- Begin with the ones digit of the multiplicand. - Separate the ones digit of that product from the amount that must be regrouped. - The empty box marks the regrouped amount used in the next column.

1. Start in the ones column: \(6\times4=24\). 2. Write \(4\) in the ones place and regroup \(2\) tens. 3. Therefore the carry in the empty box above the tens digit is \(2\).

The missing carry is \(2\), because \(6\times4=24\): write \(4\) and regroup \(2\) tens into the next column.

- Break \(32\) into tens and ones. - Multiply each place-value part by \(4\). - Add the two partial products.

1. \(30\times4=120\). 2. \(2\times4=8\). 3. Add the partial products: \(120+8=128\).

\(30\times4=120\), \(2\times4=8\), and \(120+8=128\).

- Think of \(204\) as \(200+4\). - Multiply each nonzero place by \(3\). - Regroup the ones product if needed.

1. \(200\times3=600\), \(0\times3=0\), and \(4\times3=12\). 2. Combine the place values: \(600+12=612\). 3. The tens digit of the product is \(1\) because \(12\) ones regroup as \(1\) ten and \(2\) ones.

\(612\). The tens digit comes from regrouping \(12\) ones as \(1\) ten and \(2\) ones.

- Look for ending zeros that represent factors of \(10\). - Identify the basic multiplication fact inside each product. - Which product has two factors with several nonzero digits? - Track the total number of factors of \(10\) in each mental-math product.

1. For a), use \(8 \times 6 = 48\) and account for the two factors of \(10\): \(8 \times 600 = 4800\). 2. For b), use \(4 \times 7 = 28\) and account for two factors of \(10\): \(40 \times 70 = 2800\). 3. For c), use \(5 \times 9 = 45\) and account for three factors of \(10\): \(5 \times 9000 = 45{,}000\). 4. For d), the standard algorithm is more useful because both factors have several nonzero place values. 5. For e), \(10 \times 1000 = 10{,}000\).

a) \(4800\) b) \(2800\) c) \(45{,}000\) d) The standard multiplication algorithm is more useful. e) \(10{,}000\)

- How many times is the hourly amount produced during one day? - Write a multiplication expression. - Break \(24\) into tens and ones. - Check that you used all the information in the problem.

1. Write the multiplication expression: \(375 \times 24\). 2. Use partial products: \(375 \times 20 = 7500\) and \(375 \times 4 = 1500\). 3. Add the partial products: \(7500 + 1500 = 9000\).

The factory makes \(9000\) chocolate bars in one day.

- Which amount is repeated, and how many times? - Break \(45\) into tens and ones. - Align the place values when you add the partial products. - Estimate the product before calculating exactly.

1. Write the multiplication expression: \(225 \times 45\). 2. Use partial products: \(225 \times 40 = 9000\) and \(225 \times 5 = 1125\). 3. Add the partial products: \(9000 + 1125 = 10{,}125\).

The shipment contains \(10{,}125\) notebooks.

- Compare the first and last digits of the products. - Count the zeros between those digits. - Compare the number of zeros with the number of \(1\)s in the first factor.

1. The products are \(12\times 9=108\), \(112\times 9=1008\), \(1112\times 9=10{,}008\), and \(11{,}112\times 9=100{,}008\). 2. Every product begins with \(1\) and ends with \(8\). 3. Each additional \(1\) in the first factor adds one more zero between the \(1\) and the \(8\) in the product.

The products are \(108,1008,10{,}008,100{,}008\). Each product begins with \(1\), ends with \(8\), and has one more zero in the middle than the previous product.

- Break apart a factor into a friendly multiple of \(10\) or \(100\) and a small remaining part. - A factor close to a power of ten may be handled by multiplying by the power of ten and then adjusting.

1. For a), decompose \(11\): \(26 \times 11 = 26 \times 10 + 26 \times 1 = 260 + 26 = 286\). 2. For b), use compensation: \(45 \times 9 = 45 \times 10 - 45 = 450 - 45 = 405\). 3. For c), decompose \(101\): \(12 \times 101 = 12 \times 100 + 12 \times 1 = 1200 + 12 = 1212\).

a) \(286\) b) \(405\) c) \(1212\)

- Multiply the number of plants by the harvest per plant. - Calculate the two beds separately. - Compare the two total weights.

1. Bed A produces about \(12 \times 450\,\text{g} \approx 5400\,\text{g}\). 2. Bed B produces about \(16 \times 300\,\text{g} \approx 4800\,\text{g}\). 3. Since \(5400\,\text{g} > 4800\,\text{g}\), Bed A produces more based on the estimates.

Bed A: about \(5400\,\text{g}\) Bed B: about \(4800\,\text{g}\) Bed A produces more strawberries based on the estimates.

- Break \(12\) into \(10+2\). - Find one partial product for the tens and one for the ones. - Add the partial products with their place values aligned.

1. Multiply by the tens: \(354 \times 10=3540\). 2. Multiply by the ones: \(354 \times 2=708\). 3. Add the partial products: \(3540+708=4248\).

\(4248\)

- Multiply from right to left and regroup when needed. - For a two-digit factor, split it into tens and ones. - Add the partial products using correct place-value alignment.

1. For a), \(4 \times 5 = 20\), so write \(0\) and regroup \(2\) tens. Then \(1 \times 5 + 2 = 7\), and \(6 \times 5 = 30\). Thus, \(614 \times 5 = 3070\). 2. For b), \(7 \times 3 = 21\), so write \(1\) and regroup \(2\) tens. Then \(3 \times 3 + 2 = 11\), so write \(1\) and regroup \(1\) hundred. Next, \(1 \times 3 + 1 = 4\), and \(2 \times 3 = 6\). Thus, \(2137 \times 3 = 6411\). 3. For c), \(9 \times 2 = 18\), so write \(8\) and regroup \(1\) ten. Then \(2 \times 2 + 1 = 5\), \(0 \times 2 = 0\), and \(4 \times 2 = 8\). Thus, \(4029 \times 2 = 8058\). 4. For d), use partial products: \(328 \times 10=3280\) and \(328 \times 4=1312\). Then \(3280+1312=4592\).

a) \(3070\) b) \(6411\) c) \(8058\) d) \(4592\)

- Multiply from right to left, one place at a time. - Regroup whenever a place-value product is greater than \(9\). - Keep every digit aligned with its place value.

1. For a), \(6 \times 4 = 24\), so write \(4\) and regroup \(2\). Then \(3 \times 4 + 2 = 14\), so write \(4\) and regroup \(1\). Next, \(2 \times 4 + 1 = 9\), and \(8 \times 4 = 32\). Thus, \(4 \times 8236 = 32{,}944\). 2. For b), \(2 \times 7 = 14\), so write \(4\) and regroup \(1\). Then \(5 \times 7 + 1 = 36\), so write \(6\) and regroup \(3\). Next, \(0 \times 7 + 3 = 3\). Then \(4 \times 7 = 28\), so write \(8\) and regroup \(2\). Finally, \(1 \times 7 + 2 = 9\). Thus, \(7 \times 14{,}052 = 98{,}364\). 3. For c), \(9 \times 6 = 54\), so write \(4\) and regroup \(5\). Then \(1 \times 6 + 5 = 11\), so write \(1\) and regroup \(1\). Next, \(6 \times 6 + 1 = 37\), so write \(7\) and regroup \(3\). Then \(5 \times 6 + 3 = 33\), so write \(3\) and regroup \(3\). Finally, \(3 \times 6 + 3 = 21\). Thus, \(35{,}619 \times 6 = 213{,}714\). 4. For d), \(8 \times 9 = 72\), so write \(2\) and regroup \(7\). Then \(0 \times 9 + 7 = 7\), \(1 \times 9 = 9\), and \(2 \times 9 = 18\), so write \(8\) and regroup \(1\). Finally, \(7 \times 9 + 1 = 64\). Thus, \(9 \times 72{,}108 = 648{,}972\).

a) \(32{,}944\) b) \(98{,}364\) c) \(213{,}714\) d) \(648{,}972\)

- Which operation represents repeated addition of the same number? - Break \(25\) into tens and ones. - Add the two partial products.

1. Repeated addition of the same number can be written as \(364 \times 25\). 2. Use partial products: \(364 \times 20=7280\) and \(364 \times 5=1820\). 3. Add: \(7280+1820=9100\).

The total is \(9100\).

- For part a), compare each factor with the matching benchmark factor; no exact multiplication is needed yet. - For part b), split one factor by place value so the partial products are easy to find. - Check that the exact product agrees with your benchmark comparison.

1. Both \(4508>4500\) and \(63>60\), and all factors are positive, so \(4508\times63>4500\times60=270{,}000\). 2. \(4508\times60=270{,}480\) and \(4508\times3=13{,}524\). 3. Add: \(270{,}480+13{,}524=284{,}004\), which agrees with the benchmark comparison.

a) \(4508\times63>270{,}000\) because both positive factors are larger than the benchmark factors. b) \(284{,}004\)

- Rewrite the multiple of \(10\) as a one-digit factor times \(10\). - Use a basic multiplication fact first. - Then use place value to multiply the result by \(10\).

1. \(14 \times 60 = 14 \times 6 \times 10 = 84 \times 10 = 840\). 2. \(23 \times 30 = 23 \times 3 \times 10 = 69 \times 10 = 690\). 3. \(40 \times 18 = 4 \times 18 \times 10 = 72 \times 10 = 720\). 4. \(12 \times 80 = 12 \times 8 \times 10 = 96 \times 10 = 960\).

a) \(840\) b) \(690\) c) \(720\) d) \(960\)

- Rewrite the multiple of \(10\) as a one-digit factor times \(10\). - Find the product with the one-digit factor first. - Use place value to multiply that result by \(10\).

1. \(13 \times 20 = 13 \times 2 \times 10 = 26 \times 10 = 260\). 2. \(24 \times 30 = 24 \times 3 \times 10 = 72 \times 10 = 720\). 3. \(15 \times 40 = 15 \times 4 \times 10 = 60 \times 10 = 600\). 4. \(42 \times 20 = 42 \times 2 \times 10 = 84 \times 10 = 840\). 5. \(31 \times 30 = 31 \times 3 \times 10 = 93 \times 10 = 930\).

a) \(260\) b) \(720\) c) \(600\) d) \(840\) e) \(930\)

- Look for a factor that can be decomposed using \(10\), \(20\), or \(100\). - Look for factors that combine to make \(100\). - Choose a strategy that reduces the product to easier partial products.

1. For a), decompose \(15\): \(12 \times 10 + 12 \times 5 = 120 + 60 = 180\). 2. For b), factor \(36\) as \(4 \times 9\): \((25 \times 4) \times 9 = 100 \times 9 = 900\). 3. For c), use \(20 - 1\): \(20 \times 8 - 1 \times 8 = 160 - 8 = 152\). 4. For d), use \(100 + 2\): \(100 \times 7 + 2 \times 7 = 700 + 14 = 714\).

a) \(12 \times 10 + 12 \times 5 = 180\) b) \((25 \times 4) \times 9 = 900\) c) \(20 \times 8 - 1 \times 8 = 152\) d) \(100 \times 7 + 2 \times 7 = 714\)

- Use known multiplication facts and place-value patterns. - Remember that a product is the result of multiplication. - Multiply neighboring bottom values to complete the second row, then find the top.

1. The left brick in the second row is \(2 \times 5 = 10\). 2. The right brick in the second row is \(5 \times 8 = 40\). 3. The top is \(10 \times 40 = 400\).

Second row: \(10\), \(40\) Top: \(400\)

- Start with the rightmost digit before considering the internal zero. - Determine what must be regrouped from the ones multiplication. - Then include that regrouped amount when multiplying the tens digit.

1. The ones calculation is \(7\times6=42\), so write \(2\) and carry \(4\) to the tens column. 2. The carry in the empty box is therefore \(4\). 3. In the tens column, \(0\times6+4=4\), so the tens digit of the product is \(4\).

The missing carry is \(4\), because \(7\times6=42\). The next calculation is \(0\times6+4=4\), which explains the tens digit \(4\) in the product.

- Identify which digit of \(23\) creates the row containing the empty box. - Find the one-digit product first. - Use the place value of that multiplier digit to interpret the row.

1. The units partial product is \(324\times3=972\). 2. The tens digit is \(2\), so \(324\times2=648\). The missing digit is \(4\). 3. Because the \(2\) represents \(20\), the tens partial-product row represents \(6480\). 4. \(972+6480=7452\).

Missing digit: \(4\) Tens partial-product value: \(6480\) Product: \(7452\)

- Match each partial-product row to a digit of the multiplier. - Explain the zero row from the ones digit before working on the tens row. - The tens row is shifted because its multiplier digit represents tens.

1. The ones digit of \(40\) is \(0\), so the first partial product is \(618\times0=0\). 2. The tens digit is \(4\). The unshifted digit product is \(618\times4=2472\), so the missing digit is \(7\). 3. That row represents \(24{,}720\) after the tens-place shift, so the product is \(24{,}720\).

The missing digit is \(7\). The first row is \(0\) because the ones digit of \(40\) is \(0\). The tens digit \(4\) gives \(618\times4=2472\), and the shifted row represents \(24{,}720\). The product is \(24{,}720\).

- Work through the units partial product from right to left. - Track the amount regrouped from the ones calculation into the tens calculation. - Interpret the second row using the place value of the multiplier digit \(3\).

1. The units row comes from \(205\times4\). Since \(5\times4=20\), write \(0\) and regroup \(2\) tens. Then \(0\times4+2=2\), so the missing digit is \(2\) and the row is \(820\). 2. The tens row comes from \(205\times3=615\). Because the \(3\) represents \(30\), its shifted row represents \(6150\). 3. \(820+6150=6970\), verifying the displayed product.

a) The missing digit is \(2\): \(5\times4=20\) regroups \(2\) tens, so \(0\times4+2=2\). b) The shifted tens row represents \(6150\), and \(820+6150=6970\).

- Determine whether the row containing the empty box is generated by the ones digit or tens digit of the multiplier. - Complete that one-digit multiplication first. - Then interpret the other row by place value and combine the two contributions.

1. The first row comes from the units digit \(6\). Since \(128\times6=768\), the missing digit is \(6\). 2. The second row comes from \(128\times3=384\), shifted one place left because the \(3\) represents \(30\). It therefore represents \(3840\). 3. Add \(768+3840=4608\). The school has \(4608\) pencils.

The missing digit is \(6\), because the units row comes from multiplier digit \(6\) and \(128\times6=768\). The tens row represents \(128\times30=3840\). Thus \(768+3840=4608\) pencils.

- Identify the place of the multiplier digit associated with the second row. - Distinguish \(497\times3\) from \(497\times30\). - Add the values represented by the two partial-product rows.

1. The second row comes from the digit \(3\) in \(31\). 2. That \(3\) is in the tens place, so it represents \(30\). Therefore the row represents \(14{,}910\), not \(1491\). 3. The units row represents \(497\). Adding gives \(497+14{,}910=15{,}407\).

a) The digit \(3\). b) It represents multiplication by \(30\), so the row represents \(14{,}910\). c) \(497+14{,}910=15{,}407\).

- Split \(27\) into \(20+7\). - Find one partial product for each part. - Add the partial products.

1. Multiply by the ones digit: \(34\times7=238\). 2. Multiply by the tens value: \(34\times20=680\). 3. Add: \(238+680=918\).

Partial products: \(238\) and \(680\) Product: \(918\)

- Read the width of each part from the diagram. - Both parts have the same height. - Add the two partial areas.

1. The left part is \(23\times10=230\,\text{cm}^2\). 2. The right part is \(23\times6=138\,\text{cm}^2\). 3. Add the partial areas: \(230+138=368\,\text{cm}^2\).

Partial areas: \(230\,\text{cm}^2\) and \(138\,\text{cm}^2\) Total: \(368\,\text{cm}^2\)

- Look for numbers that are easy to double mentally. - Break a number into place-value parts when that simplifies doubling. - A written algorithm can help when several places require regrouping.

1. For a), mental math is convenient: \(450{,}000 \times 2 = 900{,}000\). 2. For b), a place-value strategy works well: \(105{,}000 \times 2 = 210{,}000\) and \(50 \times 2 = 100\), giving \(210{,}100\). 3. For c), a written algorithm is a reasonable way to organize the regrouping. The product is \(643{,}368\). 4. For d), compensation is convenient: \(250{,}000 \times 2 - 200 = 499{,}800\).

a) \(900{,}000\) b) \(210{,}100\) c) \(643{,}368\) d) \(499{,}800\) Method choices may vary if they are explained and used correctly.

- How many days are in each year under the given assumption? - How many days are in \(10\) years? - How many hours are in \(1\) day? - Which operation converts the total number of days to hours?

1. Find the number of days in \(10\) years: \(10 \times 365=3650\) days. 2. Convert days to hours. Since \(1\) day has \(24\) hours, \(3650 \times 24=87{,}600\) hours.

Ava has lived \(87{,}600\,\text{hours}\).

- Identify which factor stays the same. - Determine how much the other factor increases each time. - Multiply the fixed factor by that increase. - Compare consecutive products.

1. The products are \(1000,2250,3500,4750,6000\). 2. The second factor increases by \(10\) each time. 3. Since \(125\times 10=1250\), each product is \(1250\) greater than the previous product.

The products are \(1000,2250,3500,4750,6000\). Each product increases by \(1250\).

- First find the total number of bottles delivered. - Multiply the number of bottles by the deposit per bottle. - The product is in cents; convert it to dollars. - Written multiplication may help with \(43{,}200 \times 15\).

1. Find the total number of pallets: \(6 \times 15 = 90\). 2. Find the total number of cases: \(90 \times 40 = 3600\). 3. Find the total number of bottles: \(3600 \times 12 = 43{,}200\). 4. Find the total deposit in cents: \(43{,}200 \times 15 = 648{,}000\) cents. 5. Convert cents to dollars: \(648{,}000 \div 100 = \$6480\).

The total refundable deposit is \(\$6480\).

- “Every seat is sold” means every seat in each section earns its listed price. - Find the revenue for each section separately. - Add the three large amounts carefully by place value.

1. Find the Front Orchestra revenue: \(125 \times \$45 = \$5625\). 2. Find the Rear Orchestra revenue: \(360 \times \$32 = \$11{,}520\). 3. Find the Balcony revenue: \(215 \times \$24 = \$5160\). 4. Add the three amounts: \(\$5625 + \$11{,}520 + \$5160 = \$22{,}305\).

When every seat is sold, the theater earns \(\$22{,}305\).

- First find the total number of boxes. - Then find the total number of rolls. - If you calculate the cost in cents, convert the result to dollars. - Work step by step from crates to boxes to rolls.

1. Find the total number of boxes: \(6 \times 12 = 72\). 2. Find the total number of rolls: \(72 \times 15 = 1080\). 3. Find the total cost in cents: \(1080 \times 40 = 43{,}200\) cents. 4. Convert to dollars: \(43{,}200\) cents is \(\$432.00\).

All the rolls cost \(\$432.00\).

- Look for factors with zeros that make place-value reasoning useful. - Break apart one factor when that creates easier products. - Use the standard algorithm when several partial products are difficult to track mentally.

1. For a), use mental math with the basic fact \(6 \times 7 = 42\) and the place value of \(7000\): \(6 \times 7000 = 42{,}000\). 2. For b), use partial products: \(6 \times 7050 = 6 \times 7000 + 6 \times 50 = 42{,}000 + 300 = 42{,}300\). 3. For c), use partial products by decomposing \(16\): \(16 \times 7000 = 10 \times 7000 + 6 \times 7000 = 70{,}000 + 42{,}000 = 112{,}000\). 4. For d), use partial products: \(16 \times 7058 = 10 \times 7058 + 6 \times 7058 = 70{,}580 + 42{,}348 = 112{,}928\).

a) \(42{,}000\) (mental math) b) \(42{,}300\) (partial products) c) \(112{,}000\) (partial products) d) \(112{,}928\) (partial products)

- Products with factors ending in zeros may be efficient to find mentally. - For less friendly factors, break one factor apart by place value. - Compare how much regrouping or recordkeeping each problem requires.

1. In Pair 1, \(500 \times 800\) is efficient mentally: \(5 \times 8 = 40\), and the four place-value zeros give \(400{,}000\). 2. For \(12 \times 7895\), use partial products or the standard algorithm: \(10 \times 7895 + 2 \times 7895 = 78{,}950 + 15{,}790 = 94{,}740\). 3. In Pair 2, \(24 \times 4000\) is efficient mentally because \(24 \times 4 = 96\), so the product is \(96{,}000\). 4. For \(24 \times 1069\), use partial products or the standard algorithm: \(20 \times 1069 + 4 \times 1069 = 21{,}380 + 4276 = 25{,}656\).

Pair 1: \(500 \times 800 = 400{,}000\) (mental math) and \(12 \times 7895 = 94{,}740\) (partial products or standard algorithm) Pair 2: \(24 \times 4000 = 96{,}000\) (mental math) and \(24 \times 1069 = 25{,}656\) (partial products or standard algorithm)

- Record the final result for each of at least three starting numbers. - Combine the three multipliers into one product. - Compare that combined multiplier with your examples.

1. For \(123\): \(123 \times 2 = 246\), \(246 \times 2 = 492\), and \(492 \times 25 = 12{,}300\). 2. For \(150\): \(150 \times 2 = 300\), \(300 \times 2 = 600\), and \(600 \times 25 = 15{,}000\). 3. For \(204\): \(204 \times 2 = 408\), \(408 \times 2 = 816\), and \(816 \times 25 = 20{,}400\). 4. The combined multiplier is \(2 \times 2 \times 25 = 100\), so the final result is always \(100\) times the starting number.

Example tests: \(123 \to 12{,}300\) \(150 \to 15{,}000\) \(204 \to 20{,}400\) Rule: the final result is always \(100\) times the starting number because \(2 \times 2 \times 25 = 100\).

- Calculate the product of the three multipliers in Method A. - Compare that combined multiplier with the multiplier in Method B. - Explain why regrouping factors does not change a product.

1. Method A gives \(145\times 5=725\), \(725\times 2=1450\), and \(1450\times 9=13{,}050\). 2. Method B gives \(145\times 90=13{,}050\). 3. The multipliers in Method A combine to \(5\times 2\times 9=90\). 4. Therefore, both methods multiply the starting number by the same overall factor, so they always give the same result.

Both methods give \(13{,}050\) when the starting number is \(145\). They always agree because \(5\times 2\times 9=90\).

- Find the combined mass of all the crates. - Add the cargo mass to the empty truck's mass. - Convert the stated limit to kilograms before comparing.

1. The cargo mass is \(25 \times 110\,\text{kg} = 2750\,\text{kg}\). 2. The loaded truck mass is \(4200\,\text{kg} + 2750\,\text{kg} = 6950\,\text{kg}\). 3. The limit is \(7\times1000\,\text{kg}+500\,\text{kg}=7500\,\text{kg}\). 4. Since \(6950<7500\), the truck is within the limit by \(550\,\text{kg}\).

Yes. The cargo mass is \(25\times110\,\text{kg}=2750\,\text{kg}\). The loaded mass is \(4200\,\text{kg}+2750\,\text{kg}=6950\,\text{kg}\), and the limit is \(7500\,\text{kg}\). Since \(6950<7500\), the truck is \(550\,\text{kg}\) below the limit.

- Look for a factor that is close to a multiple of \(100\). - Break apart a factor by place value when each partial product is easy. - Use the associative property to group factors that make \(100\).

1. For a), use compensation: \(14 \times 199 = 14 \times 200 - 14 = 2800 - 14 = 2786\). 2. For b), decompose \(907\): \(8 \times 907 = 8 \times 900 + 8 \times 7 = 7200 + 56 = 7256\). 3. For c), rearrange and group the factors: \(25 \times 44 = 25 \times 4 \times 11 = 100 \times 11 = 1100\).

a) \(2786\) b) \(7256\) c) \(1100\)

- Is one factor close to a friendly multiple of \(100\)? - For the two-digit factors, break one factor apart by place value or use the standard algorithm.

1. For a), use compensation: \(99 \times 7 = 100 \times 7 - 7 = 700 - 7 = 693\). 2. For b), use partial products or the standard algorithm: \(38 \times 24 = 38 \times 20 + 38 \times 4 = 760 + 152 = 912\).

a) \(693\); mentally use \(100 \times 7 - 7\). b) \(912\)

- Estimate each product first. - Calculate the exact products using a reliable multiplication method. - When the values are close, compare the digits from left to right.

1. Calculate the products: A: \(6400 \times 15=96{,}000\) B: \(4000 \times 24=96{,}000\) C: \(11{,}999 \times 8=95{,}992\) D: \(5998 \times 16=95{,}968\) E: \(3201 \times 30=96{,}030\) 2. Compare the values: \(95{,}968<95{,}992<96{,}000<96{,}030\). Choice D has the least value.

D: \(5998 \times 16=95{,}968\)

- Round the factors to estimate each product. - Then calculate each exact product carefully. - Use the exact values, not only the estimates, for the final comparison.

1. Estimate: A: \(14{,}990 \times 12 \approx 15{,}000 \times 12=180{,}000\) B: \(11{,}250 \times 16 \approx 11{,}000 \times 16=176{,}000\) C: \(8950 \times 20 \approx 9000 \times 20=180{,}000\) D: \(22{,}480 \times 8 \approx 22{,}500 \times 8=180{,}000\) 2. Calculate exactly: A is \(179{,}880\), B is \(180{,}000\), C is \(179{,}000\), and D is \(179{,}840\). 3. The greatest exact value is \(180{,}000\), from choice B.

A: estimate \(180{,}000\); exact \(179{,}880\) B: estimate \(176{,}000\); exact \(180{,}000\) C: estimate \(180{,}000\); exact \(179{,}000\) D: estimate \(180{,}000\); exact \(179{,}840\) The greatest exact product is B, \(180{,}000\).

- Recalculate each product without relying on the stated answer. - Use partial products or the standard algorithm. - Estimate each product to check whether the result is reasonable.

1. a) \(18 \times 14 = 18 \times (10 + 4) = 180 + 72 = 252\), so \(242\) is incorrect. 2. b) \(16 \times 16 = 256\), so the statement is correct. 3. c) \(21 \times 16 = 336\), so the statement is correct. 4. d) \(22 \times 14 = 22 \times (10 + 4) = 220 + 88 = 308\), so \(298\) is incorrect.

a) Incorrect; \(18 \times 14 = 252\). b) Correct. c) Correct. d) Incorrect; \(22 \times 14 = 308\).

- Calculate each product and record its value. - Estimate first to catch large calculation errors. - Compare the exact products by place value.

1. Calculate the products: A is \(258 \times 7=1806\), B is \(314 \times 6=1884\), C is \(49 \times 37=1813\), and D is \(182 \times 9=1638\). 2. Compare the values: \(1884>1813>1806>1638\). Choice B has the greatest value.

B: \(314 \times 6=1884\)

- Multiply by the tens and ones separately. - Align the partial products by place value. - Add the partial products and estimate to check the result.

1. For a), \(537 \times 20=10{,}740\) and \(537 \times 4=2148\). Then \(10{,}740+2148=12{,}888\). 2. For b), \(1284 \times 10=12{,}840\) and \(1284 \times 3=3852\). Then \(12{,}840+3852=16{,}692\). 3. For c), \(2115 \times 30=63{,}450\) and \(2115 \times 6=12{,}690\). Then \(63{,}450+12{,}690=76{,}140\).

a) \(12{,}888\) b) \(16{,}692\) c) \(76{,}140\)

- Calculate both products first. - Compare the results from left to right by place value. - Subtract the smaller product from the greater product.

1. Lucas: \(8 \times 12{,}345=98{,}760\). 2. Maya: \(6 \times 16{,}465=98{,}790\). 3. Since \(98{,}790>98{,}760\), Maya has the greater product. 4. The difference is \(98{,}790-98{,}760=30\).

Maya gets the greater product. The difference is \(30\).

- Calculate both products in each comparison. - Look for factor relationships that may simplify a comparison. - When products are close, compare every place carefully.

1. \(6 \times 12{,}450=74{,}700\) and \(3 \times 24{,}900=74{,}700\), so a) is \(=\). 2. \(7 \times 15{,}312=107{,}184\) and \(8 \times 13{,}400=107{,}200\), so b) is \(<\). 3. \(4 \times 48{,}216=192{,}864\) and \(9 \times 21{,}428=192{,}852\), so c) is \(>\).

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

- Calculate each product first. - Use the standard algorithm and regroup carefully. - Compare the products by place value.

1. \(736 \times 8=5888\). 2. \(1465 \times 4=5860\). 3. \(982 \times 6=5892\). 4. In order from least to greatest: \(5860<5888<5892\).

a) \(736 \times 8 = 5888\) b) \(1465 \times 4 = 5860\) c) \(982 \times 6 = 5892\) From least to greatest: b), a), c).

- Calculate both products first. - The difference is found by subtracting the smaller value from the greater value.

1. \(a=3407 \times 5=17{,}035\). 2. \(b=2138 \times 8=17{,}104\). 3. The difference is \(17{,}104-17{,}035=69\).

\(a=17{,}035\) \(b=17{,}104\) The difference is \(69\).

- Calculate the first product. - Compare the second factors \(38\) and \(37\). - Decide how changing one factor by \(1\) changes the product.

1. Use partial products: \(412 \times 30=12{,}360\) and \(412 \times 8=3296\). 2. Add: \(12{,}360+3296=15{,}656\). 3. The second expression has one fewer group of \(412\), because \(37\) is one less than \(38\). 4. Therefore, the difference between the products is \(412\).

\(412 \times 38=15{,}656\). The difference between the two products is \(412\).

- Break the two-digit factor into tens and ones. - Align and add the partial products. - Compare the exact products by place value.

1. \(456 \times 23=456 \times 20+456 \times 3=9120+1368=10{,}488\). 2. \(382 \times 47=382 \times 40+382 \times 7=15{,}280+2674=17{,}954\). 3. \(709 \times 18=709 \times 10+709 \times 8=7090+5672=12{,}762\). 4. Order the products: \(10{,}488<12{,}762<17{,}954\).

a) \(10{,}488\) b) \(17{,}954\) c) \(12{,}762\) Least to greatest: \(10{,}488<12{,}762<17{,}954\).

- Calculate both products using partial products. - Add each pair of partial products carefully. - The products are close, so compare every digit of the final values.

1. \(617 \times 24=617 \times 20+617 \times 4=12{,}340+2468=14{,}808\). 2. \(423 \times 35=423 \times 30+423 \times 5=12{,}690+2115=14{,}805\). 3. Since \(14{,}808>14{,}805\), the correct symbol is \(>\).

\(617 \times 24>423 \times 35\) because \(14{,}808>14{,}805\).

- Calculate both products using partial products. - Compare the exact values. - Subtract the smaller product from the greater product.

1. Product A: \(482 \times 30=14{,}460\) and \(482 \times 4=1928\), so \(482 \times 34=16{,}388\). 2. Product B: \(342 \times 40=13{,}680\) and \(342 \times 8=2736\), so \(342 \times 48=16{,}416\). 3. Product B is greater because \(16{,}416>16{,}388\). 4. The difference is \(16{,}416-16{,}388=28\).

Product B is greater. Product A is \(16{,}388\), product B is \(16{,}416\), and the difference is \(28\).

- First find the number of large building sets. - Add the two types to find the total production. - Subtract the number shipped from the number produced. - The difference is the amount left at the factory.

1. The number of large sets is \(12{,}350\times 14=172{,}900\). 2. The factory makes \(12{,}350+172{,}900=185{,}250\) sets altogether. 3. After shipping, \(185{,}250-155{,}200=30{,}050\) sets remain.

\(30{,}050\) building sets remain at the factory.

- Focus on the red pen packages asked for in the question. - Find the total number of cartons in all the containers. - Use the carton total to find the number of packages exported. - Subtract the exported amount from the red pen supply.

1. The center receives \(4800\times 18=86{,}400\) packages of red pens. 2. The containers hold \(12\times 60=720\) cartons altogether. 3. The cartons contain \(720\times 110=79{,}200\) packages of red pens. 4. The number remaining is \(86{,}400-79{,}200=7200\).

\(7200\) packages of red pens remain.

- Find each school’s total separately. - Compare the two totals. - Subtract the smaller total from the larger one.

1. Pine Grove raises \(245 \times \$14 = \$3430\). 2. Downtown raises \(218 \times \$18 = \$3924\). 3. Since \(\$3924 > \$3430\), Downtown raises more. 4. The difference is \(\$3924 - \$3430 = \$494\).

Downtown Elementary raises more money. The difference is \(\$494\).

- Scale the assumed per-minute rate to one hour. - Use the number of waking hours to find the daily estimate. - Multiply the daily estimate by the number of days in the year.

1. a) There are \(60\) minutes in an hour, so the estimate is \(15 \times 60 = 900\) blinks per hour. 2. b) In \(16\) waking hours, the estimate is \(900 \times 16 = 14{,}400\) blinks. 3. c) In \(365\) days, the estimate is \(14{,}400 \times 365 = 5{,}256{,}000\) blinks.

a) About \(900\) times b) About \(14{,}400\) times c) About \(5{,}256{,}000\) times

- Round each factor to a nearby number that is easy to multiply. - You can first find how many notebooks are in one carton. - Compare the exact result with your estimate to check that it is reasonable.

1. One reasonable estimate is \(200\times20\times10=40{,}000\) notebooks. 2. Find the number of notebooks in one carton: \(15\times12=180\). 3. Find the exact total: \(220\times180=39{,}600\).

Estimate: about \(40{,}000\) notebooks Exact total: \(39{,}600\) notebooks

- Find each room's total capacity separately. - First determine the capacity of one shelving unit. - Subtract the lesser capacity from the greater capacity.

1. Room A holds \(15\times6\times12=1080\) binders. 2. Room B holds \(12\times8\times10=960\) binders. 3. Since \(1080>960\), Room A holds more. 4. The difference is \(1080-960=120\) binders.

Room A holds \(120\) more binders than Room B.

- Find the total number of apples first. - Use the greatest possible weight for every apple to maximize the total. - Use the least possible weight for every apple to minimize the total.

1. a) The number of apples is \(6\times15\times12=1080\). 2. b) The greatest total occurs when every apple weighs \(160\,\text{g}\): \(1080\times160\,\text{g}=172{,}800\,\text{g}\). 3. c) The least total occurs when every apple weighs \(140\,\text{g}\): \(1080\times140\,\text{g}=151{,}200\,\text{g}\).

a) \(1080\) apples b) \(172{,}800\,\text{g}\) c) \(151{,}200\,\text{g}\)

- Identify which multiplier digit produces each partial-product row. - The second row comes from a tens digit, not a ones digit. - After completing both partial products, combine their place-value contributions.

1. The units partial product is \(138\times4=552\). 2. The tens digit of the multiplier is \(2\), so the unshifted digit product is \(138\times2=276\). The missing digit is \(7\). 3. Because that \(2\) means \(20\), the English written algorithm shifts this row one place left, so it represents \(2760\). 4. Add the partial products: \(552+2760=3312\).

a) \(7\) b) The row comes from \(138\times20\), so its value is \(2760\). c) \(3312\)

- Use the first partial-product row to identify the missing ones digit of the multiplier. - The second multiplier digit is in the tens place, so its row represents tens of groups. - Add the two partial-product values to find the total.

1. The units partial-product row is \(654\). Since \(327\times2=654\), the missing units digit is \(2\), so the multiplier is \(42\). 2. The second row comes from the tens digit \(4\), which represents \(40\). Since \(327\times4=1308\), that row represents \(13{,}080\). 3. Add the represented row values: \(654+13{,}080=13{,}734\).

a) \(2\) b) The second row represents \(327\times40=13{,}080\). c) \(13{,}734\) tickets

- Find one day's total for each type of item. - Account for all three basket sizes. - Multiply each daily total by \(6\).

1. Chocolate bars per day: \(45\times4+45\times8+45\times12=1080\). 2. For \(6\) days: \(1080\times6=6480\) chocolate bars. 3. Jars of jam per day: \(45\times2+45\times5+45\times8=675\). 4. For \(6\) days: \(675\times6=4050\) jars of jam.

\(6480\) chocolate bars and \(4050\) jars of jam

- Find the daily number of flags from both pack types. - Find the daily number of stickers from both pack types. - Scale both totals to \(3\) days before finding the difference.

1. The number of flags produced each day is \(250\times2+180\times5=1400\). 2. In \(3\) days, \(1400\times3=4200\) flags are produced. 3. The number of stickers produced each day is \(250\times3+180\times8=2190\). 4. In \(3\) days, \(2190\times3=6570\) stickers are produced. 5. The difference is \(6570-4200=2370\).

\(2370\) more stickers

- Use the related basic fact, then apply place value for the factor ending in zero. - Reversing the factors should not change a product. - Divide the product by one factor to recover the other factor.

1. For a), \(37 \times 40=1480\). Division check: \(1480 \div 40=37\). Commutative-property check: \(40 \times 37=1480\). 2. For b), \(215 \times 30=6450\). Division check: \(6450 \div 30=215\). Commutative-property check: \(30 \times 215=6450\). 3. For c), \(506 \times 70=35{,}420\). Division check: \(35{,}420 \div 70=506\). Commutative-property check: \(70 \times 506=35{,}420\).

a) \(37 \times 40=1480\); checks: \(1480 \div 40=37\) and \(40 \times 37=1480\) b) \(215 \times 30=6450\); checks: \(6450 \div 30=215\) and \(30 \times 215=6450\) c) \(506 \times 70=35{,}420\); checks: \(35{,}420 \div 70=506\) and \(70 \times 506=35{,}420\)

- Break \(23\) into tens and ones. - Reversing the factors should not change the product. - Division can undo multiplication.

1. Use partial products: \(54 \times 20=1080\) and \(54 \times 3=162\). Then \(1080+162=1242\). 2. Commutative-property check: \(23 \times 54=1242\). 3. Inverse-operation check: \(1242 \div 23=54\).

\(54 \times 23=1242\). Checks: \(23 \times 54=1242\) and \(1242 \div 23=54\).

- Round the ticket counts to convenient tens for an estimate. - Then find exactly how many tickets were sold altogether and how many were sold at the door. - Find the revenue from each type of ticket and add.

1. One estimate is to use about \(260\) tickets sold, with about \(190\) advance tickets and \(70\) tickets sold at the door. The estimated revenue is \(190\times\$12+70\times\$15=\$3330\). 2. Exactly \(300-42=258\) tickets are sold. 3. The number of tickets sold at the door is \(258-185=73\). 4. Advance-ticket revenue is \(185\times\$12=\$2220\). 5. Revenue from tickets sold at the door is \(73\times\$15=\$1095\). 6. The exact total revenue is \(\$2220+\$1095=\$3315\).

One reasonable estimate is \(\$3330\). The exact total ticket revenue is \(\$3315\).

- Find the total attendance for all five shows. - Then multiply the total number of tickets by the price of one ticket.

1. The total attendance is \(240+310+350+180+320=1400\). 2. The total ticket revenue is \(1400 \times \$12=\$16{,}800\).

The circus's total ticket revenue was \(\$16{,}800\).

- Find the total number of peak-season passengers and the resulting revenue. - Repeat the calculation for the off-season. - Add the two revenue amounts and compare the result with the captain's estimate.

1. Peak-season ridership is \(90 \times 12 \times 40=43{,}200\) passengers. Peak-season revenue is approximately \(43{,}200 \times \$8=\$345{,}600\). 2. Off-season ridership is \(120 \times 5 \times 20=12{,}000\) passengers. Off-season revenue is approximately \(12{,}000 \times \$8=\$96{,}000\). 3. Estimated annual revenue is \(\$345{,}600+\$96{,}000\approx\$441{,}600\). 4. Since \(\$441{,}600<\$500{,}000\), the captain’s estimate is too high. The difference is approximately \(\$500{,}000-\$441{,}600=\$58{,}400\).

No. The estimated annual revenue is approximately \(\$441{,}600\), which is approximately \(\$58{,}400\) less than \(\$500{,}000\).

- “Increase by” and “decrease by” indicate addition and subtraction, not scaling. - Find the two new factors before multiplying. - Subtract the new product from the original product.

1. The original product is \(120 \times 40 = 4800\). 2. The new factors are \(120 + 20 = 140\) and \(40 - 20 = 20\). The new product is \(140 \times 20 = 2800\). 3. The decrease is \(4800 - 2800 = 2000\).

a) \(4800\) b) \(2800\) c) The product decreased by \(2000\).

- Multiply to find the total sugar used. - Subtract the amount used from the starting amount. - Check that the remainder is reasonable.

1. Find the sugar used for all the trays: \(14 \times 165\,\text{g} = 2310\,\text{g}\). 2. Subtract the amount used from the supply: \(2500\,\text{g} - 2310\,\text{g} = 190\,\text{g}\).

\(190\,\text{g}\) of sugar remain.

- Rewrite \(2001\) as a sum involving a multiple of \(1000\). - Multiply \(432\) by each part. - Add the partial products.

1. Decompose \(2001\) as \(2000 + 1\). 2. Distribute: \((2000 + 1) \times 432 = 2000 \times 432 + 1 \times 432\). 3. Add the partial products: \(864{,}000 + 432 = 864{,}432\).

\((2000 + 1) \times 432 = 864{,}000 + 432 = 864{,}432\)

- In problem 1, start with the ones multiplication and identify what is regrouped into the tens column. - In problem 2, identify the multiplier digit responsible for the second partial-product row. - Explain the row's value using the place value of that multiplier digit.

1. \(5\times6=30\). Write \(0\) ones and regroup \(3\) tens, so the hidden carry is \(3\). 2. The units partial product is \(538\times7=3766\). The tens digit is \(4\), so \(538\times4=2152\); the missing digit is \(1\). 3. Because the \(4\) in \(47\) represents \(40\), the tens partial product represents \(21{,}520\). 4. \(3766+21{,}520=25{,}286\).

1) The hidden carry is \(3\), because \(5\times6=30\) regroups \(3\) tens. 2) The missing digit is \(1\). The tens row represents \(538\times40=21{,}520\), and the final product is \(25{,}286\).

- Use the given products to find the missing factors in the bottom row. - After completing one row, multiply neighboring values to find the row above. - For the final multiplication, decompose \(36\) as \(30 + 6\).

1. The second bottom value is found from \(2 \times x = 4\), so \(x = 2\). 2. The third bottom value satisfies \(2 \times y = 6\), so \(y = 3\). 3. The last bottom value satisfies \(3 \times z = 6\), so \(z = 2\). 4. The third row is \(4 \times 6 = 24\) and \(6 \times 6 = 36\). 5. Use partial products: \(24 \times 36 = 24 \times (30 + 6) = 720 + 144 = 864\).

Bottom row: \(2\), \(2\), \(3\), \(2\) Second row: \(4\), \(6\), \(6\) Third row: \(24\), \(36\) Top: \(864\)

- Check each partial product separately against its multiplier digit. - Remember that the second row corresponds to the tens digit. - Recalculate only the row that does not match its digit multiplication.

1. The ones partial product is correct: \(463 \times 7 = 3241\). 2. The tens digit is \(2\), so the unshifted tens partial product should be \(463 \times 2 = 926\), not \(936\). 3. The corrected tens contribution is \(9260\). 4. Add: \(3241 + 9260 = 12{,}501\).

The tens partial product is incorrect. It should come from \(463 \times 2 = 926\), giving a correct product of \(12{,}501\).

- Identify which multiplier digit is responsible for the hidden row. - Find the one-digit product first. - Use the tens place to determine the value represented by that row.

1. The visible units row is \(384\times5=1920\). 2. The tens digit is \(2\), so the hidden one-digit product is \(384\times2=768\). 3. Because the \(2\) represents \(20\), the tens row represents \(7680\). 4. \(1920+7680=9600\).

The hidden entry is \(768\). The tens row represents \(7680\), and \(1920+7680=9600\).

- Compare the ones and tens digits of the two multipliers separately. - Match each multiplier digit to its partial-product row. - Relate the change in the tens digit to the change in the whole multiplier.

1. Both multipliers have ones digit \(8\), so both units rows are \(672\times8=5376\). 2. The tens digit of \(58\) is \(5\). Since \(672\times5=3360\), the hidden digit is \(6\), and the tens row represents \(33{,}600\). 3. The multipliers differ by \(10\), so the product difference is \(672\times10=6720\).

a) The \(5376\) units row is unchanged because both multipliers have ones digit \(8\). b) \(6\) c) The multiplier increases by \(10\), so the product increases by \(672\times10=6720\).

- Treat the units and tens partial-product rows separately. - Identify the multiplier digit that generated each row. - Account for the tens-place value before adding the two represented values.

1. The units digit is \(7\), so \(4006\times7=28{,}042\). The first missing digit is \(8\). 2. The tens digit is \(3\), so \(4006\times3=12{,}018\). The second missing digit is \(1\). Because the \(3\) represents \(30\), that row represents \(120{,}180\). 3. \(28{,}042+120{,}180=148{,}222\).

The missing digits are \(8\) and \(1\). The first row comes from \(4006\times7\); the second comes from the tens digit \(3\) and represents \(4006\times30=120{,}180\). The product is \(148{,}222\).

- Which digit in the second factor makes the incomplete row? - Find the partial product before accounting for its place-value position. - Add the two partial products to check the final product.

1. The multiplier's tens digit represents \(30\). Multiply \(427\times3=1281\), so the tens partial-product row represents \(12{,}810\). 2. The missing digit in that row is \(2\). 3. The units row is \(427\times6=2562\). Add \(2562+12{,}810=15{,}372\).

The missing digit is \(2\); the product is \(15{,}372\).

- Calculate and compare both products. - Determine how \(412\) changes to \(206\). - Determine how \(16\) changes to \(32\). - Think about how opposite changes to the factors affect a product.

1. \(412 \times 16=412 \times 10+412 \times 6=4120+2472=6592\). 2. \(206 \times 32=206 \times 30+206 \times 2=6180+412=6592\). 3. The products are equal. 4. The first factor was divided by \(2\), while the second factor was multiplied by \(2\). These changes cancel: \((412 \div 2) \times (16 \times 2)=412 \times 16\).

a) Both products are \(6592\). b) The products are equal. c) Dividing one factor by \(2\) and multiplying the other factor by \(2\) keeps the product unchanged.

- List the leap-year positions beginning with Year \(1\). - Subtract the number of leap years from \(60\) to find the regular years. - Find the total days for each type of year separately. - Add the two day totals.

1. After Year \(1\), there are \(59\) years. Since \(59 \div 4 = 14\) remainder \(3\), there are \(14\) additional leap years. The total is \(1 + 14 = 15\) leap years. 2. Find the number of regular years: \(60 - 15 = 45\). 3. Find the days in leap years: \(15 \times 366 = 5490\) days. 4. Find the days in regular years: \(45 \times 365 = 16{,}425\) days. 5. Add the totals. Starting with \(16{,}425\) days and adding \(5490\) days gives \(21{,}915\) days.

a) There are \(15\) leap years and \(45\) regular years. b) The period contains \(21{,}915\) days.