 8 = 6\). 2. Complete the second row: \(6 + 7 = 13\) and \(7 + 10 = 17\). 3. Complete the third row: \(14 + 13 = 27\) and \(13 + 17 = 30\). 4. The top brick is \(27 + 30 = 57\).

Bottom row: \(8\), \(6\), \(7\), \(10\) Second row: \(14\), \(13\), \(17\) Third row: \(27\), \(30\) Top: \(57\)

- Finish one complete row before starting the next. - Compare the values in the second row before adding upward again. - Ask what happens when every neighboring pair in one row has the same sum.

1. The second row is \(50+60=110\), \(60+40=100\), \(40+70=110\), and \(70+30=100\). 2. The third row is \(110+100=210\), \(100+110=210\), and \(110+100=210\). 3. The fourth row is \(210+210=420\) and \(210+210=420\). 4. The top brick is \(420+420=840\). 5. The second row alternates \(110,100,110,100\). Every adjacent pair there sums to \(210\), so the whole third row repeats \(210\); that repetition then produces two equal \(420\) bricks.

The top value is \(840\). The second row alternates \(110\) and \(100\), so every adjacent pair sums to \(210\). Therefore, the third row is all \(210\), and the fourth row is \(420,420\).

- Start with a three-brick relationship in which the upper value and one lower value are known. - After finding one missing lower brick, look for the next relationship with only one unknown. - Check the finished wall by adding upward.

1. The bottom-left brick is \(25 - 15 = 10\). 2. The right brick in the second row is \(60 - 25 = 35\). 3. The bottom-right brick is \(35 - 15 = 20\).

Bottom row: \(10\), \(15\), \(20\) Second row: \(25\), \(35\) Top: \(60\)

- Use subtraction when an upper brick and one lower brick are known. - After completing each row, compare values that are the same distance from the left and right ends. - Think about why adding matching pairs on opposite sides makes the same mirror pattern in the row above.

1. The missing bottom values are \(7-4=3\) and \(7-4=3\), so the bottom row is \(2,3,4,3,2\). 2. The outside blanks in the second row are \(2+3=5\) and \(3+2=5\), giving \(5,7,7,5\). 3. The third row is \(5+7=12\), \(7+7=14\), and \(7+5=12\). 4. The fourth row is \(12+14=26\) and \(14+12=26\). 5. The top is \(26+26=52\). 6. Each row reads the same from left to right as from right to left. Because matching values on opposite sides are added in matching positions, the same mirror pattern appears in the row above.

Bottom row: \(2,3,4,3,2\) Second row: \(5,7,7,5\) Third row: \(12,14,12\) Fourth row: \(26,26\) Top: \(52\) Pattern: each row reads the same from left to right and right to left. Adding matching pairs on the two sides makes matching numbers in the row above.

- Start with the known \(20\) and one of its lower neighbors. - After building the left side upward, use the top to recover the right side. - Check every recovered lower value by adding upward again.

1. The second bottom brick is \(20 - 12 = 8\). 2. The left brick in the second row is \(5 + 8 = 13\). 3. The left brick in the third row is \(13 + 20 = 33\). 4. The right brick in the third row is \(72 - 33 = 39\). 5. The right brick in the second row is \(39 - 20 = 19\). 6. The last bottom brick is \(19 - 12 = 7\).

Bottom row: \(5\), \(8\), \(12\), \(7\) Second row: \(13\), \(20\), \(19\) Third row: \(33\), \(39\) Top: \(72\)

- Use the known center brick to recover a neighboring bottom value first. - Build the left side upward until the top lets you work backward on the right. - Verify the completed wall from bottom to top.

1. The second bottom brick is \(110 - 50 = 60\). 2. The left brick in the second row is \(40 + 60 = 100\). 3. The left brick in the third row is \(100 + 110 = 210\). 4. The right brick in the third row is \(440 - 210 = 230\). 5. The right brick in the second row is \(230 - 110 = 120\). 6. The last bottom brick is \(120 - 50 = 70\).

Bottom row: \(40\), \(60\), \(50\), \(70\) Second row: \(100\), \(110\), \(120\) Third row: \(210\), \(230\) Top: \(440\)

- Begin at the top, where one second-row brick is already known. - Move downward by subtracting a known lower neighbor from its upper sum. - Check the recovered bottom row by adding upward.

1. The left brick in the second row is \(750 - 420 = 330\). 2. The middle bottom brick is \(330 - 180 = 150\). 3. The right bottom brick is \(420 - 150 = 270\).

Bottom row: \(180\), \(150\), \(270\) Second row: \(330\), \(420\) Top: \(750\)

- Complete both walls first and compare the top values. - Notice which bottom brick is used in both middle-row sums. - Compare the middle bottom values in the two walls while remembering that the bottom-row totals are equal.

1. In wall a), the second row is \(5+5=10\) and \(5+5=10\), so the top is \(10+10=20\). 2. In wall b), the second row is \(4+7=11\) and \(7+4=11\), so the top is \(11+11=22\). 3. Both bottom rows total \(15\), but the middle bottom brick is used to both second-row bricks. Wall b) has middle brick \(7\) instead of \(5\), so that greater middle value is counted twice in the top and makes wall b) larger.

Wall b) has the greater top value: \(22\), compared with \(20\) for wall a). The middle bottom brick is used in both sums above it, so wall b)’s larger middle value affects the top twice even though the two bottom rows have the same total.

- The top and known right third-row brick determine the left third-row brick. - Continue through connected three-brick groups with only one unknown. - Verify all recovered values by adding upward at the end.

1. The left brick in the third row is \(40 - 22 = 18\). 2. The middle brick in the second row is \(18 - 8 = 10\). 3. The second bottom brick is \(8 - 3 = 5\). 4. The third bottom brick is \(10 - 5 = 5\). 5. The right brick in the second row is \(22 - 10 = 12\). 6. Check the last bottom brick: \(12 - 5 = 7\).

Bottom row: \(3\), \(5\), \(5\), \(7\) Second row: \(8\), \(10\), \(12\) Third row: \(18\), \(22\) Top: \(40\)

- Complete both walls using the stated subtraction rule. - Compare the two top bricks. - Track how the bottom middle number affects both bricks in the middle row.

1. In a), the middle row is \(50 - 15 = 35\) and \(15 - 5 = 10\), so the top is \(35 - 10 = 25\). 2. In b), the middle row is \(50 - 16 = 34\) and \(16 - 5 = 11\), so the top is \(34 - 11 = 23\). 3. When the bottom middle number increases by \(1\), the top decreases from \(25\) to \(23\), a decrease of \(2\).

The top is \(25\) in a) and \(23\) in b). Increasing the bottom middle number by \(1\) decreases the top by \(2\).

- Read the row and column counts from each diagram before describing the change. - Extend both dimension patterns by one step. - Compare the total dots in the third and fourth arrays.

1. The diagrams show arrays of \(2 \times 3\), \(3 \times 4\), and \(4 \times 5\). Both the number of rows and the number of columns increase by \(1\) each time. 2. The next array has dimensions \(5 \times 6\), so it contains \(5 \times 6 = 30\) dots. 3. The third array contains \(4 \times 5 = 20\) dots, so the increase is \(30 - 20 = 10\) dots.

The rows and columns each increase by \(1\). The next array is \(5 \times 6\) with \(30\) dots, an increase of \(10\) dots.

- Read the height of each labeled bar from the vertical scale. - The halfway height must have the same difference from the lower and higher values. - Do not use left-to-right position; compare the numerical heights.

1. The bar heights are \(8\), \(24\), and \(16\) for A, B, and C. 2. Bar C has height \(16\). 3. \(16-8=8\) and \(24-16=8\). 4. Because the two differences are equal, C is exactly halfway in height between A and B.

a) Bar C b) \(16-8=8\) and \(24-16=8\)

- Compare the same bar in both graphs. - Find all three increases. - Decide whether the increases are equal.

1. A increased by \(15 - 10 = 5\). 2. B increased by \(25 - 20 = 5\). 3. C increased by \(35 - 30 = 5\). 4. Every bar increased by \(5\), so the statement is true.

A increased by \(5\), B increased by \(5\), and C increased by \(5\). The statement is true.

- Order the three bars in each graph from least to greatest. - Compare the two orders. - Check whether any bar keeps the same value.

1. In a), \(A < B < C\). 2. In b), \(A > B > C\). 3. The order is reversed, and B stays at \(20\) in both graphs.

1) The order is reversed. 2) Bar B stays at \(20\).

- The two rows in each row-pair have the same number of dots. - Match dots that are in the same column. - Ask whether any dot can be left without a partner when there are four equal rows.

1. Each pair of rows has the same number of dots in both rows. Pair each dot in one row with the dot directly below it in the other row, so no dot is left without a partner. 2. Any \(4\times n\) array has four equal rows. Pair the rows two at a time. Every dot belongs to a pair, so the total is even.

a) Each dot can be paired with the dot in the same column of the matching row, so no dot is left over. b) In every \(4\times n\) array, the four equal rows can be paired two at a time. Every dot has a partner, so every product in the \(4\)s facts is even.

- Can one group of \(6\) be split completely into pairs? - For the \(3\)s facts, look at what happens when one more group of \(3\) is added. - Does adding \(3\) switch an even answer to odd and an odd answer to even?

1. Every product in the \(6\)s facts is even. Each group of \(6\) can be split into \(3\) pairs, so any number of groups still has only pairs. 2. In the \(3\)s facts, the products alternate odd, even, odd, even as the second factor increases by \(1\). 3. Each new fact adds another group of \(3\), which is odd. Adding an odd number switches a total from odd to even or from even to odd.

a) All the \(6\)s products are even because each group of \(6\) can be split into pairs with nothing left over. b) The \(3\)s products alternate odd, even, odd, even. Each new fact adds \(3\), so the total switches between odd and even each time.

- Try a multiplication fact with one even factor and one odd factor. - One example is enough to show that Mason is wrong. - After finding an example, say what must be true about at least one factor to make an even product.

1. Mason’s claim is false. For example, \(4\times3=12\). The product \(12\) is even though \(3\) is odd. 2. One even factor is enough to make a product even. The even factor can be split into pairs, and repeating complete pairs keeps the total even.

Mason is not correct. For example, \(4 \times 3=12\). A correct pattern is: if at least one factor is even, the product is even.

- Think about the value of the two tens digits and the value of the two ones digits. - If the ones digits trade places, does their combined value change? - For the least and greatest sums, decide which two digits should be in the tens places.

1. Swapping the ones digits does not change the total value of the ones digits, and the tens digits stay where they are. So the sum does not change. 2. Once the two tens digits are chosen, the other two cards must be the ones digits. That fixes both the tens total and the ones total. 3. For the least sum, put \(2\) and \(4\) in the tens places. One arrangement is \(25+48=73\). 4. For the greatest sum, put \(5\) and \(8\) in the tens places. One arrangement is \(52+84=136\). 5. There are \(6\) possible pairs of tens digits, giving the sums \(73,82,100,109,127,136\).

a) Swapping the ones digits keeps their total the same, so the sum does not change. b) After the tens digits are chosen, the other two digits must be the ones digits, so the sum is fixed. c) Least sum: \(73\); greatest sum: \(136\); \(6\) different sums.

- Subtract carefully by place value. - Compare the tens digits of the results. - Add the hundreds and ones digits in each result. - Think about why different starting numbers can produce the same difference.

1. \(421 - 124 = 297\). 2. \(623 - 326 = 297\). 3. \(825 - 528 = 297\). 4. Every result is \(297\). Its tens digit is \(9\), and its hundreds and ones digits add to \(2 + 7 = 9\).

a) \(297\) b) \(297\) c) \(297\) d) All three results are \(297\). The tens digit is \(9\), and the outer digits add to \(9\).

- Calculate all three expressions in Set A and compare the results. - Repeat for Set B. - Compare the two repeated results by finding their difference.

1. Set A gives \(654 - 456 = 198\), \(765 - 567 = 198\), and \(876 - 678 = 198\). 2. Set B gives \(642 - 246 = 396\), \(753 - 357 = 396\), and \(864 - 468 = 396\). 3. All three answers in each set are the same. Compare the two repeated answers: \(396 - 198 = 198\). 4. Therefore, every result in Set B is \(198\) greater than the result in the same position in Set A.

a) Set A: \(198\), \(198\), \(198\); Set B: \(396\), \(396\), \(396\) b) Each Set B result is \(198\) greater than the result in the same position in Set A.

- Rearrange the digits again after every subtraction. - Put the greatest digit in the hundreds place for the greatest number. - Stop as soon as a result has appeared before. - Check each subtraction carefully, especially when regrouping is needed.

1. Arrange \(8,4,1\): \(841 - 148 = 693\). 2. Arrange \(9,6,3\): \(963 - 369 = 594\). 3. Arrange \(9,5,4\): \(954 - 459 = 495\). 4. Repeating the rule with \(9,5,4\) gives \(954 - 459 = 495\) again. 5. The result repeats on the fourth subtraction step.

\(841 - 148 = 693\) \(963 - 369 = 594\) \(954 - 459 = 495\) \(954 - 459 = 495\) The first repeated result appears on step \(4\).

- Work on one chain at a time. - With a zero, the least arrangement may begin with \(0\), as in \(038\), which has value \(38\). - Look for a result that appears in both chains.

1. Maya: \(721 - 127 = 594\). Theo: the least arrangement is \(038\), whose value is \(38\), so \(830 - 38 = 792\). 2. Maya continues with \(954 - 459 = 495\), reaching \(495\) in \(2\) steps. 3. Theo continues: \(972 - 279 = 693\), then \(963 - 369 = 594\), then \(954 - 459 = 495\). 4. Both chains eventually reach \(594\) and then \(495\), although Theo's chain is longer.

a) Maya: \(721 - 127 = 594\); Theo: \(830 - 38 = 792\) b) Maya: \(594 \rightarrow 495\). Theo: \(792 \rightarrow 693 \rightarrow 594 \rightarrow 495\). c) Both chains meet at \(594\) and end at \(495\).

- Look at one corner. How many sides touch it? - Ask how many times that corner appears when all three side sums are added. - For c), the outside number \(17\) already contains the two corners on that side.

1. Each corner touches two sides, so its number is included in two of the three outside sums. Therefore the total of the outside numbers counts every corner twice. 2. The three corner numbers therefore add to \(42\div2=21\). 3. The outside number \(17\) is the sum of the two corners on that side. The opposite corner is the part of the corner total not included in that side sum: \(21-17=4\).

a) Each corner touches two sides, so each corner number is used in two outside sums. b) \(21\) c) \(4\), because \(21-17=4\)

- First find the difference after only the first number increases. - Compare that answer with \(320\). - To make a difference smaller, should the number being subtracted increase or decrease? - Test the same idea with a simple subtraction such as \(10-5=5\).

1. Increasing the first number by \(50\) would increase the difference to \(300 + 50 = 350\). 2. The target difference, \(320\), is \(350 - 320 = 30\) less than \(350\). 3. To decrease the difference by \(30\), increase the number being subtracted by \(30\).

The number being subtracted must increase by \(30\). Raising the first number by \(50\) would raise the difference from \(300\) to \(350\). To get \(320\), subtract \(30\) more.

- Find a place where an upper brick and one brick below it are known. - Use subtraction to find missing bricks in lower rows. - Use each new value to solve neighboring bricks. - Then add upward to complete the upper rows.

1. Find the second bottom brick: \(222 - 128 = 94\). 2. Find the third bottom brick: \(209 - 62 = 147\). 3. Find the middle brick in the second row: \(94 + 147 = 241\). 4. Find the third-row bricks: \(222 + 241 = 463\) and \(241 + 209 = 450\). 5. Find the top: \(463 + 450 = 913\).

Bottom row: \(128\), \(94\), \(147\), \(62\) Second row: \(222\), \(241\), \(209\) Third row: \(463\), \(450\) Top: \(913\)

- Start at the top. If an upper brick and one brick below it are known, subtract to find the other lower brick. - Continue one row at a time. - Look for groups of three bricks with two known values. - Check each result by adding upward.

1. Wall a): \(500 - 240 = 260\); \(260 - 160 = 100\); \(160 - 60 = 100\); \(100 - 60 = 40\); \(240 - 100 = 140\); and \(140 - 40 = 100\). 2. Wall b): \(600 - 330 = 270\); \(330 - 130 = 200\); \(270 - 130 = 140\); \(200 - 120 = 80\); \(130 - 80 = 50\); and \(140 - 50 = 90\).

a) Bottom row: \(100\), \(60\), \(40\), \(100\); second row: \(160\), \(100\), \(140\); third row: \(260\), \(240\); top: \(500\) b) Bottom row: \(120\), \(80\), \(50\), \(90\); second row: \(200\), \(130\), \(140\); third row: \(330\), \(270\); top: \(600\)

- Start at the top. Use the top and one known brick below it to find the other brick in that row. - Continue working downward one row at a time. - When an upper brick and one lower brick are known, subtract to find the other lower brick.

1. Find the left brick in the third row: \(670 - 308 = 362\). 2. Find the middle brick in the second row: \(362 - 189 = 173\). 3. Find the third bottom brick: \(173 - 101 = 72\). 4. Find the right brick in the second row: \(308 - 173 = 135\). 5. Find the fourth bottom brick: \(135 - 72 = 63\). 6. Find the first bottom brick: \(189 - 101 = 88\).

First bottom brick: \(88\) Third bottom brick: \(72\) Fourth bottom brick: \(63\)

- Begin at the top and find the missing brick next to \(430\). - Use subtraction when working from an upper brick to a missing lower brick. - Use each new value to continue downward. - Check by adding from the bottom upward.

1. Find the right brick below the top: \(905 - 430 = 475\). 2. Find the left brick in the second row: \(430 - 230 = 200\). 3. Find the second bottom brick: \(200 - 120 = 80\). 4. The remaining values are \(230 - 80 = 150\), \(475 - 230 = 245\), and \(245 - 150 = 95\).

Right brick in the third row: \(475\) Left brick in the second row: \(200\) Second bottom brick: \(80\)

- Start with a connected group that has two known values. - When an upper brick and one lower brick are known, subtract to find the other lower brick. - Enter each new value before choosing the next step. - You may need to work backward from the top.

1. Find the second bottom brick: \(250 - 140 = 110\). 2. Find the left brick in the second row: \(80 + 110 = 190\). 3. Find the left brick in the third row: \(190 + 250 = 440\). 4. Find the right brick in the third row: \(900 - 440 = 460\). 5. Find the right brick in the second row: \(460 - 250 = 210\). 6. Find the rightmost bottom brick: \(210 - 140 = 70\).

\(70\)

- The number-wall rule does not change when the values are less familiar. - Use written side calculations if the arithmetic is difficult to do mentally.

1. Find the third bottom brick: \(196 - 104 = 92\). 2. Complete the second row: \(123 + 85 = 208\) and \(85 + 92 = 177\). 3. Complete the third row: \(208 + 177 = 385\) and \(177 + 196 = 373\). 4. Find the top: \(385 + 373 = 758\).

Bottom row: \(123\), \(85\), \(92\), \(104\) Second row: \(208\), \(177\), \(196\) Third row: \(385\), \(373\) Top: \(758\)

- Begin near the top, where \(800\) and \(450\) determine the other third-row value. - When an upper sum and one lower addend are known, subtract to recover the other lower value. - Check the completed wall by adding upward from the bottom row.

1. Find the left brick in the third row: \(800 - 450 = 350\). 2. Find the left and right bricks in the second row: \(350 - 200 = 150\) and \(450 - 200 = 250\). 3. Complete the bottom row: \(150 - 60 = 90\), \(200 - 90 = 110\), and \(250 - 110 = 140\). 4. Check by adding upward: \(60 + 90 = 150\), \(90 + 110 = 200\), and \(110 + 140 = 250\).

Bottom row: \(60\), \(90\), \(110\), \(140\) Second row: \(150\), \(200\), \(250\) Third row: \(350\), \(450\) Top: \(800\)

- Trace which bricks in the middle row use each bottom-row value. - The middle bottom brick is used in both sums directly above the bottom row. - Predict the effect of the changes on the top before calculating any new wall bricks.

1. The middle row is \(1+4=5\) and \(4+2=6\), so the original top is \(11\). 2. In a three-brick bottom row, the middle bottom brick is used in both middle-row sums, while each outer bottom brick is used once. 3. Increasing the left brick by \(3\) adds \(3\) to the top. Increasing the middle brick by \(3\) adds \(6\). Increasing the right brick by \(3\) adds \(3\). 4. The total change is \(3+6+3=12\), so the new top would be \(23\).

a) Original top: \(11\) b) The top increases by \(12\), to \(23\). c) The middle bottom brick is used in both bricks above it, so its change is counted twice; each outer bottom brick is used only once.

- Look for given numbers that occur at equal distances to the left and right of the center. - After finding one unknown on the left, check whether the same relationship appears at the matching position on the right. - The explanation should name the mirror relationship, not only list the completed numbers.

1. The bottom outside bricks are each \(75-25=50\), giving bottom row \(50,25,100,25,50\). 2. The next row is \(75,125,125,75\); then \(200,250,200\); then \(450,450\); and the top is \(900\). 3. Mirror pairs are equal: bottom positions \(1\) and \(5\), bottom positions \(2\) and \(4\); second-row positions \(1\) and \(4\), positions \(2\) and \(3\); third-row positions \(1\) and \(3\); and the two fourth-row bricks. 4. Because the givens mirror each other around the center, once a value is found on one side, its matching position on the other side has the same value.

a) Bottom: \(50,25,100,25,50\) Second row: \(75,125,125,75\) Third row: \(200,250,200\) Fourth row: \(450,450\) Top: \(900\) b) Matching positions the same distance from the center have equal numbers. c) The starting numbers match on the two sides, so the same calculations repeat. Once you find a number on one side, the matching place on the other side has the same number.

- Use the top and known left third-row value to find the right third-row value. - Continue downward where an upper value and one lower value are known. - Confirm the completed wall by recomputing each row upward.

1. The right brick in the third row is \(800 - 380 = 420\). 2. The middle brick in the second row is \(380 - 180 = 200\). 3. The first bottom brick is \(180 - 100 = 80\). 4. The third bottom brick is \(200 - 100 = 100\). 5. The right brick in the second row is \(420 - 200 = 220\). 6. The last bottom brick is \(220 - 100 = 120\).

Bottom row: \(80\), \(100\), \(100\), \(120\) Second row: \(180\), \(200\), \(220\) Third row: \(380\), \(420\) Top: \(800\)

- Complete each wall upward before comparing the tops. - Track how the \(+20\) change in the left outer brick reaches the top. - Track how the \(-10\) change in the middle brick affects both second-row bricks.

1. In wall 1, the second row is \(200+100=300\) and \(100+200=300\), so the top is \(600\). 2. In wall 2, the second row is \(220+90=310\) and \(90+200=290\), so the top is \(600\). 3. Increasing the left outer brick by \(20\) increases the top by \(20\), because an outer brick is used once. Decreasing the middle brick by \(10\) decreases both second-row bricks by \(10\), lowering the top by \(20\). The two changes cancel.

Both top values are \(600\). The \(+20\) change in an outer bottom brick adds \(20\) to the top, while the \(-10\) change in the middle bottom brick subtracts \(20\) from the top because the middle brick is used twice. The overall change is \(0\).

- Work downward from the top one row at a time. - Check the bottom row by adding upward to reproduce every given brick.

1. Find the left brick in the third row: \(999 - 555 = 444\). 2. Find the left and right bricks in the second row: \(444 - 244 = 200\) and \(555 - 244 = 311\). 3. Complete the bottom row: \(200 - 80 = 120\), \(244 - 120 = 124\), and \(311 - 124 = 187\).

Bottom row: \(80\), \(120\), \(124\), \(187\) Second row: \(200\), \(244\), \(311\) Third row: \(444\), \(555\) Top: \(999\)

- Begin at the highest connected set with one missing value. - Work downward by subtraction until a complete lower row emerges. - Verify the finished challenge by rebuilding the sums upward.

1. The right brick in the third row is \(950 - 450 = 500\). 2. The left and right bricks in the second row are \(450 - 240 = 210\) and \(500 - 240 = 260\). 3. Complete the bottom row: \(210 - 100 = 110\), \(240 - 110 = 130\), and \(260 - 130 = 130\).

Bottom row: \(100\), \(110\), \(130\), \(130\) Second row: \(210\), \(240\), \(260\) Third row: \(450\), \(500\) Top: \(950\)