Free math worksheets

- Begin with the ones digit of the multiplicand. - Separate the ones digit of that product from the amount that must be regrouped. - The star 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 starred carry 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.

- 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 \(9100\).

- Round each factor to a nearby number that is easy to multiply. - Find the tens-place and ones-place partial products. - Align the partial products before adding.

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

a) \(4508\times63>270{,}000\) because both factors are greater than the corresponding positive benchmark factors \(4500\) and \(60\). 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 starred carry 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 star. - First find the digit product before interpreting its place value. - The row shift must match the place value of the multiplier digit.

1. The units partial product is \(324\times3=972\). 2. The tens digit is \(2\), so the unshifted digit product is \(324\times2=648\). The missing digit is \(4\). 3. Because the \(2\) means \(20\), the rendered row is shifted one place left and represents \(6480\). 4. Add \(972+6480=7452\).

Missing digit: \(4\) Shifted-row 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 starred row 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.

- Read the multiplier directly from the displayed layout and identify the place of the digit associated with the second row. - Distinguish the raw one-digit product from the place value represented by its rendered position. - Verify the final line by combining the values represented by the two rows.

1. The second row comes from the digit \(3\) in the multiplier \(31\). 2. That \(3\) is in the tens place, so it represents \(30\). The raw digit product \(497\times3=1491\) is therefore shifted one place left in the English written algorithm and displayed as \(14{,}910\). 3. The units row is \(497\). Adding the displayed row values gives \(497+14{,}910=15{,}407\), matching the final product.

a) The digit \(3\). b) The row represents multiplication by \(30\), so it is shifted one place left and represents \(14{,}910\). c) \(497+14{,}910=15{,}407\), so the displayed product is verified.

- 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 \(1250\).

- Compare the first and last digits of the products. - Count the \(9\)s in each product. - Compare that count with the number of \(6\)s in the first factor.

1. The products are \(6\times 12=72\), \(66\times 12=792\), \(666\times 12=7992\), and \(6666\times 12=79{,}992\). 2. Each product begins with \(7\) and ends with \(2\). 3. The number of \(9\)s between them is one less than the number of \(6\)s in the first factor.

The products are \(72,792,7992,79{,}992\). Each product begins with \(7\), ends with \(2\), and has one more \(9\) in the middle than the previous product.

- Compare each second factor with \(13\). - Calculate the first product carefully. - Use multiples of the first product to predict the others. - Check the digit pattern in the products.

1. The second factors are \(1,2,3,4,5\) times \(13\). 2. Since \(77\times 13=1001\), the remaining products are \(2,3,4,5\) times \(1001\). 3. The products are \(1001,2002,3003,4004,5005\).

\(77\times 13=1001\) \(77\times 26=2002\) \(77\times 39=3003\) \(77\times 52=4004\) \(77\times 65=5005\) The products are consecutive multiples of \(1001\).

- Determine how the second factor changes. - Calculate the first product. - Relate each later second factor to \(7\). - Use the first product to predict the later products.

1. The first product is \(15{,}873\times 7=111{,}111\). 2. The second factor increases by \(7\), so the next two second factors are \(28\) and \(35\). 3. Because the second factors are \(1,2,3,4,5\) times \(7\), the products are \(1,2,3,4,5\) times \(111{,}111\). 4. The complete pattern ends with \(15{,}873\times 28=444{,}444\) and \(15{,}873\times 35=555{,}555\).

\(15{,}873\times 7=111{,}111\) \(15{,}873\times 14=222{,}222\) \(15{,}873\times 21=333{,}333\) \(15{,}873\times 28=444{,}444\) \(15{,}873\times 35=555{,}555\)

- Calculate the first three products. - Compare the first and last digits of the products. - Count the \(9\)s in the middle. - Use the pattern to predict the last two products.

1. The products are \(81,891,8991,89{,}991,899{,}991\). 2. Every product begins with \(8\) and ends with \(1\). 3. The number of \(9\)s between them is one less than the number of \(9\)s in the first factor.

\(9\times 9=81\) \(99\times 9=891\) \(999\times 9=8991\) \(9999\times 9=89{,}991\) \(99{,}999\times 9=899{,}991\) Each product begins with \(8\), ends with \(1\), and gains one more \(9\) in the middle.

- 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 to the empty truck's mass. - Convert the limit to kilograms and compare.

1. Find the cargo's mass: \(25 \times 110\,\text{kg} = 2750\,\text{kg}\). 2. Find the loaded truck's mass: \(4200\,\text{kg} + 2750\,\text{kg} = 6950\,\text{kg}\). 3. Convert the limit: \(7\) metric tons \(500\,\text{kg} = 7500\,\text{kg}\). 4. Since \(6950\,\text{kg} < 7500\,\text{kg}\), the truck stays within the limit. 5. Find the remaining capacity: \(7500\,\text{kg} - 6950\,\text{kg} = 550\,\text{kg}\).

Yes. The loaded truck has a mass of \(6950\,\text{kg}\), which 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) \(912\)

- Keep each two-digit block together when forming the partner. - Rewrite \(101\) as \(100+1\) if that makes the multiplication easier. - For part c, use two blocks from \(10\) through \(99\). - Compare both results for your example.

1. Part a, ones: \(2+7=9\). Tens: \(1+3=4\). Hundreds: \(7+2=9\). Thousands: \(3+1=4\). Therefore, \(3712+1237=4949\). 2. For part b, \(37+12=49\), and \(49\times 101=49\times(100+1)=4900+49=4949\). 3. The two results are equal. 4. For example, using the blocks \(54\) and \(28\) gives \(5428+2854=8282\), while \((54+28)\times 101=82\times 101=8282\).

a) \(4949\) b) \(4949\); it equals the result from part a). c) Answers will vary. One example is \(5428+2854=8282\) and \((54+28)\times 101=8282\).

- Swap the two-digit blocks, not the individual digits within each block. - Use careful regrouping in the subtraction. - To multiply by \(99\), consider multiplying by \(100\) and subtracting one copy of the number.

1. Swapping the blocks \(86\) and \(34\) gives the partner \(3486\). 2. Use the standard subtraction algorithm. Ones: regroup and calculate \(14-6=8\). Tens: after the regrouping, \(2<8\), so regroup one hundred and calculate \(12-8=4\). Hundreds: \(5-4=1\). Thousands: \(8-3=5\). Thus, \(8634-3486=5148\). 3. The difference between the blocks is \(86-34=52\). 4. The check gives \(52\times 99=52\times 100-52=5200-52=5148\), which matches the subtraction.

a) \(3486\) b) \(5148\) c) \(52\times 99=5148\), so the results agree.

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

- 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 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 determine which one-digit multiplication produced it. - The second multiplier digit is in the tens place, so interpret its row accordingly. - Read the completed product only after identifying the hidden multiplier digit.

1. The units partial-product row is \(654\). Since \(327\times2=654\), the missing units digit of the multiplier is \(2\). The multiplier is \(42\). 2. The second row comes from the tens digit \(4\), which represents \(40\). The unshifted digit product is \(327\times4=1308\), and the English written layout shifts it one place left to represent \(13{,}080\). 3. The displayed final product is \(13{,}734\), so there are \(13{,}734\) tickets.

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. The number of chocolate bars used each day is \(45\times4+45\times8+45\times12=1080\). 2. For \(6\) days, the company needs \(1080\times6=6480\) chocolate bars. 3. The number of jars of jam used each day is \(45\times2+45\times5+45\times8=675\). 4. For \(6\) days, the company needs \(675\times6=4050\) jars of jam.

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

- 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 must be regrouped into the tens column. - In problem 2, identify the multiplier digit responsible for the second partial-product row. - Explain the row shift using the place value of that multiplier digit.

1. In \(8245\times6\), the ones calculation is \(5\times6=30\). Write \(0\) in the ones place and carry \(3\) to the tens column, so the starred carry is \(3\). 2. For \(538\times47\), 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 renderer shifts that row one place left, so it represents \(21{,}520\). 4. Add \(3766+21{,}520=25{,}286\).

1) The starred carry is \(3\), because \(5\times6=30\) writes \(0\) ones and regroups \(3\) tens. 2) The missing digit is \(1\), since \(538\times4=2152\). Because that \(4\) represents \(40\), the rendered row represents \(21{,}520\). 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 \(12{,}501\).

- Use the place of the multiplier digit associated with the fully hidden row. - Reconstruct the raw digit product before accounting for the renderer's row shift. - Check the reconstructed row against the visible final product.

1. The visible units row is \(384\times5=1920\). 2. The tens digit is \(2\), so the hidden unshifted entry is \(384\times2=768\). 3. Because that digit represents \(20\), the English renderer shifts the row one place left, so it represents \(7680\). 4. The rows combine as \(1920+7680=9600\), matching the displayed final product.

The hidden unshifted entry is \(768\), because \(384\times2=768\). Since that \(2\) represents \(20\), the shifted row represents \(7680\). Together with \(384\times5=1920\), the rows give \(1920+7680=9600\).

- Compare the ones digits 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 partial-product rows are \(672\times8=5376\). 2. In the second layout, the tens digit is \(5\). The unshifted digit product is \(672\times5=3360\), so the missing digit is \(6\). The rendered row represents \(33{,}600\). 3. The multipliers differ by \(10\), so the product difference is \(672\times10=6720\). This change occurs entirely in the tens partial-product contribution.

a) The \(5376\) units row is unchanged because both multipliers have ones digit \(8\). b) The missing digit is \(6\), since \(672\times5=3360\). c) The tens digit increases from \(4\) to \(5\), an increase of \(10\) in the multiplier, so the product increases by \(672\times10=6720\).

- Treat the units and tens partial-product rows separately. - For each row, identify the multiplier digit that generated it. - Account for the tens-place shift 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 the unshifted digit product is \(4006\times3=12{,}018\). The second missing digit is \(1\). The rendered row represents \(120{,}180\). 3. Add \(28{,}042+120{,}180=148{,}222\).

The first missing digit is \(8\): the row comes from \(4006\times7=28{,}042\). The second is \(1\): the unshifted tens row comes from \(4006\times3=12{,}018\) and represents multiplication by \(30\) after shifting. The product is \(148{,}222\).

- Calculate each product. - Compare how far each factor is from \(40\). - Subtract each product from \(1600\). - Look for a familiar number pattern in those differences.

1. The products are \(1600,1599,1596,1591,1584\). 2. Compared with \(1600\), the decreases are \(0,1,4,9,16\). 3. The factors move equally far above and below \(40\). When they are \(n\) away from \(40\), the product is \(n\times n\) less than \(1600\).

The products are \(1600,1599,1596,1591,1584\). Their differences from \(1600\) are \(0,1,4,9,16\). As the factors move equally far above and below \(40\), the product decreases by the square of that distance.

- 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 \(\$22{,}305\).

- Determine the two-digit number that must be multiplied by \(101\) to make \(9191\). - Subtract the known block \(50\) from the sum of the two blocks. - Put the two blocks together in their original order. - Check by adding the block-swapped partner.

1. Since \(91\times 101=9191\), the two blocks must satisfy \(a+b=91\). 2. The first block is \(a=50\), so the second block is \(b=91-50=41\). 3. The original number is \(5041\). 4. Check: its partner is \(4150\), and \(5041+4150=9191\).

The original number is \(5041\).

- 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, \(180{,}000\).

- Rewrite \(808\) as \(8\times 101\). - Consider the two-digit form \(08\) for the factor \(1\). - Find \(8\times 12\) before predicting the repeated block. - Verify the prediction by multiplication.

1. The products are \(808\times 2=1616\), \(808\times 4=3232\), and \(808\times 6=4848\). 2. Since \(808=8\times 101\), multiplying by \(n\) gives \((8\times n)\times 101\). For \(n=1\), the block is \(08\), so the repeated-block form would be \(0808\), which is normally written as \(808\) without the leading zero. 3. Since \(8\times 12=96\), the pattern predicts \(9696\). Direct calculation confirms that \(808\times 12=9696\).

1. \(1616,3232,4848\) 2. The repeated block would be \(08\), but the leading zero in \(0808\) is not written. 3. The prediction is \(9696\), and \(808\times 12=9696\).

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

1. For a), \(6 \times 12{,}450=74{,}700\) and \(3 \times 24{,}900=74{,}700\), so the products are equal. 2. For b), \(7 \times 15{,}312=107{,}184\) and \(8 \times 13{,}400=107{,}200\), so the left product is less. 3. For c), \(4 \times 48{,}216=192{,}864\) and \(9 \times 21{,}428=192{,}852\), so the left product is greater.

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

- 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.

- 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