Free math worksheets

- First find the total number of notebooks in the new shipment. - Then add the notebooks that were already on the shelves. - Line up the place values when you add.

1. Add the notebooks in the new shipment: \(145 + 128 + 64 = 337\). 2. Add the notebooks already in the store: \(337 + 35 = 372\).

The store has \(372\) notebooks in all.

- Break \(47\) into tens and ones. - Find the total after adding only the tens. - Use that intermediate total as the starting point for adding the ones.

1. Add the tens: \(384+40=424\). 2. Add the remaining ones: \(424+7=431\).

\(384+40=424\), then \(424+7=431\).

- Find how much the first addend needs to reach the next convenient ten or hundred. - Use that amount as the first part of the split. - The two parts of the split must add back to the original second addend.

1. a) Split \(7\) as \(4+3\): \(146+4=150\), then \(150+3=153\). 2. b) Split \(5\) as \(2+3\): \(378+2=380\), then \(380+3=383\). 3. c) Split \(8\) as \(6+2\): \(594+6=600\), then \(600+2=602\). 4. d) Split \(9\) as \(5+4\): \(895+5=900\), then \(900+4=904\).

a) \(7=4+3\); \(146+4=150\); \(150+3=153\) b) \(5=2+3\); \(378+2=380\); \(380+3=383\) c) \(8=6+2\); \(594+6=600\); \(600+2=602\) d) \(9=5+4\); \(895+5=900\); \(900+4=904\)

- Ask how many tens are needed to reach the next hundred. - Split the second addend into that amount and the amount left over. - Show the hundred you make before adding the remainder.

1. a) Split \(60\) as \(40+20\): \(260+40=300\), then \(300+20=320\). 2. b) Split \(40\) as \(20+20\): \(480+20=500\), then \(500+20=520\). 3. c) Split \(70\) as \(50+20\): \(750+50=800\), then \(800+20=820\). 4. d) Split \(30\) as \(10+20\): \(890+10=900\), then \(900+20=920\).

a) \(60=40+20\); \(260+40=300\); \(300+20=320\) b) \(40=20+20\); \(480+20=500\); \(500+20=520\) c) \(70=50+20\); \(750+50=800\); \(800+20=820\) d) \(30=10+20\); \(890+10=900\); \(900+20=920\)

- Follow each jump sequence from its starting value. - Compare the total size of the jumps in the two panels. - Think about whether splitting an addend changes its value.

1. In a), \(497+3=500\), then \(500+3=503\). 2. In b), \(497+6=503\). 3. Both methods give \(503\) because \(3+3=6\); splitting an addend into parts does not change the total amount added.

a) First sequence: \(497, 500, 503\). Second sequence: \(497, 503\). b) Yes. \(3+3=6\), so both methods add the same total amount and end at \(503\).

- Work with the tens and ones first. - After solving the related fact, identify which hundreds are unchanged. - Your answer should explicitly connect the two-digit fact to the three-digit sum.

1. a) \(48+7=55\). The tens-and-ones part becomes \(55\), while the \(2\) hundreds remain, so \(248+7=255\). 2. b) \(15+6=21\). The tens-and-ones part becomes \(21\), while the \(6\) hundreds remain, so \(615+6=621\). 3. c) \(79+4=83\). The tens-and-ones part becomes \(83\), while the \(4\) hundreds remain, so \(479+4=483\).

a) \(48+7=55\); keep the \(2\) hundreds, so \(248+7=255\). b) \(15+6=21\); keep the \(6\) hundreds, so \(615+6=621\). c) \(79+4=83\); keep the \(4\) hundreds, so \(479+4=483\).

- 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 \(673-245=428\).

- 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 \(358 + 421 = 779\).

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

- 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) \(725\), \(760\), \(765\)

- Look for two addends that make a multiple of \(100\). - The commutative property lets you change the order of addends. - The associative property lets you change how the addends are grouped.

1. Use the commutative property to place \(327\) and \(473\) together: \(327 + 473 + 158\). 2. Use the associative property to regroup: \((327 + 473) + 158\). 3. Calculate: \(800 + 158 = 958\).

\((327 + 473) + 158 = 800 + 158 = 958\)

- Look for pairs of addends that make \(100\). - You may reorder and regroup addends without changing the sum. - Add the two partial sums.

1. Reorder and regroup the addends: \((64 + 36) + (19 + 81)\). 2. Calculate the partial sums: \(100 + 100 = 200\).

\(200\)

- Look for addends that combine to make \(100\) or another multiple of \(100\). - Reorder and regroup the addends to place convenient pairs together. - Remember to include any addend that is not part of a pair.

1. Reorder and regroup the addends: \((235 + 165) + (88 + 12) + 50\). 2. Calculate the partial sums: \(400 + 100 + 50 = 550\).

\(550\)

- Look for pairs that make multiples of \(100\). - Reorder and regroup the addends to place those pairs together. - Add the partial sums.

1. Reorder and regroup the addends: \((125 + 75) + (430 + 170)\). 2. Calculate the partial sums: \(200 + 600 = 800\).

\(800\)

- Look for two addends that make a multiple of \(100\). - Think about which property lets you change how addends are grouped. - Regroup before calculating.

1. Use the associative property to regroup: \((357 + 143) + 89\). 2. Calculate: \(500 + 89 = 589\).

\(589\); associative property

- First identify whether the order of the addends changes. - Then identify whether the grouping changes. - Match each change to the name of a property.

1. The student changes \(15 + 38 + 85\) to \(15 + 85 + 38\) by using the commutative property to reorder addends. 2. The student then uses the associative property to group \(15\) and \(85\) so they are added first.

The commutative property changes the order of \(38\) and \(85\); then the associative property groups \(15\) and \(85\) to be added first.

- Can you break an addend into smaller parts? - Which place values change in each sum? - Break the numbers into hundreds, tens, and ones. - Add the hundreds first and then the tens when helpful.

1. \(560 + 30 = 590\). 2. \(240 + 500 = 740\). 3. \(310 + 280 = 590\). 4. \(470 + 6 = 476\). 5. \(620 + 150 = 770\).

a) \(590\) b) \(740\) c) \(590\) d) \(476\) e) \(770\)

- Evaluate both sides of each comparison. - You may also compare how the addends change from one side to the other. - Use \(<\) or \(>\) when one sum is smaller or greater. - Use \(=\) when the two sums have the same value.

1. For a), \(340 + 50 = 390\) and \(320 + 70 = 390\), so the values are equal. 2. For b), \(600 + 230 = 830\) and \(500 + 340 = 840\), so \(830 < 840\). 3. For c), \(180 + 400 = 580\) and \(200 + 380 = 580\), so the values are equal. 4. For d), \(450 + 120 = 570\) and \(460 + 100 = 560\), so \(570 > 560\).

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

- How far is the changed addend from the next hundred? - If you use a larger addend, subtract the extra amount afterward. - Which addend is close to a multiple of \(100\)?

1. Use \(400\) instead of \(398\), which adds \(2\) too much: \(400 + 154 - 2 = 554 - 2 = 552\). 2. Use \(500\) instead of \(499\), which adds \(1\) too much: \(270 + 500 - 1 = 770 - 1 = 769\). 3. Use \(600\) instead of \(595\), which adds \(5\) too much: \(600 + 130 - 5 = 730 - 5 = 725\).

a) \(398 + 154 = 400 + 154 - 2 = 552\) b) \(270 + 499 = 270 + 500 - 1 = 769\) c) \(595 + 130 = 600 + 130 - 5 = 725\)

- 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.

- The numbers being added are called addends. - The result of addition is called the sum. - Use a place-value addition strategy.

1. Part a: \(470 + 360 = 830\). 2. Part b: \(280 + 540 = 820\).

a) The sum is \(830\). b) The sum is \(820\).

- Break each target sum into hundreds and tens. - Think of numbers that appear when counting by tens. - Choose one simple multiple of \(10\), then subtract it from the target. - Many answers are possible, but the two addends must be different.

1. a) One possible pair is \(200\) and \(250\), because \(200 + 250 = 450\). 2. b) One possible pair is \(400\) and \(420\), because \(400 + 420 = 820\). 3. c) One possible pair is \(250\) and \(350\), because \(250 + 350 = 600\). 4. Other pairs of different multiples of \(10\) are also possible.

a) One possible answer is \(200\) and \(250\). b) One possible answer is \(400\) and \(420\). c) One possible answer is \(250\) and \(350\).

- Break the numbers into hundreds and tens. - What happens if you add the tens first? - Could you first make the next hundred? - Break the calculation into two addition steps.

1. Add the first two numbers: \(350 + 70 = 420\). 2. Add the third number: \(420 + 120 = 540\).

The sum is \(540\).

- Find both sums separately. - Compare the two results. - You can estimate first to predict which sum will be greater. - Pay attention when an addition crosses a hundred.

1. The first sum is \(470 + 80 = 550\). 2. The second sum is \(390 + 150 = 540\). 3. Since \(550 > 540\), the first sum is greater.

\(470 + 80\) is greater. It equals \(550\), while \(390 + 150 = 540\).

- 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: \(740\)

- Identify the one route whose number changes. - Add the extra trips to that route's original value. - Keep every other table value unchanged.

1. Only the Island route changes: \(18 + 7 = 25\). 2. The other route values remain unchanged.

North: \(25\) South: \(30\) Island: \(25\) Harbor: \(22\)

- Find the three-day total for each kind of equipment. - Compare the three totals.

1. Balls: \(6 + 8 + 5 = 19\). 2. Jump ropes: \(4 + 3 + 6 = 13\). 3. Hula hoops: \(5 + 7 + 4 = 16\). 4. The greatest total is \(19\).

Balls were checked out most often, with \(19\) checkouts.

- Start with the ones column in the vertical addition. - If the ones make at least \(10\), regroup \(10\) ones as \(1\) ten. - Remember to include any regrouped ten when you add the tens column.

1. Add the ones: \(7+6=13\). Write \(3\) ones and regroup \(10\) ones as \(1\) ten. 2. Add the tens, including the regrouped ten: \(4+3+1=8\) tens. 3. Add the hundreds: \(2+1=3\) hundreds. 4. The sum is \(383\).

The sum is \(383\). Regrouping \(13\) ones makes \(3\) ones and \(1\) extra ten, so that ten is included in the tens column.

- Keep the ones, tens, and hundreds in their own columns. - Add one place-value column at a time, beginning with the ones. - Ask whether any column total reaches \(10\).

1. Add the ones: \(3+6=9\). 2. Add the tens: \(4+1=5\) tens. 3. Add the hundreds: \(2+5=7\) hundreds. 4. Every column sum is less than \(10\), so no regrouping is needed. The sum is \(759\).

\(759\). No regrouping is needed because each place-value column has a sum less than \(10\).

- Begin with the ones column and then move to the tens. - If a column has \(10\) or more of one place-value unit, regroup \(10\) of them as \(1\) of the next unit. - Include any regrouped hundred when you add the hundreds column.

1. Add the ones: \(2+3=5\). 2. Add the tens: \(5+7=12\) tens. Write \(2\) tens and regroup \(10\) tens as \(1\) hundred. 3. Add the hundreds, including the regrouped hundred: \(3+4+1=8\) hundreds. 4. The sum is \(825\).

\(825\). The \(12\) tens are regrouped as \(2\) tens and \(1\) hundred.

- Break the second addend into hundreds, tens, and ones. - Add the largest place-value part first, then the smaller parts. - Record each new total before adding the next place-value part.

1. For \(435 + 258\): \(435 + 200 = 635\), \(635 + 50 = 685\), and \(685 + 8 = 693\). 2. For \(167 + 544\): \(167 + 500 = 667\), \(667 + 40 = 707\), and \(707 + 4 = 711\). 3. For \(329 + 482\): \(329 + 400 = 729\), \(729 + 80 = 809\), and \(809 + 2 = 811\).

a) \(435+200=635\), \(635+50=685\), \(685+8=693\) b) \(167+500=667\), \(667+40=707\), \(707+4=711\) c) \(329+400=729\), \(729+80=809\), \(809+2=811\)

- Each result becomes the starting number in the next line. - Notice whether each step adds hundreds, tens, or ones.

1. For part a, \(374 + 200 = 574\), \(574 + 40 = 614\), and \(614 + 7 = 621\). 2. For part b, \(586 + 100 = 686\), \(686 + 30 = 716\), and \(716 + 5 = 721\).

a) \(574\), \(614\), \(621\) b) \(686\), \(716\), \(721\)

- Look for an addend that is close to a multiple of \(100\). - If you add too much at first, think about how to undo the extra amount. - For an addend just above a hundred, consider splitting off the hundreds first.

1. For a), replace \(399\) with \(400\): \(256+400=656\). Since \(400\) is \(1\) too much, subtract \(1\): \(656-1=655\). 2. For b), replace \(198\) with \(200\): \(437+200=637\). Since \(200\) is \(2\) too much, subtract \(2\): \(637-2=635\). 3. For c), break \(202\) into \(200+2\): \(524+200=724\), then \(724+2=726\).

a) One mental strategy: \(256+400=656\), then \(656-1=655\). b) One mental strategy: \(437+200=637\), then \(637-2=635\). c) One mental strategy: \(524+200=724\), then \(724+2=726\).

- Read each row by hundreds, tens, and ones before doing any addition. - Combine chips in matching place-value columns. - After regrouping one column, remember that the next column changes.

1. The first row shows \(4\) hundreds, \(5\) tens, and \(6\) ones, so it represents \(456\). The second row represents \(265\). 2. The ones have \(6+5=11\) ones, so the ones place must be regrouped first. 3. Regroup \(10\) ones as \(1\) ten, leaving \(1\) one. The tens then total \(5+6+1=12\) tens, so regroup \(10\) tens as \(1\) hundred. 4. The hundreds total \(4+2+1=7\), so \(456+265=721\).

a) \(456\) and \(265\) b) The ones place. c) Regroup \(10\) ones as \(1\) ten, then regroup \(10\) tens as \(1\) hundred. d) \(721\)

- Compare \(198\) with a nearby hundred. - The first jump is intentionally a little too large. - The second jump corrects the extra amount.

1. Add \(200\): \(647+200=847\). 2. Since \(198\) is \(2\) less than \(200\), subtract \(2\): \(847-2=845\). 3. The jumps add \(200\) and then subtract \(2\), for a net change of \(+198\).

\(647+198=845\). The jumps have a net change of \(+198\).

- Add the tens of the second addend first, then the ones. - Check whether adding the tens changes the hundreds digit. - Use the first intermediate result as the starting number for the second step.

1. \(367+20=387\), then \(387+5=392\). 2. \(584+30=614\), then \(614+8=622\). 3. \(749+60=809\), then \(809+3=812\).

a) \(367+20=387\), then \(387+5=392\) b) \(584+30=614\), then \(614+8=622\) c) \(749+60=809\), then \(809+3=812\)

- Read the starting value and the signed jump size from the diagram. - Each landing becomes the start of the next jump. - Check the jump that crosses \(400\) carefully.

1. The diagram starts at \(386\) and shows a jump of \(+7\), landing at \(393\). 2. The next \(+7\) lands at \(400\). 3. The next \(+7\) lands at \(407\). 4. The final \(+7\) lands at \(414\).

\(393\), \(400\), \(407\), \(414\)

- First find how many ones are needed to reach the next hundred. - Split \(8\) or \(9\) so that amount comes first. - Check that the two split parts still total the original addend.

1. a) \(594+8=594+6+2=602\) and \(594+9=594+6+3=603\). 2. b) \(296+8=296+4+4=304\) and \(296+9=296+4+5=305\). 3. c) \(795+8=795+5+3=803\) and \(795+9=795+5+4=804\).

a) \(594+8=594+6+2=602\); \(594+9=594+6+3=603\) b) \(296+8=296+4+4=304\); \(296+9=296+4+5=305\) c) \(795+8=795+5+3=803\); \(795+9=795+5+4=804\)

- Read the starting value and the signed jump sizes from the image. - Each landing becomes the starting value for the next jump. - Combine the jump sizes to determine the addend represented by the whole path.

1. The line starts at \(345\). The first jump is \(+200\), so the first landing is \(545\). 2. The next jump is \(+80\), giving \(625\). 3. The last jump is \(+7\), giving \(632\). 4. The jumps add \(200+80+7=287\), so the represented equation is \(345+287=632\).

a) \(545\), \(625\), \(632\) b) \(345+287=632\)

- Compare the two numbers on either side of the box. - Think about which place-value part of \(354\) has not yet been used. - Check that all three added parts recombine to the original addend.

1. The missing step changes \(768\) to \(818\), so the missing addend is \(50\). 2. The added parts are \(300\), \(50\), and \(4\). 3. Since \(300+50+4=354\), the steps add exactly the second addend.

a) \(50\) b) \(300+50+4=354\).

- Check the total change made by each method. - Compare \(159\) with the nearby number Noah used. - For part c), count the arithmetic steps in each shown method.

1. Mia's place-value steps give \(641\). 2. Noah adds \(160\), which is \(1\) more than \(159\), and then subtracts \(1\), so his net change is \(+159\). His result is also \(641\). 3. Both methods are correct. Noah's method uses two arithmetic steps, while Mia's uses three, so Noah's method uses fewer steps for this sum.

a) Yes. Both methods give \(641\). b) Noah adds \(1\) too much and then subtracts \(1\). c) Noah's method uses fewer arithmetic steps.

- 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 \(100\).

- Look for an addend that is very close to a multiple of \(100\). - Decide whether the easier calculation adds a little too much or too little. - Correct only the amount by which the addend was changed.

1. \(398\) is only \(2\) less than \(400\), so using a nearby hundred is especially efficient. 2. Compute \(386+400=786\). 3. Subtract the extra \(2\): \(786-2=784\).

Using a nearby hundred is especially efficient. \(386+398=784\).

- Break Class 3B's amount into hundreds, tens, and ones. - Add one place-value part at a time and keep each intermediate total. - Check that the final total answers the “altogether” question.

1. Break \(385\) into \(300+80+5\). 2. Add the hundreds: \(429+300=729\). 3. Add the tens: \(729+80=809\). 4. Add the ones: \(809+5=814\).

\(429+300=729\), \(729+80=809\), \(809+5=814\). The two classes have collected \(814\) bottles altogether.

- Check each step against the place value of the amount being added. - Ask whether the second step changes the tens place by the correct amount. - Once the first incorrect step is fixed, continue from that corrected value.

1. Ava's first step is correct: \(574+300=874\). 2. In the second step, \(40\) means \(4\) tens, not \(4\) ones. The correct step is \(874+40=914\). 3. Then \(914+8=922\). 4. The correct sum is \(922\).

Ava first treats \(40\) as if it were \(4\). The correct sum is \(922\).

- Replace the second addend with a nearby multiple of \(100\). - Decide how much extra was added by that replacement. - Undo exactly that extra amount after finding the easier sum.

1. Replace \(199\) with \(200\): \(236+200=436\). Since \(200\) is \(1\) too much, subtract \(1\): \(436-1=435\). 2. Replace \(298\) with \(300\): \(458+300=758\). Since \(300\) is \(2\) too much, subtract \(2\): \(758-2=756\). 3. Replace \(399\) with \(400\): \(375+400=775\). Since \(400\) is \(1\) too much, subtract \(1\): \(775-1=774\).

a) \(236+200=436\), then \(436-1=435\) b) \(458+300=758\), then \(758-2=756\) c) \(375+400=775\), then \(775-1=774\)

- 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 \(784-538=246\).

- Look for an addend that is close to a multiple of \(100\). - If you move part of one addend to the other, keep the total amount unchanged. - If you replace an addend with a larger nearby number, account for the extra amount afterward.

1. For a), move \(5\) from \(165\) to \(295\): \(300+160=460\). 2. For b), use \(400+244=644\). Since \(400\) is \(2\) more than \(398\), subtract \(2\): \(644-2=642\). 3. For c), move \(1\) from \(301\) to \(499\): \(500+300=800\).

a) Move \(5\): \(295+165=300+160=460\). b) Use \(400\): \(400+244=644\), then \(644-2=642\). c) Move \(1\): \(499+301=500+300=800\).

- Compare each second addend with a nearby multiple of \(100\). - A very small adjustment can make a mental strategy efficient. - Use a place-value strategy when no nearby hundred gives a simple adjustment.

1. In B, \(299\) is only \(1\) less than \(300\), so compensation is especially efficient: \(457+300-1=756\). 2. For A, add by place value: \(457+200=657\), \(657+80=737\), and \(737+6=743\).

B is especially efficient with a nearby hundred. A: \(743\) B: \(756\)

- Read each row by hundreds, tens, and ones before adding. - Combine corresponding place-value columns beginning with the ones. - After regrouping, include the new unit in the next column.

1. The rows represent \(524\) and \(387\). 2. The ones total \(4+7=11\), so regroup \(10\) ones as \(1\) ten first. 3. The tens then total \(2+8+1=11\), so regroup \(10\) tens as \(1\) hundred. 4. The hundreds total \(5+3+1=9\), leaving \(1\) ten and \(1\) one, so \(524+387=911\).

a) \(524\) and \(387\) b) The ones place. c) The tens place. d) \(911\)

- Look for an addend close to a hundred, parts that make a hundred, or place values that combine cleanly. - Name the strategy you choose before giving the final total. - Include an intermediate equation that would not appear if you had only calculated the final sum.

1. a) Compensation: \(400+256=656\), then \(656-2=654\). 2. b) Make a hundred: \(420+80=500\), then \(500+200=700\). 3. c) Place-value decomposition: \(500+200=700\), \(40+10=50\), and \(3+5=8\); then \(700+50+8=758\). 4. d) Friendly pairs: \((200+100)+(75+25)=300+100=400\).

a) Compensation: \(400+256=656\); \(656-2=654\). b) Make a hundred: \(420+80=500\); \(500+200=700\). c) Place value: \(500+200=700\), \(40+10=50\), \(3+5=8\); total \(758\). d) Friendly pairs: \((200+100)+(75+25)=300+100=400\).

- 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>

- Multiply the number of rows by the sum of each row. - Think about whether regrouping the same addends can change their total. - Each number appears once when you add by rows and once when you add by columns.

1. There are three row sums of \(15\), so the total is \(3 \times 15 = 45\). 2. Regrouping the same nine addends by columns does not change their sum. Therefore, the three column sums also have a total of \(45\).

a) \(45\) b) \(45\). The same nine numbers are being added, only grouped differently.

- Look for two addends that combine to make a multiple of \(100\). - The commutative property lets you change the order of addends. - The associative property lets you change how addends are grouped.

1. For a), use the commutative property to reorder the addends: \((245+355)+178=600+178=778\). 2. For b), use the associative property to regroup: \((64+136)+482=200+482=682\). 3. For c), use the commutative and associative properties: \((312+288)+399=600+399=999\).

a) \(778\); commutative property b) \(682\); associative property c) \(999\); commutative and associative properties

- Find pairs of addends that total \(100\). - Reorder the addends so each pair is together. - Add the three partial sums.

1. Pair addends that make \(100\): \(13 + 87 = 100\), \(26 + 74 = 100\), and \(39 + 61 = 100\). 2. Reorder and regroup the addends: \((13 + 87) + (26 + 74) + (39 + 61)\). 3. Add the partial sums: \(100 + 100 + 100 = 300\).

\(300\)

- Look for pairs that make multiples of \(100\). - Reorder the addends to place convenient pairs together. - Regroup the addends before calculating.

1. Reorder and regroup the addends: \((225+375)+(187+113)+100\). 2. Calculate the partial sums: \(600+300+100=1000\). 3. Reordering uses the commutative property, and regrouping uses the associative property.

\(1000\); commutative and associative properties

- Calculate the total using each grouping. - Think about whether addition changes when addends are reordered. - Think about whether addition changes when addends are regrouped.

1. Adding in the given order gives \(456 + 23 + 102 + 304 = 885\). 2. Lisa groups the two largest numbers and the two smallest numbers: \(456 + 304 = 760\) and \(23 + 102 = 125\). Then \(760 + 125 = 885\). 3. Both methods give the same result because the commutative and associative properties allow addends to be reordered and regrouped without changing the sum.

The sum is \(885\). Both methods give the same result because changing the order and grouping of addends does not change their sum.

- Compare the number Lucas used with the original addend. - If the easier number is larger, determine how much extra was included. - For the two new sums, replace the second addend with the next multiple of \(10\), then undo the extra amount.

1. Lucas used \(300\) instead of \(295\), so he added \(5\) too much. Subtract \(5\) from \(348\): \(348-5=343\). 2. For \(57+99\), use \(57+100=157\), then subtract \(1\): \(157-1=156\). 3. For \(146+19\), use \(146+20=166\), then subtract \(1\): \(166-1=165\).

a) Lucas must subtract \(5\): \(348-5=343\). b) \(57+100=157\), then \(157-1=156\). \(146+20=166\), then \(166-1=165\).

- Look for a pair of addends whose sum is a multiple of \(100\). - Compare the remaining addition after that friendly partial sum is made. - In each new expression, identify the friendly pair before adding the third number.

1. Method B first makes a multiple of \(100\) because \(34+66=100\). The remaining calculation is then \(100+57\). 2. For a), first add \(123+77=200\). Then \(200+49=249\). 3. For b), first add \(250+150=400\). Then \(400+368=768\).

Method B first makes a multiple of \(100\), leaving \(100+57\). a) \(123+77=200\), then \(200+49=249\) b) \(250+150=400\), then \(400+368=768\)

- 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 \(230+370=600\)

- Evaluate the left sum first. - Evaluate the right sum next. - Compare the two results. - Before calculating exactly, look for compensating changes in the addends.

1. For a), \(430 + 280 = 710\) and \(520 + 190 = 710\), so the sums are equal. 2. For b), \(360 + 470 = 830\) and \(250 + 590 = 840\), so \(830 < 840\). 3. For c), \(180 + 740 = 920\) and \(630 + 290 = 920\), so the sums are equal.

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

- Subtract the known addend from the sum to find the missing addend. - Calculate the original sum before changing the addends. - Account for the increase in both addends.

1. Part a: Subtract the known addend from the sum: \(820 - 550 = 270\). 2. Part b: The original sum is \(190 + 430 = 620\). 3. Increasing both addends by \(10\) gives \(200 + 440 = 640\). 4. The sum increases by \(20\) because each of the two addends increases by \(10\).

a) The other addend is \(270\). b) The original sum is \(620\). The new sum is \(640\), so the sum increases by \(20\).

- Search for pairs that make \(100\), \(200\), or \(300\). - Use every addend exactly once in the two pairs. - Add the two pair sums only after you have written both pairs.

1. a) Pair \(57+43=100\) and \(34+66=100\). Then \(100+100=200\). 2. b) Pair \(215+85=300\) and \(88+12=100\). Then \(300+100=400\). 3. c) Pair \(46+54=100\) and \(127+73=200\). Then \(100+200=300\).

a) \(57+43=100\), \(34+66=100\); total \(200\) b) \(215+85=300\), \(88+12=100\); total \(400\) c) \(46+54=100\), \(127+73=200\); total \(300\)

- 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>

- 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 \(390\).

- Recall the names of the numbers and result in an addition equation. - Calculate both the original and changed equations. - Compare the equal-sized changes to the two addends.

1. Part a: The numbers being added are called addends. 2. Part b: \(340 + 180 = 520\). The result of addition is called the sum. 3. Part c: The new addends are \(340 + 20 = 360\) and \(180 - 20 = 160\). 4. The new sum is \(360 + 160 = 520\). 5. The sum stays the same because increasing one addend by \(20\) and decreasing the other by \(20\) are equal and opposite changes.

a) The numbers are addends. b) The result is \(520\), and it is called the sum. c) The sum remains \(520\) because the increase of \(20\) in one addend is offset by the decrease of \(20\) in the other addend.

- Compare how the two addends change from one expression to another. - Look for equal changes in opposite directions. - Verify the predicted pairs only after identifying the change pattern.

1. In \(270+50\) and \(260+60\), one addend decreases by \(10\) while the other increases by \(10\), so the sum stays the same. Both equal \(320\). 2. In \(480+60\) and \(500+40\), one addend increases by \(20\) while the other decreases by \(20\), so the sum stays the same. Both equal \(540\).

\(270+50\) and \(260+60\) are a pair: one addend decreases by \(10\) while the other increases by \(10\), so both sums are \(320\). \(480+60\) and \(500+40\) are a pair: one addend increases by \(20\) while the other decreases by \(20\), so both sums are \(540\).

- 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: \(25\), \(28\)

- 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 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: \(970\)

- 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: \(44 + 45 = 89\).

\(89\)

- 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: \(360\)

- 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: \(66\)

- 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: \(51\)

- 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: \(890\)

- 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: \(57\)

- 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: \(60\)

- Add by place value, and record your work if the sums are not easy to do mentally. - Check any regrouping in the ones and tens places.

1. The left brick in the second row is \(245+138=383\). 2. The right brick in the second row is \(138+304=442\). 3. The top brick is \(383+442=825\).

Second row: \(383\), \(442\) Top: \(825\)

- 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: \(440\)

- 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: \(112\)

- 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: \(850\)

- Read all four bar values. - Add the values in convenient pairs. - Check that the total is reasonable compared with the bars.

1. The four values are \(120\), \(200\), \(160\), and \(160\). 2. Add them: \(120 + 200 + 160 + 160 = 640\).

The graph shows \(640\) rolls altogether.

- Read all three bar values. - Add the values. - Compare the sum with \(200\).

1. The bars show \(70\), \(50\), and \(90\) tickets. 2. Their sum is \(70 + 50 + 90 = 210\). 3. Since \(210 \ne 200\), Leah is not correct.

No. The three booths sold \(210\) raffle tickets altogether.

- Use all six numbers in the table. - Find the total for each round first. - Add the two round totals.

1. In Round 1, the teams made \(12 + 15 + 9 = 36\) airplanes. 2. In Round 2, the teams made \(8 + 11 + 14 = 33\) airplanes. 3. Altogether, they made \(36 + 33 = 69\) airplanes.

The teams made \(69\) paper airplanes altogether.

- Find the total for each day. - Compare the four totals.

1. Monday's total is \(14 + 9 = 23\). 2. Tuesday's total is \(12 + 11 = 23\). 3. Wednesday's total is \(16 + 8 = 24\). 4. Thursday's total is \(10 + 15 = 25\). 5. Monday and Tuesday have the same total.

Monday and Tuesday

- Work from the ones place toward the hundreds place. - After regrouping, record the new unit in the next place before adding that column. - Check every column again after including a regrouped unit.

1. Add the ones: \(8+5=13\). Write \(3\) ones and regroup \(1\) ten. 2. Add the tens, including the regrouped ten: \(7+6+1=14\) tens. Write \(4\) tens and regroup \(1\) hundred. 3. Add the hundreds, including the regrouped hundred: \(4+3+1=8\) hundreds. 4. The sum is \(843\).

\(843\). Regroup \(13\) ones as \(3\) ones and \(1\) ten, then regroup \(14\) tens as \(4\) tens and \(1\) hundred.

- A digit in the hundreds place affects the sum more than a digit in the tens or ones place. - A three-digit number cannot have \(0\) in the hundreds place. - For the least sum, place the least possible digits in the highest-value places; reverse that idea for the greatest sum.

1. To minimize the sum, place the least nonzero digits, \(1\) and \(2\), in the hundreds places. Place the next least digits, \(0\) and \(3\), in the tens places, and place \(4\) and \(5\) in the ones places. 2. One possible arrangement is \(104+235=339\). Any arrangement with the same pairs of place-value digits has the same least sum. 3. To maximize the sum, place \(5\) and \(4\) in the hundreds places, \(3\) and \(2\) in the tens places, and \(1\) and \(0\) in the ones places. 4. One possible arrangement is \(531+420=951\).

a) \(339\) b) \(951\)

- Start with the ones digit of the target sum. - Ask whether two available digits could make \(19\) in one column. - Once the carry question is settled, look for pairs of digits that make \(9\).

1. In the ones place, the two digits must produce a ones digit of \(9\). The greatest possible sum of two distinct cards is \(8+7=15\), so a sum of \(19\) is impossible. Therefore, the ones digits must sum to \(9\), and there is no regrouped ten. 2. The same reasoning applies in the tens place, so the tens digits also sum to \(9\) with no regrouping. The hundreds digits then sum to \(9\). 3. The available pairs that sum to \(9\) are \((1,8)\), \((2,7)\), \((3,6)\), and \((4,5)\). 4. Any three pairs can fill the three places, leaving the fourth pair unused. 5. For example, \(123+876=999\), leaving \(4\) and \(5\).

The possible leftover pairs are \(1\) and \(8\), \(2\) and \(7\), \(3\) and \(6\), or \(4\) and \(5\).

- Work backward from the zeros in \(500\). - If the ones digits make \(10\), remember the regrouped ten when choosing tens digits. - If the tens total \(10\), remember the regrouped hundred when choosing hundreds digits.

1. To end with \(0\) ones while using distinct nonzero ones in this example, choose ones digits totaling \(10\), such as \(6+4\). This regroups \(1\) ten. 2. The tens column must total \(10\) after including that regrouped ten, so the two tens digits total \(9\), such as \(2+7\). 3. The hundreds column must total \(5\) after including the regrouped hundred, so the two hundreds digits total \(4\), such as \(1+3\). 4. One valid construction is \(126+374=500\).

One valid answer is \(126+374=500\). Ones digits: \(6+4=10\), so regroup \(1\) ten. Tens digits before the regrouped ten: \(2+7=9\); with the regrouped ten, the tens total \(10\). Hundreds digits before the regrouped hundred: \(1+3=4\); with the regrouped hundred, the hundreds total \(5\).

- 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: \(600\)

- 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: \(758\)

- Start in the ones place and remember that you may need to regroup. - Then work one place at a time from right to left.

1. Ones place: \(* + 8 = 17\), so the missing ones digit is \(9\). Regroup \(1\) ten. 2. Tens place: \(* + 8 + 1 = 12\), so the missing tens digit is \(3\). Regroup \(1\) hundred. 3. The regrouped \(1\) becomes the hundreds digit of the sum. The completed equation is \(39 + 88 = 127\).

The missing digits are \(3\) and \(9\). The completed equation is \(39 + 88 = 127\).