- Count the rows in the rectangle. - Count the columns in each part separately. - Think of each part as rows of equal groups.

1. Each part has \(10\) rows. 2. The wide part has \(10\) columns, giving \(10\times10\). The narrow part has \(2\) columns, giving \(10\times2\). 3. Both expressions start with \(10\) because the two parts have the same number of rows.

Wide part: \(10\times10\). Narrow part: \(10\times2\). Both parts have \(10\) rows.

- How could equal rows help you count without pointing at every square? - What is the same in each row of the rectangle? - Compare the number of rows in your multiplication with Mara's claim.

1. The array has \(10\) rows with \(11\) squares in each row. 2. The total is \(10\times11=110\) squares, which is \(11\) more than Mara's count. 3. Since \(9\times11=99\), Mara might have counted only \(9\) rows and missed one row's worth of squares.

No. There are \(110\) squares: \(10\times11=110\). Mara might have counted only \(9\) rows, since \(9\times11=99\).

- Break \(26\) into tens and ones. - Use the costs for \(20\) sets and \(6\) sets to find the cost for \(26\) sets. - Calculate the easier products first.

1. For a), \(6 \times \$14=\$84\). 2. For b), \(20 \times \$14=\$280\). 3. Since \(26=20+6\), add the two costs: \(\$280+\$84=\$364\).

a) \(6\) sets cost \(\$84\). b) \(20\) sets cost \(\$280\). c) \(26\) sets cost \(\$364\). Add the answers from parts a) and b) because \(26=20+6\).

- Break one factor into tens and ones. - Find the partial products. - Add the partial products using correct place-value alignment.

1. For a), \(23 \times 10=230\) and \(23 \times 2=46\). Then \(230+46=276\). 2. For b), \(31 \times 10=310\) and \(31 \times 4=124\). Then \(310+124=434\). 3. For c), \(15 \times 20=300\) and \(15 \times 2=30\). Then \(300+30=330\).

a) \(276\) b) \(434\) c) \(330\)

- Multiplication represents equal groups or repeated quantities. - Decide whether each story combines equal groups or finds an amount left. - Break \(12\) into \(10 + 2\) to calculate.

1. The expression \(15 \times 12\) represents \(15\) equal groups of \(12\). 2. Lukas’s story describes equal groups, so it matches multiplication. Marie’s story describes subtraction. 3. Use partial products: \(15 \times 12 = 15 \times 10 + 15 \times 2 = 150 + 30 = 180\).

Lukas wrote the matching story. The product is \(180\).

- Multiply the number sold by the price of each car. - Break \(24\) into tens and ones. - Add the partial products with their place values aligned.

1. June: \(35 \times \$24\). Use partial products: \(35 \times \$20=\$700\) and \(35 \times \$4=\$140\). The June revenue is \(\$840\). 2. July: \(53 \times \$24\). Use partial products: \(53 \times \$20=\$1060\) and \(53 \times \$4=\$212\). The July revenue is \(\$1272\).

The June revenue is \(\$840\), and the July revenue is \(\$1272\).

- Use the required decomposition \(16=10+6\). - Find the cost for each part of the decomposition separately. - Add the two partial products to get the total.

1. \(10\times18=180\), so ten books cost \(\$180\). 2. \(6\times18=108\), so six books cost \(\$108\). 3. Add the partial products: \(180+108=288\). 4. The total cost is \(\$288\).

\(10\times18=180\) \(6\times18=108\) Total: \(\$288\)

- Break \(11\) into two numbers that are easy to multiply by. - Start with ten times the number. - Determine what must be added to that product.

1. Use \(n \times 11 = n \times 10 + n\). 2. Part a: \(16 \times 11 = 160 + 16 = 176\). 3. Part b: \(27 \times 11 = 270 + 27 = 297\). 4. Part c: \(35 \times 11 = 350 + 35 = 385\). 5. Part d: \(52 \times 11 = 520 + 52 = 572\).

a) \(176\) b) \(297\) c) \(385\) d) \(572\)

- Split one factor into tens and ones. - Multiply the other factor by each part. - Add the partial products.

1. For a), \(12 \times (10 + 3) = 120 + 36 = 156\). 2. For b), \(16 \times (10 + 2) = 160 + 32 = 192\). 3. For c), \(14 \times (10 + 5) = 140 + 70 = 210\).

a) \(12 \times (10 + 3) = 120 + 36 = 156\) b) \(16 \times (10 + 2) = 160 + 32 = 192\) c) \(14 \times (10 + 5) = 140 + 70 = 210\)

- Read the multiplier by place value, not just as separate digits. - The first row comes from the ones digit; the second comes from the tens digit. - Notice how the written layout shifts the tens partial product one place left.

1. The calculation is \(23\times14\). 2. The ones partial product is \(23\times4=92\). 3. The tens digit \(1\) in \(14\) represents \(10\), so the second partial product represents \(23\times10=230\). 4. \(92+230=322\).

a) \(322\) b) The \(1\) is in the tens place, so that row has a value ten times \(23\).

- Rewrite \(21\) by place value before using the rows. - Match each written row to one term of the distributive equation.

1. The multiplier \(21\) is \(20+1\). 2. Therefore, \(42\times21=(42\times20)+(42\times1)\). 3. The two place-value partial products are \(840\) and \(42\). 4. \(840+42=882\).

\(42\times21=(42\times20)+(42\times1)=840+42=882\).

- Read both digits of the multiplier, including the zero. - A zero digit contributes a zero partial product. - The \(2\) is in the tens place, so its row has tens-place value.

1. The calculation is \(36\times20\). 2. The ones digit is \(0\), so \(36\times0=0\); that is why a real zero row appears. 3. The tens digit \(2\) represents \(20\), so the other row represents \(36\times20=720\). 4. The product is \(720\).

a) It represents \(36\times0=0\). b) \(720\)

- Think of \(23\) as a number of groups. Would two groups and twenty groups give the same total? - What does the place of a digit tell you about its value? - Check whether the two smaller products together count all \(23\) groups.

1. Split \(23\) into \(20+3\). 2. Find both partial products: \(14\times20=280\) and \(14\times3=42\). 3. Add them: \(280+42=322\). The \(2\) in \(23\) means two tens, so that part is \(14\times20\), not \(14\times2\).

\(14\times20=280\); \(14\times3=42\); \(14\times23=322\). The \(2\) is in the tens place.

- Each heading stands for a part of a two-digit number. Which parts belong to the same factor? - To check a box, look at both headings that meet there. - Make sure the total includes every box exactly once.

1. The four boxes are \(10\times10=100\), \(10\times2=20\), \(6\times10=60\), and \(6\times2=12\). 2. Add all four parts: \(100+20+60+12=192\). 3. So \(16\times12=192\).

Top row: \(100\), \(20\). Bottom row: \(60\), \(12\). Product: \(16\times12=192\).

- What value does the \(1\) have in the number \(13\)? - Rewrite \(13\) as a sum of tens and ones. - Multiply \(24\) by each part before adding.

1. In \(13\), the digit \(1\) means one ten, not one one. 2. The two partial products are \(24\times3=72\) and \(24\times10=240\). 3. Add them: \(72+240=312\). The correct product is \(312\).

Jo multiplied by \(1\) instead of \(10\) for the tens digit. \(24\times13=72+240=312\).

- Name the value of the digit \(3\) in \(34\). - Start with the related product \(26\times3\). - Account for the tens place before giving the partial product.

1. The \(3\) in \(34\) represents \(30\), not \(3\). 2. Multiply \(26\times3=78\), then make this ten times as great: \(26\times30=780\).

\(780\). The tens digit represents \(30\), so its partial product is \(26\times30\).

- Imagine covering each part of the rectangle with rows of counters. - What remains the same in the two smaller rectangles? - Add the two areas and check against the whole rectangle.

1. Each of the \(12\) rows has \(10\) squares in one part, so that part has \(12\times10=120\) squares. 2. Each row also has \(1\) extra square, so the other part has \(12\times1=12\) squares. 3. Add the two parts: \(120+12=132\) squares.

\(12\times10=120\) and \(12\times1=12\); \(120+12=132\) squares.

- Read the two digits of the multiplier separately. - The second row comes from a tens digit, not a ones digit. - Add the actual place values of both rows.

1. The ones row is \(46\times8=368\). 2. The second row is the basic product \(46\times2=92\), but its \(2\) is in the tens place. The row is worth \(920\). 3. Add the shifted partial products: \(368+920=1288\).

Second row value: \(920\); complete product: \(46\times28=1288\).

- Which number tells the number of equal groups, and which tells the size of each group? - Can a two-digit number of seats be split into easier parts? - Check that your two products together count every seat.

1. There are \(15\) rows of \(26\) seats, so calculate \(15\times26\). 2. Split \(26\) into \(20+6\). The partial products are \(15\times20=300\) and \(15\times6=90\). 3. Add \(300+90=390\). There are \(390\) seats.

\(15\times20+15\times6=300+90=390\) seats.

- Start by finding the total for one shelf at a time. - How could a multiplication fact involving \(2\) help you work with \(20\) boxes? - Check the final total against the number of boxes on both shelves.

1. The first shelf has \(20\times17=340\) bookmarks. 2. The second shelf has \(8\times17=136\) bookmarks. 3. Together there are \(340+136=476\) bookmarks. The two shelves hold all \(28\) boxes, so adding their products counts every bookmark once.

First shelf: \(20\times17=340\) bookmarks. Second shelf: \(8\times17=136\) bookmarks. Altogether: \(340+136=476\) bookmarks. The two products count all the boxes on both shelves, without counting any box twice.

- Look at how the first factor in the third line is made from the first factors in the two lines above it. - Can you add the two partial products? - For \(7 \times 17\), break \(17\) into tens and ones.

1. For a), \(10 \times 15 = 150\) and \(3 \times 15 = 45\). Since \(13 = 10 + 3\), add the partial products: \(150 + 45 = 195\). 2. For b), \(10 \times 17 = 170\). Also, \(7 \times 17 = 7 \times 10 + 7 \times 7 = 70 + 49 = 119\). Since \(17 = 10 + 7\), add the partial products: \(170 + 119 = 289\).

a) \(10 \times 15 = 150\), \(3 \times 15 = 45\), \(13 \times 15 = 195\) b) \(10 \times 17 = 170\), \(7 \times 17 = 119\), \(17 \times 17 = 289\)

- Break \(12\) and \(24\) into tens and ones. - Use the decomposed parts to label the rows and columns of an area model. - Add all four partial products.

1. Decompose the factors: \(12 = 10 + 2\) and \(24 = 20 + 4\). 2. Find the four partial products in the area model: \(10 \times 20 = 200\), \(10 \times 4 = 40\), \(2 \times 20 = 40\), and \(2 \times 4 = 8\). 3. Add the partial products: \(200 + 40 + 40 + 8 = 288\).

A total of \(288\) people can sit on the benches.

- Reversing the order of the factors uses the commutative property of multiplication. - Break one factor into tens and ones to find partial products. - The original expression and its reversed expression should have the same value.

1. For a), \(24 \times 56 = 24 \times 50 + 24 \times 6 = 1200 + 144 = 1344\). Reversing the factors gives \(56 \times 24 = 56 \times 20 + 56 \times 4 = 1120 + 224 = 1344\). 2. For b), \(37 \times 42 = 37 \times 40 + 37 \times 2 = 1480 + 74 = 1554\). Reversing the factors gives \(42 \times 37 = 42 \times 30 + 42 \times 7 = 1260 + 294 = 1554\). 3. For c), \(58 \times 19 = 58 \times 10 + 58 \times 9 = 580 + 522 = 1102\). Reversing the factors gives \(19 \times 58 = 19 \times 50 + 19 \times 8 = 950 + 152 = 1102\).

a) \(1344\) b) \(1554\) c) \(1102\)

- Calculate each product carefully using partial products. - Reverse the order of the factors to check a result. - Compare all five products.

1. For A, \(28 \times 45 = 28 \times 40 + 28 \times 5 = 1120 + 140 = 1260\). Reversing the factors gives \(45 \times 28 = 45 \times 20 + 45 \times 8 = 900 + 360 = 1260\). 2. For B, \(35 \times 36 = 35 \times 30 + 35 \times 6 = 1050 + 210 = 1260\). Reversing the factors gives \(36 \times 35 = 36 \times 30 + 36 \times 5 = 1080 + 180 = 1260\). 3. For C, \(20 \times 63 = 20 \times 60 + 20 \times 3 = 1200 + 60 = 1260\). Reversing the factors gives \(63 \times 20 = 1260\). 4. For D, \(42 \times 30 = 1260\). Reversing the factors gives \(30 \times 42 = 30 \times 40 + 30 \times 2 = 1200 + 60 = 1260\). 5. For E, \(32 \times 40 = 1280\). Reversing the factors gives \(40 \times 32 = 40 \times 30 + 40 \times 2 = 1200 + 80 = 1280\). 6. Expressions A, B, C, and D equal \(1260\), but expression E equals \(1280\).

E: \(32 \times 40 = 1280\) does not belong because the other four products equal \(1260\).

- Write a multiplication expression for each company. - Compare the factors in the two expressions. - What property says that reversing the factors does not change a product?

1. Green Orchard Produce: \(18\times24=18\times20+18\times4=360+72=432\). 2. Market Lane Produce: \(24\times18=24\times10+24\times8=240+192=432\). 3. The totals are equal because multiplication is commutative: \(18\times24=24\times18\).

a) Each company packs \(432\) apples. b) The totals are equal because the factors are reversed, and multiplication has the commutative property.

- How can you write \(22\) and \(44\) as sums of tens and ones? - Use the distributive property to split a product into easier partial products. - What happens to a product when one factor is doubled?

1. For a), \(24 \times 2 = 48\) and \(24 \times 20 = 480\). 2. For b), use the distributive property: \(24 \times 22 = 24 \times (20+2)=480+48=528\). 3. For c), \(24 \times 44 = 1056\). Because \(44\) is twice \(22\), the product \(1056\) is twice \(528\).

a) \(48\) and \(480\) b) Add the two products: \(480+48=528\), so \(24 \times 22=528\). c) \(24 \times 44=1056\). This is twice \(528\) because \(44\) is twice \(22\).

- Decide which place value has the greatest effect on a two-digit number. - Put the two greatest digits in those places. - Calculate both possible arrangements of the remaining ones digits. - Compare the two products.

1. To make the factors large, place the two greatest digits, \(6\) and \(5\), in the tens places. 2. The remaining arrangements are \(62\times 53\) and \(63\times 52\). 3. Their products are \(3286\) and \(3276\), respectively. 4. Since \(3286>3276\), the greatest product is \(62\times 53=3286\).

\(62\times 53=3286\) gives the greatest product. The digits \(6\) and \(5\) belong in the tens places because those places have the greatest effect on the values of the factors.

- Decide whether the tens digits should be large or small. - Place the two smallest digits in the tens places. - Test both arrangements of the remaining digits. - Compare the products.

1. To make the product small, place the two smallest digits, \(1\) and \(2\), in the tens places. 2. The remaining arrangements are \(14\times 26\) and \(16\times 24\). 3. Their products are \(364\) and \(384\). 4. Therefore, the smallest product is \(14\times 26=364\).

\(14\times 26=364\) gives the smallest product. The reversed equation \(26\times 14=364\) is equivalent.

- Try doubling one factor and halving the other. - Try multiplying one factor by \(4\) and dividing the other by \(4\). - Check each new product.

1. If one factor is divided by a number while the other factor is multiplied by the same number, the product stays unchanged. 2. Halving \(12\) and doubling \(40\) gives \(6\times 80=480\). 3. Doubling \(12\) and halving \(40\) gives \(24\times 20=480\). 4. Multiplying \(12\) by \(4\) and dividing \(40\) by \(4\) gives \(48\times 10=480\).

One possible answer is: \(6\times 80=480\) \(24\times 20=480\) \(48\times 10=480\)

- Compare each new first factor with \(28\). - Compare each new second factor with \(15\). - Decide whether the two factor changes compensate for each other. - Remember that changing both factors in the same direction does not keep the product fixed.

1. In part a, one factor is halved and the other is doubled, so the product stays unchanged. 2. In part b, both factors are doubled, so the product becomes four times as large. 3. In part c, one factor is divided by \(4\) and the other is multiplied by \(4\), so the product stays unchanged. 4. In part d, one factor is multiplied by \(5\) and the other is divided by \(5\), so the product stays unchanged. 5. Therefore, parts a, c, and d have the same value as \(28\times 15\).

a), c), and d) have the same value as \(28\times 15\).

- Find each team's total separately. - Break apart the two-digit factors if helpful. - Compare the products and find their difference.

1. Find the Green Team's total: \(12 \times 25 = 300\). 2. Find the Yellow Team's total: \(14 \times 22 = 308\). 3. Since \(308 > 300\), the Yellow Team plants more. 4. Find the difference: \(308 - 300 = 8\).

The Yellow Team plants \(8\) more carrots than the Green Team.

- Break \(12\) into \(10 + 2\). - Find each partial product. - Add the partial products to find the total.

1. Decompose \(12\) as \(10 + 2\): \(15 \times 12 = 15 \times (10 + 2)\). 2. Find the partial products: \(15 \times 10 = 150\) and \(15 \times 2 = 30\). 3. Add them: \(150 + 30 = 180\). The total cost is \(\$180\).

\(15 \times 12 = 15 \times 10 + 15 \times 2 = 150 + 30 = 180\). She pays \(\$180\).

- Find the total number of each kind of prize. - Find the cost of each kind separately. - Add the two costs.

1. The committee buys \(15 \times 24 = 360\) stuffed animals. 2. The stuffed animals cost \(360 \times \$3 = \$1080\). 3. The committee buys \(5 \times 120 = 600\) colored pencils. 4. The colored pencils cost \(600 \times \$1 = \$600\). 5. The total cost is \(\$1080 + \$600 = \$1680\).

All the prizes cost \(\$1680\).

- Find the total number of each type of book. - Break apart the factors if helpful. - Compare the products and subtract to find the difference.

1. Find the number of nonfiction books: \(12 \times 15 = 180\). 2. Find the number of fiction books: \(8 \times 24 = 192\). 3. Find the difference: \(192 - 180 = 12\).

The library receives \(12\) more fiction books than nonfiction books.

- Find each class's total separately. - Break apart the factors if helpful. - Compare the products and subtract.

1. Find Class A's total: \(12 \times 12 = 144\). 2. Find Class B's total: \(9 \times 15 = 135\). 3. Since \(144 > 135\), Class A prepares more bottles. 4. Find the difference: \(144 - 135 = 9\).

Class A prepares \(9\) more bottles than Class B.

- Multiply the number of packages by the number of notebooks in each package. - For the total cost, multiply the number of packages by the price of one package. - Break \(12\) into \(10 + 2\) if that helps.

1. Find the number of notebooks: \(15 \times 10 = 150\). 2. Find the total cost: \(15 \times \$12 = 15 \times (\$10 + \$2) = \$150 + \$30 = \$180\).

a) \(150\) notebooks b) \(\$180\)

- Find the cost of each product group. - Add the two costs. - Compare the total with the budget and subtract to find the difference.

1. The helmets cost \(45 \times \$38 = \$1710\). 2. The bike locks cost \(62 \times \$24 = \$1488\). 3. The total order costs \(\$1710 + \$1488 = \$3198\). 4. Since \(\$3198 < \$5000\), the budget is enough. 5. The amount left is \(\$5000 - \$3198 = \$1802\).

Yes. The order costs \(\$3198\), and \(\$1802\) remains.

- Calculate each product before comparing. - Compare the exact values. - Recall the meanings of \(<\), \(>\), and \(=\).

1. For a), \(36 \times 25=900\) and \(45 \times 20=900\), so the products are equal. 2. For b), \(58 \times 14=812\) and \(42 \times 19=798\), so the left product is greater.

a) \(=\) b) \(>\)

- Use partial products for each multiplication. - Keep the place values aligned. - Add all three products after calculating them. - Estimate to check whether the total is reasonable.

1. For a), use partial products: \(27 \times 40=1080\) and \(27 \times 3=81\). Then \(1080+81=1161\). 2. For b), use partial products: \(52 \times 10=520\) and \(52 \times 9=468\). Then \(520+468=988\). 3. For c), use partial products: \(36 \times 30=1080\) and \(36 \times 6=216\). Then \(1080+216=1296\). 4. Add the products: \(1161+988+1296=3445\).

a) \(1161\) b) \(988\) c) \(1296\) The total is \(3445\).

- First use the given relationship to find the second factor. - Then multiply the two factors. - This problem requires two multiplication steps.

1. Find the second factor: \(14 \times 6=84\). 2. Multiply the two factors: \(14 \times 84=1176\).

The product is \(1176\).

- Multiply the number of rows by the number of seats in each row. - Compare the total number of seats with \(400\). - If there are not enough seats, subtract to find how many are missing.

1. Find the number of seats: \(24 \times 16\). 2. Use partial products: \(24 \times 10=240\) and \(24 \times 6=144\). 3. Add: \(240+144=384\) seats. 4. Compare: \(384<400\). There are \(400-384=16\) fewer seats than students.

No. There are \(384\) seats, so \(16\) more seats are needed for \(400\) students.

- Find the total number of bottles of each kind. - You can also find the difference per case and multiply it by \(20\). - Compare the two totals.

1. Find the number of apple-juice bottles: \(20 \times 12 = 240\). 2. Find the number of orange-juice bottles: \(20 \times 15 = 300\). 3. Find the difference: \(300 - 240 = 60\).

The school receives \(60\) more bottles of orange juice.

- Break \(12\) into \(10 + 2\). - Break \(22\) into \(20 + 2\). - Add the useful results from part a.

1. For part a, \(2 \times 14 = 28\), \(10 \times 14 = 140\), and \(20 \times 14 = 280\). 2. For part b, combine the amounts for \(10\) boxes and \(2\) boxes: \(140 + 28 = 168\). 3. For part c, combine the amounts for \(20\) boxes and \(2\) boxes: \(280 + 28 = 308\).

a) \(28\), \(140\), and \(280\) flowers b) \(168\) flowers c) \(308\) flowers

- Use the required decomposition \(24=20+4\). - Find the product for each part of the decomposition. - Add the two partial products.

1. \(14\times20=280\). 2. \(14\times4=56\). 3. Add the partial products: \(280+56=336\). 4. The garden has \(336\) seedlings.

\(14\times20=280\) \(14\times4=56\) Total: \(336\) seedlings

- Multiply the number of rows by the seats in each row. - Split \(14\) into tens and ones. - Add the two partial products.

1. Write the product \(17 \times 14\). 2. Decompose \(14\) as \(10 + 4\): \(17 \times 10 + 17 \times 4\). 3. Add the partial products: \(170 + 68 = 238\).

\(17 \times 14 = 17 \times 10 + 17 \times 4 = 170 + 68 = 238\) seats

- Evaluate Lukas’s two partial products. - Evaluate Marie’s two partial products. - Compare how each method decomposes one of the factors.

1. Lukas calculates \(130 + 78 = 208\). 2. Marie calculates \(160 + 48 = 208\). 3. Both methods give the same product because each decomposes one factor and applies the distributive property to \(13 \times 16\).

Yes. Both methods give \(208\) because each method decomposes one factor and applies the distributive property.

- Use place value to multiply by multiples of \(10\). - Find both products in each part. - Compare the final values.

1. For part a, \(15 \times 40 = 600\) and \(12 \times 50 = 600\), so the products are equal. 2. For part b, \(21 \times 30 = 630\) and \(32 \times 20 = 640\), so \(630 < 640\). 3. For part c, \(18 \times 50 = 900\) and \(44 \times 20 = 880\), so \(900 > 880\).

a) \(15 \times 40 = 12 \times 50\) b) \(21 \times 30 < 32 \times 20\) c) \(18 \times 50 > 44 \times 20\)

- Estimate each product before calculating. - Use place value to multiply by multiples of \(10\). - Compare the final products.

1. For part a, \(12 \times 40 = 480\) and \(14 \times 30 = 420\), so \(480 > 420\). 2. For part b, \(25 \times 20 = 500\) and \(50 \times 10 = 500\), so the products are equal. 3. For part c, \(18 \times 30 = 540\) and \(20 \times 27 = 540\), so the products are equal. 4. For part d, \(16 \times 50 = 800\) and \(15 \times 60 = 900\), so \(800 < 900\).

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

- Calculate each product or compare how the factors change from one expression to the other. - Ask whether a change in one factor is balanced by the change in the other factor. - Compare exact products if the factor relationship is not enough to decide.

1. a) \(16\times25=400\) and \(8\times50=400\), so the products are equal. 2. b) \(20\times25=500\) and \(12\times40=480\), so the left product is greater. 3. c) \(32\times25=800\) and \(16\times60=960\), so the left product is less. 4. d) \(44\times25=1100\) and \(22\times50=1100\), so the products are equal.

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

- Each factor is one less than a multiple of \(10\) or \(100\). - Multiply by the nearby round number first. - Subtract one group of the other factor.

1. \(63 \times 9=63 \times 10-63=630-63=567\). 2. \(24 \times 19=24 \times 20-24=480-24=456\). 3. \(15 \times 49=15 \times 50-15=750-15=735\). 4. \(32 \times 99=32 \times 100-32=3200-32=3168\).

a) \(567\) b) \(456\) c) \(735\) d) \(3168\)

- Use the product for \(63 \times 14\) as your starting point. - Increasing a factor by \(1\) adds one group of the other factor.

1. The partial products are \(63 \times 10=630\) and \(63 \times 4=252\). 2. Add: \(630+252=882\). 3. Increasing the second factor from \(14\) to \(15\) adds one more group of \(63\). Therefore, \(882+63=945\).

1. \(630\) and \(252\) 2. \(882\) 3. The product increases by \(63\) to \(945\) because there is one additional group of \(63\).

- Use the known middle product to find the missing bottom factor. - Multiply the two neighboring bottom numbers on the left. - Break one of the two-digit factors into tens and ones to find the top product.

1. The right middle brick is \(21\) and lies above a missing number and \(3\). The missing bottom number is \(7\), since \(7\times3=21\). 2. The left middle brick is \(2\times7=14\). 3. To find the top, multiply \(14\times21\). Split \(21\) into \(20+1\): \(14\times20=280\) and \(14\times1=14\). 4. Add the partial products: \(280+14=294\).

Bottom row: \(2,7,3\) Second row: \(14,21\) Top: \(294\) Top multiplication: \(14\times21=280+14=294\)

- Calculate the two partial products separately. - Add them because the two decomposed parts combine to make the original factor.

1. Decompose \(17\) as \(10+7\). 2. \(48\times10=480\) and \(48\times7=336\). 3. Add the partial products: \(480+336=816\). 4. The decomposition works because \(10+7=17\).

\(48\times17=48\times10+48\times7=480+336=816\). The decomposition works because \(17=10+7\).

- Check each partial product separately. - When multiplying \(36\) by \(4\), account for both the tens and ones.

1. Decomposing \(14\) as \(10+4\) is correct. 2. The second partial product is incorrect: \(36\times4=144\), not \(44\). 3. Add the correct partial products: \(360+144=504\). 4. Therefore, the \(36\) crates hold \(504\) bottles.

The error is \(36\times4=44\). The correct calculation is \(36\times14=360+144=504\), so there are \(504\) bottles.

- Round both factors to convenient multiples of ten for the estimate. - For the exact product, use a decomposition of \(18\) that makes multiplication efficient.

1. Estimate: \(63\approx60\) and \(18\approx20\), so \(60\times20=1200\). A result of \(1134\) is reasonably close to \(1200\). 2. Calculate exactly using \(18=20-2\): \(63\times18=63\times20-63\times2=1260-126=1134\). 3. The exact calculation confirms the stated total.

The estimate is \(1200\), so \(1134\) is reasonable. The exact calculation \(63\times18=1134\) confirms the result.

- Ask what place value the \(3\) has in \(34\). - Distinguish the unshifted basic product from the value represented by its position in the written layout.

1. The digit \(3\) in \(34\) is in the tens place, so it represents \(30\), not \(3\). 2. The basic product \(27\times3=81\) is shifted one place left in the written algorithm, so the row represents \(810\). 3. The ones partial product is \(27\times4=108\). 4. \(810+108=918\), which verifies the product.

The statement is wrong because the \(3\) represents \(30\). The second row represents \(810\), not \(81\). The row values are \(810\) and \(108\), and \(810+108=918\).

- Focus on the multiplier's tens digit. - Compute the basic product before applying the place-value shift. - Distinguish the digits shown in the row from the row's full place-value amount.

1. The calculation is \(46\times23\). 2. The tens digit of the multiplier is \(2\), so the unshifted basic product is \(46\times2=92\). The hidden digit is \(2\). 3. Because this is the tens row, its value in the full multiplication is \(920\). 4. The other row is \(46\times3=138\), and \(138+920=1058\).

The missing digit is \(2\), and the full tens partial product is \(920\).

- The first row comes from the ones digit of the multiplier. - The second row comes from a tens digit, so its rendered place value is ten times its basic product. - Add the two row values, not just the unshifted strings.

1. The ones multiplier digit is \(6\), so the first row is \(58\times6=348\). Its missing digit is \(4\). 2. The second basic row is \(58\), but it comes from the tens digit \(1\), so the rendered shift makes its value \(580\). 3. Add the written row values: \(348+580=928\). 4. Therefore the missing final-product digit is \(2\).

Missing ones-row digit: \(4\), giving \(348\). Second-row value: \(580\). Row addition: \(348+580=928\). Missing final-product digit: \(2\).

- Check each partial product separately before checking the final sum. - Begin with the multiplier's ones digit. - After correcting that row, combine it with the shifted tens row.

1. The calculation is intended to be \(32\times24\). 2. The ones partial product should be \(32\times4=128\), not \(118\). 3. The tens partial product is based on \(32\times2=64\), representing \(640\) after the shift. 4. The correct total is \(128+640=768\).

The first partial product is wrong. It should be \(128\), and the correct product is \(768\).

- Identify the value of the \(2\) in \(27\). - The tens row is not multiplying by \(2\) ones; it is multiplying by \(2\) tens. - Add the place-value amounts represented by both rows.

1. The calculation is \(50\times27\). 2. The ones partial product is \(50\times7=350\). 3. The \(2\) in \(27\) represents \(20\), so \(50\times20=1000\). The written layout shifts the basic product \(100\) one place left. 4. \(350+1000=1350\).

The tens row represents \(1000\), and the product is \(1350\).

- Do not treat the second visible row as only \(67\); use its tens-place shift. - Add the represented values of the two rows before using the estimate.

1. The units multiplier digit \(8\) gives the first partial product \(536\). 2. The tens multiplier digit \(1\) gives a basic partial product \(67\), which the English layout shifts to represent \(670\). 3. Add the represented row values: \(536+670=1206\). 4. The estimate \(70\times20=1400\) is in the same general range.

The first row represents \(67\times8=536\). The second row \(67\) is shifted and represents \(670\). Exact product: \(536+670=1206\). Estimate: \(70\times20=1400\), so \(1206\) is reasonable.

- Separate the multiplier into ones and tens. - Translate the second row from its basic product into its shifted place-value amount. - Add the two row values.

1. The ones row is \(35\times2=70\), so its value is \(70\). 2. The tens row is based on \(35\times4=140\), but the \(4\) represents \(40\), so that row has value \(1400\). 3. \(70+1400=1470\).

The partial-product values are \(70\) and \(1400\), and the product is \(1470\).

- Which one-digit factor of \(24\) gives the second basic partial product? - That factor occupies the tens place. How does that change the row's value? - Add the shifted row to the ones row.

1. The second basic partial product is \(96\). Since \(24\times4=96\), the missing tens digit of the multiplier is \(4\). 2. This \(4\) represents \(40\), so its row is worth \(960\), not \(96\). 3. The ones row is worth \(24\times3=72\). Add \(960+72=1032\). 4. The hidden final-product digit is \(3\).

Hidden multiplier digit: \(4\). Shifted partial product: \(960\). Completed product: \(24\times43=1032\), so the hidden result digit is \(3\).

- Check the ones and tens basic products separately. - The tens row should use the multiplier digit \(3\). - Correct the row before applying its tens-place shift and adding.

1. The calculation is \(47\times32\). 2. The ones partial product is \(47\times2=94\), which is correct. 3. The tens partial product should be based on \(47\times3=141\), not \(131\). Because it is the tens row, it represents \(1410\). 4. \(94+1410=1504\).

The tens basic partial product should be \(141\), and the correct product is \(1504\).

- First find which single digit gives the ones-row product. - Remember that the second row is worth a multiple of ten. - Add the two partial products after identifying both multiplier digits.

1. Since \(24\times4=96\), the multiplier has \(4\) in the ones place. 2. Since \(24\times20=480\), the multiplier has \(2\) in the tens place. 3. The multiplier is \(24\), and the product is \(96+480=576\).

Multiplier: \(24\). Complete product: \(24\times24=576\).

- Multiply the two labeled side lengths of each rectangle. - Match sides that have the same length when assembling the larger rectangle. - The written method splits one factor into its tens and ones.

1. The four areas are a) \(10\times10=100\), b) \(3\times10=30\), c) \(10\times2=20\), and d) \(3\times2=6\). 2. Place the pieces into a larger rectangle: its sides are \(10+2=12\) and \(10+3=13\). Its area is \(100+30+20+6=156\), so \(12\times13=156\). 3. In standard written multiplication, \(12\times3=36\) is the ones row and \(12\times10=120\) is the tens row. Their sum is also \(156\).

Areas: a) \(100\), b) \(30\), c) \(20\), d) \(6\) square units. Whole rectangle: \(12\) by \(13\); \(12\times13=156\) square units. Written rows: \(30+6=36\) for multiplying by \(3\), and \(100+20=120\) for multiplying by \(10\). The two written rows add to \(156\), the same as the four rectangle pieces.

- Calculate the original product first. - Write the changed factors for each part. - Compare each new product with the original product. - Think about how multiplying or dividing a factor changes a product.

1. For a), \(25 \times 12=300\). 2. For b), the new factors are \(25 \times 2=50\) and \(12 \div 2=6\). Their product is \(50 \times 6=300\), so the product is unchanged. 3. For c), the new factors are \(50\) and \(24\). Their product is \(50 \times 24=1200\), which is \(4\) times the original product. Doubling both factors multiplies the product by \(2 \times 2=4\).

a) The product is \(300\). b) The product stays the same: \(300\). c) The product becomes \(4\) times as great: \(1200\).

- Use a multiplication fact to find the price of one bag of sand. - Use the multiplicative comparison to find the soil price per bag. - Find the cost of all soil bags, then add delivery.

1. Since \(40 \times \$4 = \$160\), one bag of sand costs \(\$4\). 2. One bag of specialty soil costs \(12 \times \$4 = \$48\). 3. The specialty soil costs \(35 \times \$48 = \$1680\). 4. With delivery, the total is \(\$1680 + \$45 = \$1725\).

The specialty soil and delivery cost \(\$1725\) altogether.

- Start with the ones digit of the product. - Find digits whose product with \(4\) ends in \(2\). - Test each possible two-digit factor completely.

1. The ones digit of the product is \(2\). The missing ones digit in \(2\square\), when multiplied by \(4\), must therefore produce a product ending in \(2\). The possible digits are \(3\) and \(8\). 2. Test \(3\): \(23 \times 14 = 322\), which has the required form. 3. Test \(8\): \(28 \times 14 = 392\), which does not have the required form. 4. Both missing digits are \(3\).

\(23 \times 14 = 322\)

- Start with the ones partial product because it is not shifted. - Use the final digit \(6\) and the multiplier digit \(4\) to constrain the hidden multiplicand digit. - After recovering the multiplicand, recompute both partial products and the sum.

1. The multiplicand is \(6\square\), and the multiplier is \(24\). 2. The ones partial product comes from multiplying by \(4\). The pattern \(2\square6\) must be \(256\), so the multiplicand is \(64\). 3. Then \(64\times2=128\), so the second basic partial product is \(128\), representing \(1280\) in the tens row. 4. \(256+1280=1536\).

The completed calculation is \(64\times24=1536\), with basic partial products \(256\) and \(128\).

- Use the tens partial product first because its multiplier digit \(2\) is visible. - Recover the multiplicand before solving for the hidden ones digit of the multiplier. - Check both partial products against the completed factors.

1. The tens basic partial product is \(96\), and it comes from multiplying the multiplicand by \(2\). Therefore, the multiplicand is \(48\). 2. The ones basic partial product is \(240\). Since \(48\times5=240\), the hidden ones digit of the multiplier is \(5\). 3. The completed factors are \(48\) and \(25\), and \(48\times25=1200\).

The missing digits are \(4\) and \(5\). The completed calculation is \(48\times25=1200\).

- Check that each split still adds to the original factor. - Each partial product counts a different part of the same rectangular array. - Explain why neither arrangement leaves out or repeats a part.

1. Maya gets \(18\times20=360\) and \(18\times4=72\), totaling \(432\). 2. Eli gets \(10\times24=240\) and \(8\times24=192\), also totaling \(432\). 3. Maya splits the \(24\) columns into \(20\) and \(4\). Eli splits the \(18\) rows into \(10\) and \(8\). Both methods count the same \(18\times24\) array without missing or repeating squares.

Maya: \(360+72=432\). Eli: \(240+192=432\). Both splits count every square in the same \(18\times24\) rectangle, just in different directions.

- Use the ones digit of the product to constrain the missing ones digit in the second factor. - Then separate the contribution from the visible \(7\) in the first factor. - Ask what multiple of \(10\) times the completed second factor supplies the rest of the product.

1. The ones digit of the product is \(1\). Since the first factor ends in \(7\), the second factor must end in \(3\), because \(7\times3\) ends in \(1\). So the second factor is \(23\). 2. The contribution from the ones digit of the first factor is \(7\times23=161\). 3. The remaining contribution is \(851-161=690\). Since \(30\times23=690\), the tens part of the first factor is \(30\). 4. Therefore the first factor is \(37\), and \(37\times23=851\).

\(37\times23=851\)