Free math worksheets

- Find how much must be subtracted first to reach the previous multiple of \(10\). - Use that amount as the first part of the split. - The split parts must total the original subtrahend.

1. a) Split \(5\) as \(2+3\): \(232-2=230\), then \(230-3=227\). 2. b) Split \(7\) as \(3+4\): \(463-3=460\), then \(460-4=456\). 3. c) Split \(6\) as \(1+5\): \(851-1=850\), then \(850-5=845\). 4. d) Split \(8\) as \(4+4\): \(544-4=540\), then \(540-4=536\).

a) \(5=2+3\); \(232-2=230\); \(230-3=227\) b) \(7=3+4\); \(463-3=460\); \(460-4=456\) c) \(6=1+5\); \(851-1=850\); \(850-5=845\) d) \(8=4+4\); \(544-4=540\); \(540-4=536\)

- Find the small first subtraction that lands exactly on the previous hundred. - Split the subtrahend into that amount and what remains. - Show both subtraction steps in the answer.

1. a) Split \(4\) as \(2+2\): \(302-2=300\), then \(300-2=298\). 2. b) Split \(8\) as \(5+3\): \(605-5=600\), then \(600-3=597\). 3. c) Split \(6\) as \(3+3\): \(103-3=100\), then \(100-3=97\). 4. d) Split \(5\) as \(1+4\): \(901-1=900\), then \(900-4=896\).

a) \(4=2+2\); \(302-2=300\); \(300-2=298\) b) \(8=5+3\); \(605-5=600\); \(600-3=597\) c) \(6=3+3\); \(103-3=100\); \(100-3=97\) d) \(5=1+4\); \(901-1=900\); \(900-4=896\)

- In each group, all expressions have the same minuend. - Find one subtraction that reaches the previous ten and use it for the whole group. - For each original subtrahend, determine how much remains after that common first step.

1. a) The common first step is \(453-3=450\). Then subtract the remainders \(2,4,6,1\): the differences are \(448,446,444,449\). 2. b) The common first step is \(814-4=810\). Then subtract the remainders \(2,4,1,5\): the differences are \(808,806,809,805\).

a) First \(453-3=450\): \(450-2=448\), \(450-4=446\), \(450-6=444\), \(450-1=449\). b) First \(814-4=810\): \(810-2=808\), \(810-4=806\), \(810-1=809\), \(810-5=805\).

- Find how many tens are needed to reach the previous hundred. - Use that same first subtraction for every expression in the group. - Subtract only the remaining tens in the second step.

1. a) The common first step is \(320-20=300\). The remaining amounts are \(20,40,70\), giving \(280,260,230\). 2. b) The common first step is \(610-10=600\). The remaining amounts are \(20,40,70\), giving \(580,560,530\).

a) First \(320-20=300\): \(300-20=280\), \(300-40=260\), \(300-70=230\). b) First \(610-10=600\): \(600-20=580\), \(600-40=560\), \(600-70=530\).

- First find the value on the left. - Then compare it with the number on the right. - Choose the symbol that shows whether the left side is less than, greater than, or equal to the right side.

1. For a), \(504 - 7 = 497\), so \(497 = 497\). 2. For b), \(712 - 6 = 706\), so \(706 < 708\). 3. For c), \(230 - 50 = 180\), so \(180 = 180\). 4. For d), \(941 - 9 = 932\), so \(932 < 933\).

a) \(=\) b) \(<\) c) \(=\) d) \(<\)

- Separate the hundreds from the tens-and-ones part. - Solve the tens-and-ones subtraction first. - State explicitly how that related fact determines the three-digit difference.

1. a) \(24-8=16\); keeping the \(5\) hundreds gives \(524-8=516\). 2. b) \(11-5=6\); keeping the \(3\) hundreds gives \(311-5=306\). 3. c) \(52-6=46\); keeping the \(8\) hundreds gives \(852-6=846\). 4. d) \(33-7=26\); keeping the \(1\) hundred gives \(133-7=126\).

a) \(24-8=16\); keep \(5\) hundreds: \(524-8=516\). b) \(11-5=6\); keep \(3\) hundreds: \(311-5=306\). c) \(52-6=46\); keep \(8\) hundreds: \(852-6=846\). d) \(33-7=26\); keep \(1\) hundred: \(133-7=126\).

- A subtraction result can be checked with the inverse operation. - Combine the claimed difference with the number that was subtracted. - Compare that total with the starting number.

1. Add the claimed difference to the number subtracted: \(428+245=673\). 2. The sum equals the starting number, so the subtraction is correct.

Yes. \(428+245=673\), so \(673-245=428\).

- Which parts of \(457\) have already been subtracted? - Which place value has not yet been used? - Continue from the last intermediate result.

1. Paul has already subtracted the \(4\) hundreds and \(5\) tens in \(457\). 2. He still needs to subtract the \(7\) ones. 3. The final step is \(374 - 7 = 367\).

The final step is \(374 - 7 = 367\). The difference is \(367\).

- 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 \(879 - 345 = 534\).

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

- Decide which method gives a short, clear path for each expression. - When subtracting in parts, separate tens and ones. - When counting up, record the jumps and add their sizes.

1. a) Subtract in parts: \(400-70=330\), then \(330-2=328\). 2. b) Subtract in parts: \(900-30=870\), then \(870-6=864\). 3. c) Subtract in parts: \(600-50=550\), then \(550-8=542\). 4. d) Count up: \(19+1=20\), then \(20+180=200\); the total increase is \(1+180=181\).

a) Subtract in parts: \(400-70=330\); \(330-2=328\). b) Subtract in parts: \(900-30=870\); \(870-6=864\). c) Subtract in parts: \(600-50=550\); \(550-8=542\). d) Count up: \(19+1=20\); \(20+180=200\); \(1+180=181\).

- Find each difference separately. - Compare the hundreds, tens, and ones digits of the results. - Check whether any two results are equal.

1. Lucas: \(900 - 450 = 450\). 2. Sarah: \(1000 - 555 = 445\). 3. Max: \(812 - 362 = 450\). 4. Compare the results: \(445 < 450 = 450\).

Sarah: \(445\); Lucas: \(450\); Max: \(450\). \(445 < 450 = 450\)

- 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) \(320\), \(388\), \(318\)

- Identify the total and the part that rides the bus. - Subtract the known part from the total. - You can decompose \(280\) into hundreds and tens. - Check that your answer is less than \(630\).

1. Subtract the number of bus riders from the total: \(630 - 280\). 2. Subtract in parts: \(630 - 200 = 430\), then \(430 - 80 = 350\).

\(350\) students walk to school.

- Use subtraction to find the difference between the capacity and the current number of books. - Count up to the next hundred first. - Then count the remaining hundreds to the target.

1. Subtract the number of books on the shelf from its capacity: \(600 - 380 = 220\).

\(220\) more books can fit on the shelf.

- Decide whether the number of daffodils must be less than or greater than the number of tulips. - The phrase “fewer than” indicates subtraction. - Decompose \(370\) into hundreds and tens.

1. Subtract the difference from the number of tulips: \(850 - 370\). 2. Subtract in parts: \(850 - 300 = 550\), then \(550 - 70 = 480\).

The garden center planted \(480\) daffodils.

- What is the total distance of the trail? - How far has the class already walked? - Use subtraction to find the remaining distance. - Count up to the next hundred or subtract in parts.

1. Subtract the distance already walked from the total distance: \(940\,\text{m} - 460\,\text{m}\). 2. Subtract in parts: \(940 - 400 = 540\), then \(540 - 60 = 480\). 3. The remaining distance is \(480\,\text{m}\).

\(480\,\text{m}\) remain.

- Subtract the known part from the whole. - Decompose \(245\) into hundreds, tens, and ones. - Check your answer with a related addition equation.

1. Subtract the number of bears from the total: \(730 - 245\). 2. Decompose \(245\): \(730 - 200 = 530\), \(530 - 40 = 490\), and \(490 - 5 = 485\).

There are \(485\) stuffed dogs.

- Notice whether you are subtracting ones or tens. - Could you first go back to a multiple of \(10\) or \(100\)? - Compare parts a and b. How does subtracting \(4\) differ from subtracting \(40\)?

1. \(670 - 4 = 666\). 2. \(670 - 40 = 630\). 3. \(910 - 7 = 903\). 4. \(910 - 70 = 840\). 5. \(340 - 6 = 334\).

a) \(666\) b) \(630\) c) \(903\) d) \(840\) e) \(334\)

- Subtract the tens first, then the ones. - For a problem with \(1000\), you can count up from the smaller number to \(1000\). - When subtracting the ones, check whether you cross a multiple of \(10\).

1. \(1000 - 430 = 570\). 2. Count up from \(760\) to \(1000\), or subtract in parts: \(1000 - 760 = 240\). 3. Subtract \(5\) tens: \(684 - 50 = 634\). 4. Subtract in parts: \(457 - 30 = 427\), then \(427 - 8 = 419\). 5. Subtract in parts: \(832 - 20 = 812\), then \(812 - 5 = 807\).

a) \(570\) b) \(240\) c) \(634\) d) \(419\) e) \(807\)

- Find the difference on the left first. - Compare that result with the number on the right. - The open side of the comparison symbol faces the greater value.

1. \(430 - 70 = 360\), and \(360 > 350\). 2. \(650 - 80 = 570\), and \(570 < 580\). 3. \(910 - 50 = 860\), and \(860 = 860\). 4. \(500 - 33 = 467\), and \(467 < 477\). 5. \(1000 - 12 = 988\), and \(988 > 978\).

a) \(>\) b) \(<\) c) \(=\) d) \(<\) e) \(>\)

- Keep the same three numbers in each fact family. - In a related subtraction fact, the addition sum becomes the starting number. - Explain what subtraction does to the addition relationship.

1. a) The related subtraction equations are \(360-356=4\) and \(360-4=356\). 2. b) The related subtraction equations are \(500-445=55\) and \(500-55=445\). 3. In each family, subtraction undoes the addition: the sum is the whole, and subtracting either addend gives the other addend.

a) \(360-356=4\) and \(360-4=356\) b) \(500-445=55\) and \(500-55=445\) The subtraction facts undo the addition: subtracting either addend from the sum gives the other addend.

- Identify which class collected more. - Decide whether the unknown amount must be less than or greater than \(385\). - Subtract the difference from the greater amount.

1. The other class collected fewer acorns, so subtract the difference: \(385 - 120 = 265\).

The other class collected \(265\) acorns.

- 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.

- 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.

- Find each difference separately. - Regroup carefully when subtracting tens. - Compare all four results.

1. For a), \(320 - 50 = 270\). 2. For b), \(350 - 80 = 270\). 3. For c), \(310 - 60 = 250\). 4. For d), \(440 - 70 = 370\). 5. Problems a) and b) both have a result of \(270\).

a) \(320 - 50 = 270\) b) \(350 - 80 = 270\) c) \(310 - 60 = 250\) d) \(440 - 70 = 370\) Problems a) and b) have the same result.

- Write the situation as a subtraction equation with one missing value. - The amount sold is the distance from the remaining amount back to the starting amount. - Use addition to check that the two parts make the original total.

1. The number sold is the missing subtrahend in \(820-\square=550\). 2. \(820-550=270\), so \(270\) notebooks were sold. 3. Check: \(550+270=820\).

a) \(270\) notebooks b) \(550+270=820\)

- Write the starting number before the minus sign. - Subtract to find the result. - Recall the name of the result of subtraction.

1. Write the subtraction expression: \(900 - 560\). 2. Calculate: \(900 - 560 = 340\). 3. The result of a subtraction equation is called the difference.

\(900 - 560 = 340\). The result is called the difference.

- Read the ending marker and the size and direction of the jump from the diagram. - Use the inverse operation to work backward to the starting marker.

1. The diagram shows an ending marker of \(810\,\text{m}\) after a forward jump of \(345\,\text{m}\). 2. Subtract the distance traveled from the ending marker: \(810-345\). 3. Subtract in steps: \(810-300=510\), \(510-40=470\), and \(470-5=465\). 4. The starting marker was \(465\,\text{m}\).

The hiker started at the \(465\,\text{m}\) marker because \(810\,\text{m}-345\,\text{m}=465\,\text{m}\).

- Use both the total number of muffins and the number already decorated. - Think of a question that asks about the part still left. - Choose an operation that compares the whole amount with the part already decorated.

1. One possible question is, “How many muffins are not decorated yet?” 2. Subtract: \(42 - 9 = 33\).

Example question: “How many muffins are not decorated yet?” Answer: \(33\) muffins are not decorated yet.

- Match each statement to the two numbers for that club. - Check all three statements before choosing.

1. Statement A is false because \(11 > 8\). 2. Statement B is true because \(12 - 10 = 2\). 3. Statement C is false because both values are \(9\).

Statement B is true.

- Find the difference between the two days for each instrument. - Compare the three differences.

1. The flute difference is \(12 - 9 = 3\). 2. The guitar difference is \(13 - 8 = 5\). 3. The drum difference is \(10 - 10 = 0\). 4. The greatest difference is \(5\), for guitar.

Guitar

- Subtract the amount used from the amount available for each material. - Compare the three remaining amounts.

1. Cardboard remaining: \(25 - 8 = 17\). 2. Yarn remaining: \(18 - 6 = 12\). 3. Craft sticks remaining: \(30 - 12 = 18\). 4. The least remaining amount is \(12\), for yarn.

Yarn has the least amount remaining.

- Start with the ones column and compare the number of ones available with the number being subtracted. - If needed, unbundle \(1\) ten into \(10\) ones. - After unbundling, update both the ones and tens columns before subtracting.

1. There are not enough ones to subtract \(8\) ones from \(2\) ones, so unbundle \(1\) ten as \(10\) ones. 2. The ones become \(12\), and the tens decrease from \(6\) tens to \(5\) tens. Then \(12-8=4\). 3. Subtract the tens: \(5-3=2\). 4. Subtract the hundreds: \(4-1=3\). The difference is \(324\).

The difference is \(324\). Unbundling \(1\) ten changes \(2\) ones to \(12\) ones and leaves \(5\) tens before the subtraction continues.

- Keep corresponding place values aligned. - Start with the ones and compare the top and bottom digit in each column. - Unbundling is needed only when a place does not contain enough units to subtract the digit below it.

1. Subtract the ones: \(6-3=3\). 2. Subtract the tens: \(8-4=4\) tens. 3. Subtract the hundreds: \(7-2=5\) hundreds. 4. In every column, the top digit is at least as large as the digit below it, so no unbundling is needed. The difference is \(543\).

\(543\). No unbundling is needed because each place can be subtracted directly.

- A three-digit number cannot have \(0\) in the hundreds place. - Compare possible hundreds digits before arranging tens and ones. - Once the hundreds digits differ, they decide which three-digit number is greater.

1. A three-digit number cannot start with \(0\), so the smallest possible hundreds digit is \(1\). Putting the remaining digits in increasing order gives \(104\). 2. The greatest possible hundreds digit is \(4\). Putting the remaining digits in decreasing order gives \(410\). 3. Since \(1\) hundred is less than \(4\) hundreds, \(104<410\).

a) Least: \(104\); greatest: \(410\) b) The least number must use \(1\) hundred because \(0\) cannot begin a three-digit number; the greatest uses \(4\) hundreds because \(4\) is the largest digit. c) \(104<410\)

- Break the second number into tens and ones. - Subtract the tens first and record the new value. - Use that intermediate value before subtracting the ones.

1. For \(263-47\), subtract the tens and then the ones: \(263-40=223\), then \(223-7=216\). 2. For \(514-38\), subtract the tens and then the ones: \(514-30=484\), then \(484-8=476\). 3. For \(821-56\), subtract the tens and then the ones: \(821-50=771\), then \(771-6=765\).

a) \(263-40=223\), then \(223-7=216\) b) \(514-30=484\), then \(484-8=476\) c) \(821-50=771\), then \(771-6=765\)

- Break the subtrahend into hundreds, tens, and ones. - Subtract the hundreds first, then the tens, and finally the ones. - Record each intermediate result before moving to the next place-value part.

1. For \(452-168\): \(452-100=352\), \(352-60=292\), and \(292-8=284\). 2. For \(731-354\): \(731-300=431\), \(431-50=381\), and \(381-4=377\). 3. For \(615-279\): \(615-200=415\), \(415-70=345\), and \(345-9=336\).

a) \(452-100=352\), \(352-60=292\), \(292-8=284\) b) \(731-300=431\), \(431-50=381\), \(381-4=377\) c) \(615-200=415\), \(415-70=345\), \(345-9=336\)

- Read the starting value and signed jump sizes from the image. - Apply each jump in order and record the landing. - Combine the two negative jump sizes to identify the total amount subtracted.

1. The line starts at \(452\). The first jump is \(-70\), so the first landing is \(382\). 2. The next jump is \(-8\), giving \(374\). 3. The jumps subtract \(70+8=78\), so the represented equation is \(452-78=374\).

a) \(382\), \(374\) b) \(452-78=374\)

- Check the steps in order rather than starting the subtraction again. - Compare each operation with the change in the next number. - Once you find the first incorrect step, continue from the last correct value.

1. The first two steps are correct. 2. The last step is incorrect because \(474-1=473\), not \(475\). 3. Therefore, \(824-351=473\).

Ben's first incorrect step is \(474-1=475\). It should be \(474-1=473\), so the difference is \(473\).

- Break each subtrahend into hundreds, tens, and ones. - Keep each intermediate result because the next subtraction starts there. - Compare the final differences only after completing the step calculations.

1. For a), \(903-500=403\), \(403-40=363\), and \(363-6=357\). 2. For b), \(721-300=421\), \(421-80=341\), and \(341-4=337\). 3. For c), \(612-200=412\), \(412-70=342\), and \(342-5=337\). 4. The differences in parts b and c are equal.

a) \(903-500=403\), \(403-40=363\), \(363-6=357\) b) \(721-300=421\), \(421-80=341\), \(341-4=337\) c) \(612-200=412\), \(412-70=342\), \(342-5=337\) The differences in b) and c) are equal: both are \(337\).

- Keep the starting number fixed and think about what happens when more is subtracted. - Calculate each difference after making a prediction. - Compare the order of the subtrahends with the order of the differences.

1. With the same minuend, subtracting a larger number should produce a smaller difference. 2. The differences are \(424\), \(337\), \(215\), and \(106\). 3. They decrease in the same order that the subtrahends increase, so the results match the prediction.

a) The differences should decrease. b) \(424\), \(337\), \(215\), \(106\) c) Yes.

- Read the numbers and subtraction signs directly from the tree. - Start with the deepest operation and use each result in the next box. - After completing the boxes, read the nested operations as one left-to-right subtraction sequence.

1. The deepest subtraction is \(815-240=575\). 2. The next result is \(575-87=488\). 3. The final result is \(488-356=132\). 4. The tree represents starting at \(815\) and subtracting \(240\), then \(87\), then \(356\), ending at \(132\).

Missing results: \(575\), \(488\), \(132\). Sequential subtraction: \(815-240-87-356=132\).

- Read each chart by its hundreds, tens, and ones columns. - Treat each question as a missing-addend equation. - Compare each two-digit chart with its ten-times-as-large chart.

1. Panel a) shows \(60\), and \(60+40=100\). 2. Panel b) shows \(600\), and \(600+400=1000\). 3. Panel c) shows \(45\), and \(45+55=100\). 4. Panel d) shows \(450\), and \(450+550=1000\). 5. \(400\) is ten times \(40\), and \(550\) is ten times \(55\).

a) \(60\); add \(40\) b) \(600\); add \(400\) c) \(45\); add \(55\) d) \(450\); add \(550\) e) Each missing addend for \(1000\) is ten times the corresponding missing addend for \(100\).

- Start by subtracting the \(300\) in the subtrahend. - From \(218\), determine exactly how much must be subtracted to reach \(200\). - Keep track of how much of the original \(362\) remains after each step.

1. Subtract the hundreds: \(518-300=218\). 2. Of the remaining \(62\), subtract \(18\) first to land on \(200\): \(218-18=200\). 3. There are \(44\) left to subtract. Subtract \(42\) to get \(158\), then subtract the final \(2\): \(158-2=156\). 4. The subtracted parts total \(300+18+42+2=362\), so the difference is \(156\).

\(518-300=218\), \(218-18=200\), \(200-42=158\), then \(158-2=156\).

- A correct sum can be checked by undoing one addend. - Compare the result of the check with the other addend. - If the check fails, adjust the claimed sum and verify again.

1. If \(774\) were the correct sum, then \(774-538\) would equal \(246\). 2. \(774-538=236\), so the claim fails the inverse-operation check. 3. The correct sum is \(538+246=784\). 4. Check: \(784-538=246\).

The claim is wrong. The correct sum is \(784\), and \(784-538=246\).

- Replace the subtrahend with a nearby multiple of \(100\). - Decide whether that easier subtraction removes too much or too little. - Correct the difference by exactly the amount of the replacement.

1. Subtract \(500\): \(832-500=332\). Since this subtracts \(1\) too much, add \(1\): \(332+1=333\). 2. Subtract \(300\): \(546-300=246\). Since this subtracts \(2\) too much, add \(2\): \(246+2=248\). 3. Subtract \(200\): \(715-200=515\). Since this subtracts \(3\) too much, add \(3\): \(515+3=518\).

a) \(832-500=332\), then \(332+1=333\) b) \(546-300=246\), then \(246+2=248\) c) \(715-200=515\), then \(515+3=518\)

- Compare \(198\) with \(200\). - Decide whether subtracting \(200\) removes too much or too little. - Show the exact correction after the easier subtraction.

1. Subtract the nearby hundred: \(462-200=262\). 2. This subtracts \(2\) too many, so add \(2\): \(262+2=264\). 3. There are \(264\) books left.

\(462-200=262\); \(262+2=264\). There are \(264\) books left.

- Compare the subtrahend with the multiple of \(100\) shown in the equation. - What is the difference between those two numbers? - Add back the amount that was subtracted in excess.

1. The difference between \(399\) and \(400\) is \(1\). Since \(924 - 400 = 524\), add \(1\): \(524 + 1 = 525\). 2. The difference between \(298\) and \(300\) is \(2\). Since \(637 - 300 = 337\), add \(2\): \(337 + 2 = 339\). 3. The difference between \(196\) and \(200\) is \(4\). Since \(512 - 200 = 312\), add \(4\): \(312 + 4 = 316\).

a) \(924 - 399 = 924 - 400 + 1 = 525\) b) \(637 - 298 = 637 - 300 + 2 = 339\) c) \(512 - 196 = 512 - 200 + 4 = 316\)

- Compare each changed number with the original one. - Think about whether the distance between two numbers changes when both move by the same amount. - Verify both methods numerically.

1. Maya's method gives \(453-200+1=253+1=254\). 2. Eli adds \(1\) to both the minuend and the subtrahend: \(453-199=454-200=254\). 3. Adding the same amount to both numbers does not change their difference.

a) Both methods give \(254\). b) Maya corrects for subtracting \(1\) too much; Eli adds \(1\) to both numbers, which keeps the difference unchanged.

- Read the starting number from the hundreds, tens, and ones columns. - Compare the shown ones with the \(8\) ones that must be subtracted. - After unbundling for the ones place, check the tens place again.

1. The chart shows \(6\) hundreds, \(2\) tens, and \(3\) ones, so the starting number is \(623\). 2. There are only \(3\) ones, so unbundle \(1\) ten first: \(623\) becomes \(6\) hundreds, \(1\) ten, and \(13\) ones. 3. There is then only \(1\) ten, so unbundle \(1\) hundred. The number is now \(5\) hundreds, \(11\) tens, and \(13\) ones. 4. Subtract \(3\) hundreds, \(5\) tens, and \(8\) ones to get \(265\).

a) \(623\) b) Unbundle a ten first. c) Rename the number as \(5\) hundreds, \(11\) tens, and \(13\) ones. d) \(265\)

- Look for a subtrahend close to a multiple of \(100\). - If you use compensation, show both the easier subtraction and the correction. - If you subtract in parts, show how the subtrahend is decomposed.

1. a) Compensation: \(754-400=354\), then \(354+1=355\). 2. b) Subtract in parts: \(600-100=500\), \(500-20=480\), then \(480-6=474\). 3. c) Compensation: \(432-300=132\), then \(132+2=134\).

a) Compensation: \(754-400=354\); \(354+1=355\). b) Subtract in parts: \(600-100=500\); \(500-20=480\); \(480-6=474\). c) Compensation: \(432-300=132\); \(132+2=134\).

- Use addition when the first number is missing. - To find a missing number being subtracted, find the difference between the first number and the known result. - Check each completed equation.

1. For a), subtract in parts: \(950 - 200 = 750\), then \(750 - 70 = 680\). 2. For b), use the inverse operation: \(445 + 155 = 600\), so \(600 - 155 = 445\). 3. For c), find the missing number being subtracted: \(824 - 549 = 275\), so \(824 - 275 = 549\).

a) \(680\) b) \(600\) c) \(275\)

- You are looking for pairs with one fixed difference. - For each number, look for another number that is \(7\) greater or \(7\) less. - Numbers near the same multiple of \(10\) or \(100\) may be useful to compare. - Compare two numbers by thinking about their distance on a number line.

1. Compare numbers that are close in value. 2. The matching pairs give \(135-128=7\), \(676-669=7\), \(403-396=7\), \(812-805=7\), and \(250-243=7\). 3. No other pair from the list has a difference of \(7\).

\(135-128=7\) \(676-669=7\) \(403-396=7\) \(812-805=7\) \(250-243=7\)

- Read the starting number and each subtracted part from the tree. - Work from the deepest subtraction outward. - Combine the three subtracted parts to identify the single subtrahend represented by the tree.

1. The deepest subtraction is \(632-200=432\). 2. Next, \(432-70=362\). 3. Finally, \(362-5=357\). 4. The three subtracted parts total \(200+70+5=275\), so the full equation is \(632-275=357\).

Missing results: \(432\), \(362\), \(357\). Full equation: \(632-275=357\).

- \(199\) is close to which multiple of \(100\)? - If you subtract too much, how do you correct the result? - When two numbers are close, count up from the smaller number to the larger number.

1. For part a, subtract \(200\): \(753 - 200 = 553\). Since \(200\) is \(1\) more than \(199\), add \(1\): \(553 + 1 = 554\). 2. For part b, count up \(2\) from \(398\) to \(400\), then \(2\) more from \(400\) to \(402\). The total difference is \(2 + 2 = 4\).

a) \(753 - 200 + 1 = 554\) b) \(398 + 2 = 400\), \(400 + 2 = 402\); the difference is \(4\).

- Choose a step-by-step method that is easy to track. - Breaking the subtrahend into place-value parts is one valid approach. - Record each intermediate value so your method can be checked.

1. For a), \(876-300=576\), \(576-40=536\), and \(536-2=534\). 2. For b), \(524-60=464\), then \(464-7=457\). 3. For c), \(911-400=511\), \(511-50=461\), and \(461-8=453\).

a) One valid method: \(876-300=576\), \(576-40=536\), \(536-2=534\) b) One valid method: \(524-60=464\), then \(464-7=457\) c) One valid method: \(911-400=511\), \(511-50=461\), \(461-8=453\)

- Look for a number close to a multiple of \(100\) or two numbers close together. - Name your method and include the step that makes that method identifiable. - Check that any compensation correction goes in the right direction.

1. a) Subtract in parts: \(800-70=730\), then \(730-2=728\). 2. b) Compensation: \(534-200=334\), then \(334+1=335\). 3. c) Count up: \(395+5=400\), then \(400+12=412\); \(5+12=17\). 4. d) Place-value subtraction: \(900-400=500\), \(60-50=10\), and \(7-2=5\); total \(515\).

a) Subtract in parts: \(800-70=730\); \(730-2=728\). b) Compensation: \(534-200=334\); \(334+1=335\). c) Count up: \(395+5=400\); \(400+12=412\); \(5+12=17\). d) Place value: \(900-400=500\), \(60-50=10\), \(7-2=5\); total \(515\).

- Read the operation, known values, and missing position from each tree. - When the number being subtracted is missing, work backward from the known difference. - Check each completed tree by reading the subtraction from start to finish.

1. a) The missing number is \(1000-720=280\). 2. b) The missing number is \(720-485=235\). 3. c) \(485-139=346\).

a) \(280\) b) \(235\) c) \(346\)

- Look for pairs of digits whose difference is \(3\). - Try to make the difference \(3\) in the hundreds, tens, and ones places. - Check that no digit repeats within an equation.

1. To make a difference of \(333\) without regrouping, choose a pair of digits with a difference of \(3\) in each place. 2. Using the pairs \((4,1)\), \((5,2)\), and \((6,3)\) gives \(456-123=333\). 3. Using the pairs \((7,4)\), \((8,5)\), and \((9,6)\) gives \(987-654=333\).

\(456-123=333\) \(987-654=333\)

- Find the nearest multiple of \(100\) to each subtrahend. - Compare how far each subtrahend is from that nearby hundred. - If you replace a subtrahend by a nearby hundred, remember to adjust the difference to compensate.

1. In problem A, \(395\) is \(5\) away from \(400\). In problem B, \(278\) is \(22\) away from \(300\). Therefore, problem A has the subtrahend closer to a multiple of \(100\). 2. For A, use compensation: \(700-400+5=305\). 3. For B, compensation also works: \(642-300+22=364\).

Problem A, because \(395\) is only \(5\) away from \(400\), while \(278\) is \(22\) away from \(300\). A: \(305\) B: \(364\)

- First find how many pages Klara has already read. - Subtract the pages she has left from the total number of pages. - Compare Klara's result with Max's \(240\) pages.

1. Find the number of pages Klara has read: \(560 - 310 = 250\). 2. Compare the amounts read: \(250 > 240\), so Klara has read more pages.

Klara has read more. She has read \(250\) pages, while Max has read \(240\) pages.

- Find all four differences first. - Use a subtraction strategy such as counting back to a nearby hundred. - Compare the hundreds digits of the results before comparing tens and ones.

1. Find the differences: a) \(920 - 350 = 570\), b) \(640 - 180 = 460\), c) \(810 - 270 = 540\), and d) \(758 - 490 = 268\). 2. Order the results: \(268 < 460 < 540 < 570\).

a) \(570\) b) \(460\) c) \(540\) d) \(268\) Order: \(268 < 460 < 540 < 570\)

- Split the amount being subtracted into two useful parts. - For some items, reaching a multiple of \(100\) first can make the next step easier. - Keep the first landing value before doing the second subtraction.

1. \(520-20=500\), then \(500-40=460\). 2. \(340-40=300\), then \(300-40=260\). 3. \(600-40=560\), then \(560-2=558\). 4. \(900-70=830\), then \(830-5=825\). 5. \(1000-10=990\), then \(990-3=987\).

a) \(520-20=500\), then \(500-40=460\) b) \(340-40=300\), then \(300-40=260\) c) \(600-40=560\), then \(560-2=558\) d) \(900-70=830\), then \(830-5=825\) e) \(1000-10=990\), then \(990-3=987\)

- Read the start, end, and signed jump sizes from the image. - Each landing becomes the starting point for the next jump. - The sum of the positive jump sizes is the distance between the endpoints.

1. Starting at \(780\), the \(+20\) jump lands at \(800\). 2. The \(+100\) jump lands at \(900\), and the \(+50\) jump ends at \(950\). 3. The total distance is \(20+100+50=170\). 4. The diagram represents \(950-780=170\).

a) \(800\), \(900\) b) \(20+100+50=170\) c) \(950-780=170\)

- Order the three numbers from least to greatest. - When a question asks how many more or how many are needed, find a difference. - Pay attention when the subtraction crosses a ten or a hundred. - Check each answer by adding the difference to the smaller amount.

1. Order the amounts: \(820 > 640 > 560\). Lisa has the most, and Tim has the fewest. 2. Lisa and Tim: \(820 - 560 = 260\). 3. Lisa and Paul: \(820 - 640 = 180\). 4. Paul and Tim: \(640 - 560 = 80\).

a) Most: Lisa; fewest: Tim b) \(260\) marbles c) \(180\) marbles d) \(80\) marbles

- Read the starting value and each signed jump directly from the diagram. - A negative jump moves to a smaller number. - Use each landing as the starting value for the next jump.

1. a) From \(432\), jump \(-10\) to \(422\), then \(-5\) to \(417\). 2. b) From \(432\), jump \(-100\) to \(332\), then \(-50\) to \(282\). 3. c) From \(432\), jump \(-200\) to \(232\), then \(-10\) to \(222\), then \(-5\) to \(217\).

a) \(422\), \(417\) b) \(332\), \(282\) c) \(232\), \(222\), \(217\)

- Look at how each category changes from September to October. - Subtract the September number from the October number in each row. - Compare the four increases after calculating them.

1. Nonfiction increased by \(210 - 145 = 65\). 2. Fiction increased by \(312 - 238 = 74\). 3. Comics increased by \(254 - 189 = 65\). 4. Picture books increased by \(167 - 92 = 75\). 5. Since \(75\) is greatest, picture books had the greatest increase.

Nonfiction: \(65\) Fiction: \(74\) Comics: \(65\) Picture books: \(75\) Picture books had the greatest increase.

- Decide which inverse operation will isolate each missing number. - In part 4, combine the two known addends first. - Check each value in the original equation.

1. Subtract to find the missing addend: \(610 - 480 = 130\). 2. Find the missing number being subtracted: \(820 - 740 = 80\). 3. Add to find the missing first number: \(410 + 190 = 600\). 4. First combine the known addends: \(350 + 100 = 450\). Then subtract: \(700 - 450 = 250\).

1. \(130\) 2. \(80\) 3. \(600\) 4. \(250\)

- Decide how adding apples changes the total. - Decide how removing pears changes the total. - Combine the increase and decrease.

1. Adding the apples changes the total by \(+55\). 2. Removing the pears changes the total by \(-20\). 3. Combine the changes: \(55 - 20 = 35\). The positive result means the total increases.

The total number of pieces of fruit increases by \(35\).

- 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) \(290\,\text{L}\)

- Use the relationship: starting number minus number subtracted equals difference. - When the starting number is missing, add the number subtracted and the difference. - When the number subtracted is missing, subtract the difference from the starting number.

1. For a), subtract to find the difference: \(840 - 365 = 475\). 2. For b), add the number subtracted and the difference to find the starting number: \(210 + 450 = 660\). 3. For c), subtract the difference from the starting number to find the number subtracted: \(910 - 280 = 630\).

a) \(475\) b) \(660\) c) \(630\)

- Identify the starting amount as the whole. - The amount sold and the amount left are two parts of that whole. - Use the related addition equation to check the subtraction.

1. The starting amount is the number sold plus the number remaining: \(370+230=600\). 2. The subtraction equation is \(600-370=230\). 3. The related addition check is \(230+370=600\).

a) \(600\) pencils b) \(600-370=230\) and \(230+370=600\)

- Compare \(396\) with \(400\). - Decide whether the first subtraction removes too much or too little. - The correction should undo that extra change.

1. Replacing \(396\) with \(400\) subtracts \(4\) too much. 2. Because too much was subtracted, the adjustment must add \(4\), not subtract \(4\). 3. \(502-400+4=102+4=106\).

Liam subtracts the adjustment instead of adding it back. The correct difference is \(106\).

- Decompose the numbers into hundreds and tens if helpful. - Notice whether the minuend and subtrahend change by the same amount. - Check a difference with a related addition equation.

1. For a), \(450 - 180 = 270\) and \(540 - 270 = 270\), so the values are equal. 2. For b), \(720 - 350 = 370\) and \(810 - 440 = 370\), so the values are equal. 3. For c), \(620 - 280 = 340\) and \(750 - 390 = 360\), so \(340 < 360\).

a) \(=\) b) \(=\) c) \(<\)

- Calculate and record each difference. - Regroup carefully across tens and hundreds when needed. - Compare all five results.

1. Calculate the differences: A) \(724 - 385 = 339\), B) \(613 - 274 = 339\), C) \(801 - 462 = 339\), D) \(542 - 193 = 349\), and E) \(930 - 591 = 339\). 2. A, B, C, and E all have a result of \(339\). D has a result of \(349\).

D) \(542 - 193 = 349\)

- Which subtrahend is closest to a multiple of \(100\)? - Why are multiples of \(100\) easier to subtract mentally? - If you subtract \(200\) instead of \(199\), how must you adjust?

1. The subtrahends \(312\) and \(245\) are not especially close to a multiple of \(100\). The subtrahend \(199\) is only \(1\) less than \(200\). 2. Choose C. Subtract \(200\): \(631 - 200 = 431\). 3. Since \(200\) is \(1\) more than \(199\), add \(1\): \(431 + 1 = 432\).

C is best because \(199\) is close to \(200\). \(631 - 200 + 1 = 432\).

- Decide which number is closest to a convenient hundred. - State which term you changed before calculating. - The correction must exactly undo the amount of the adjustment.

1. a) Adjust the minuend to \(300\): \(300-65=235\). Add back \(2\): \(235+2=237\). 2. b) Adjust the subtrahend to \(100\): \(514-100=414\). Add back \(3\): \(414+3=417\). 3. c) Adjust the subtrahend to \(200\): \(705-200=505\). Add back \(2\): \(505+2=507\). 4. d) Adjust the minuend to \(600\): \(600-47=553\). Add back \(3\): \(553+3=556\).

a) Minuend: \(300-65=235\); add back \(2\): \(237\). b) Subtrahend: \(514-100=414\); add back \(3\): \(417\). c) Subtrahend: \(705-200=505\); add back \(2\): \(507\). d) Minuend: \(600-47=553\); add back \(3\): \(556\).

- Find the original difference first. - Determine the new number being subtracted. - Consider what happens when less is subtracted from the same starting number.

1. Part a: \(950 - 380 = 570\). 2. Part b: The new number subtracted is \(380 - 20 = 360\). 3. The new difference is \(950 - 360 = 590\). 4. Decreasing the number subtracted by \(20\) increases the difference by \(20\).

a) The difference is \(570\). b) The new difference is \(590\), which is \(20\) greater than the original difference.

- 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) \(300\)

- Read the two endpoints of each marked interval. - Subtract the smaller endpoint from the larger endpoint. - Compare the three differences after calculating them.

1. \(A\): \(580-220=360\). 2. \(B\): \(460-140=320\). 3. \(C\): \(820-400=420\). 4. Since \(320<360<420\), the order is \(B, A, C\).

a) \(360\) b) \(320\) c) \(420\) d) \(B, A, C\)

- 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: \(3\)

- 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 \(33 + 38 = 71\).

\(20\)

- 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: \(62\)

- 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: \(63\)

- 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 \(59\).

- 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: \(750\)

- 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: \(40\)

- Work downward with the inverse relationship. - Start by asking, “What number minus \(10\) equals \(25\)?”

1. For the left value in the second row, solve \(x - 10 = 25\), so \(x = 35\). 2. For the bottom-left value, solve \(y - 15 = 35\), so \(y = 50\). 3. For the bottom-right value, solve \(15 - z = 10\), so \(z = 5\).

Bottom row: \(50\), \(15\), \(5\) Second row: \(35\), \(10\) Top: \(25\)

- Start with the connected bricks that include the known \(23\). - Work backward when a lower brick is missing. - Then continue upward using the addition rule.

1. The missing middle bottom value satisfies \(x+12=23\), so \(x=11\). 2. The left brick in the second row is \(78+11=89\). 3. The top is \(89+23=112\).

Bottom row: \(78\), \(11\), \(12\) Second row: \(89\), \(23\) Top: \(112\)

- Read the starting number from the chart before checking Jordan's claim. - Compare the ones shown with the \(7\) ones that must be subtracted. - If there are no tens available to unbundle directly, look to the hundreds place first.

1. The chart shows \(6\) hundreds, \(0\) tens, and \(4\) ones, so the starting number is \(604\). 2. There are not enough ones to subtract \(7\) ones. Reversing the order within a column changes the subtraction and is not a valid regrouping method. 3. Rename \(604\) as \(5\) hundreds, \(10\) tens, and \(4\) ones, then as \(5\) hundreds, \(9\) tens, and \(14\) ones. 4. Subtract by place: \(14-7=7\), \(9-4=5\), and \(5-2=3\), so \(604-247=357\).

a) \(604\) b) There are only \(4\) ones, so \(7\) ones cannot be subtracted without unbundling; reversing the digits is not valid. c) \(5\) hundreds, \(9\) tens, and \(14\) ones d) \(357\)

- A correct difference should combine with the number subtracted to make the starting number. - Test the claimed result in the related addition equation. - If the check fails, correct the difference and verify again.

1. Check Sam's claim: \(458+283=741\), not \(731\), so \(283\) is not correct. 2. The correct difference is \(731-458=273\). 3. Check: \(458+273=731\).

Sam is not correct. The difference is \(273\), and \(458+273=731\).

- Add neighboring bricks to build the next row. - Work upward one row at a time. - Keep checking that each new value stays within \(1000\).

1. The second row is \(280\), \(120\), \(60\), and \(30\). 2. The third row is \(400\), \(180\), and \(90\). 3. The fourth row is \(580\) and \(270\). 4. The top is \(850\).

Second row: \(280\), \(120\), \(60\), \(30\) Third row: \(400\), \(180\), \(90\) Fourth row: \(580\), \(270\) Top: \(850\)

- Complete each tree from its deepest operation outward. - Compare the total change made by the operations in each method. - In the shorter method, ask how much too much is subtracted first.

1. In a), \(642-200=442\), \(442-70=372\), and \(372-8=364\). 2. In b), \(642-300=342\), then \(342+22=364\). 3. Method b) uses fewer arithmetic steps. 4. Method b) subtracts \(300\) and adds back \(22\), which has a net effect of subtracting \(278\).

a) Method a) boxes: \(442\), \(372\), \(364\). Method b) boxes: \(342\), \(364\). b) Method b) uses fewer arithmetic steps. c) Subtracting \(300\) and then adding back \(22\) subtracts \(278\) altogether.

- The curved arrows show the signed jump sizes. - Use each jump to determine the next hidden landing. - Add the jump sizes only after you have read them from the diagram.

1. The jumps are \(+2\), \(+50\), \(+200\), and \(+3\). 2. Starting at \(548\), the missing landings are \(550\), \(600\), and \(800\), before the diagram ends at \(803\). 3. The total distance is \(2+50+200+3=255\). 4. Thus the endpoints differ by \(255\).

Jumps: \(+2\), \(+50\), \(+200\), \(+3\) Missing landings: \(550\), \(600\), \(800\) Total distance: \(255\)

- Compare the Before and After values for each day. - Find each increase, then compare the increases.

1. Monday's increase is \(14 - 11 = 3\). 2. Tuesday's value decreased from \(15\) to \(14\). 3. Wednesday's increase is \(18 - 13 = 5\). 4. The greatest increase is \(5\), on Wednesday.

Wednesday

- Identify the two values used in the answer. - Write a question that asks for their difference.

1. The answer compares David's \(17\) leaves with Alma's \(12\) leaves. 2. Their difference is \(17 - 12 = 5\).

One possible question is, “How many more leaves did David collect than Alma?”

- Start with a row or column that has only one missing value. - Check each result against the other totals.

1. Group A has \(15 - 8 = 7\) blue cards. 2. Group B has \(14 - 8 = 6\) red cards. 3. These values also satisfy the remaining row and column totals.

Group A: \(7\) blue cards Group B: \(6\) red cards

- Begin with a row or column that has only one missing value. - Enter each result before using another total. - Check the completed row and column totals against the table.

1. The number of large silver stars is \(14-9=5\). 2. The number of small gold stars is \(16-9=7\). 3. The number of large gold stars is \(15-7=8\).

Large silver stars: \(5\) Small gold stars: \(7\) Large gold stars: \(8\)

- Begin in the ones place and decide whether there are enough ones to subtract. - After changing the tens to help the ones, check the tens column again before subtracting there. - When a place does not have enough units, unbundle \(1\) unit from the place to its left.

1. In the ones place, \(2\) ones are not enough to subtract \(8\) ones. Unbundle \(1\) ten, giving \(12\) ones and leaving \(2\) tens. Then \(12-8=4\). 2. In the tens place, \(2\) tens are not enough to subtract \(5\) tens. Unbundle \(1\) hundred, giving \(12\) tens and leaving \(6\) hundreds. Then \(12-5=7\). 3. Subtract the hundreds: \(6-4=2\). 4. The difference is \(274\).

\(274\). First unbundle a ten for the ones place, then unbundle a hundred for the tens place.

- The result must satisfy two inequalities: it is greater than \(5\) and less than \(10\). - List the whole-number results that are greater than \(5\) and less than \(10\). - Look for pairs of numbers that are close together. - Sorting the numbers by size may make the pairs easier to find.

1. The possible results are \(6, 7, 8,\) or \(9\). 2. Comparing nearby numbers gives \(605-597=8\), \(411-403=8\), \(904-895=9\), and \(715-706=9\). 3. The pair \(302\) and \(298\) has a difference of \(4\), so it does not meet the condition. 4. No other pair from the list has a difference from \(6\) through \(9\).

\(605-597=8\) \(411-403=8\) \(904-895=9\) \(715-706=9\)

- Sorting the numbers can help with the least positive difference. - The greatest difference uses the greatest and least values. - For the last part, think about which larger number could be paired with a smaller number to make a difference near \(500\). - Compare the distance from \(500\) for the best candidate from each possible larger number.

1. Sort the numbers: \(34, 98, 156, 207, 441, 523, 615, 782\). The least adjacent difference is \(207-156=51\), so no other positive difference can be smaller. 2. The greatest difference uses the greatest and least numbers: \(782-34=748\). 3. For part c, \(441\) or any smaller minuend gives a difference at most \(441-34=407\), which is at least \(93\) away from \(500\). 4. For each larger possible minuend, choose the available subtrahend closest to making a difference of \(500\): \(523-34=489\), which is \(11\) away; \(615-98=517\), which is \(17\) away; and \(782-207=575\), which is \(75\) away. 5. Therefore, \(489\) is the difference closest to \(500\).

a) \(207-156=51\) b) \(782-34=748\) c) \(523-34=489\)

- 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. 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: \(210 - 140 = 70\).

\(70\)

- 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: \(800\)

- 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: \(800\)

- Find a connected group in which two of the three values are known. - To recover a missing lower-left value, you may need to add even though the wall rule uses subtraction.

1. For the third bottom value, solve \(x - 30 = 70\), so \(x = 100\). 2. The middle value in the second row is \(250 - 100 = 150\). 3. The two values in the third row are \(250 - 150 = 100\) and \(150 - 70 = 80\). 4. The top is \(100 - 80 = 20\). 5. For the first bottom value, solve \(y - 250 = 250\), so \(y = 500\).

Bottom row: \(500\), \(250\), \(100\), \(30\) Second row: \(250\), \(150\), \(70\) Third row: \(100\), \(80\) Top: \(20\)

- Check every row total and column total. - The corrected value must satisfy both its row total and its column total.

1. Team B's Round 2 value must combine with \(7\) to make the row total \(16\). 2. The correct value is \(16 - 7 = 9\). 3. This also makes the Round 2 column total correct because \(8 + 9 = 17\).

The Team B, Round 2 entry should be \(9\), not \(10\).

- Start in the ones place and check whether there are enough ones to subtract. - If the tens place has no tens to unbundle, look one place farther left. - After unbundling a hundred, update the tens place before unbundling a ten. - Keep track of every place-value change before subtracting.

1. In the ones place, \(3\) ones are not enough to subtract \(8\) ones, and there are \(0\) tens available to unbundle directly. 2. Unbundle \(1\) hundred. The \(7\) hundreds become \(6\) hundreds, and the tens place becomes \(10\) tens. 3. Unbundle \(1\) of those tens. That leaves \(9\) tens and gives \(13\) ones. 4. Subtract by place: \(13-8=5\), \(9-6=3\), and \(6-4=2\). 5. The difference is \(235\).

\(235\). Because the tens digit is \(0\), first unbundle \(1\) hundred into \(10\) tens, then unbundle \(1\) ten into \(10\) ones.

- For the greatest difference, maximize the first number and minimize the second. - For the least positive difference, begin with hundreds digits that are as close as possible. - Compare every possible neighboring pair of hundreds digits before deciding which construction is least.

1. For the greatest difference, make the greatest possible minuend and the least possible subtrahend: \(654-123=531\). 2. For the least positive difference, the hundreds digits must differ by \(1\). If they differed by \(2\) or more, the hundreds-place difference would be at least \(200\), and the two-digit endings could not reduce the difference below \(100\). 3. Check each possible neighboring pair of hundreds digits, making the greater number's last two digits as small as possible and the smaller number's last two digits as large as possible: \(234-165=69\), \(314-265=49\), \(412-365=47\), \(512-463=49\), and \(612-543=69\). 4. The least of these differences is \(47\), so \(412-365=47\) is the least possible positive difference.

a) \(654-123=531\) b) \(412-365=47\)

- 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: \(950\)