Free math worksheets

- Use the distributive property to break a factor into easier parts. - When you begin with a factor that is too large, subtract the product for the extra part. - How much greater is \(20\) than \(18\)?

1. Method 1: Use the distributive property. \(10 \times 6 = 60\) and \(8 \times 6 = 48\). Then \(60 + 48 = 108\). 2. Method 2: Use a nearby multiple of ten. \(20 \times 6 = 120\). The extra amount is \(2 \times 6 = 12\), so \(120 - 12 = 108\).

Method 1: \(10 \times 6 + 8 \times 6 = 60 + 48 = 108\) Method 2: \(20 \times 6 - 2 \times 6 = 120 - 12 = 108\)

- What happens if you double a number and then double the result? - Can you break the number into thousands, hundreds, tens, and ones? - Is there a nearby number that makes the multiplication easier?

Four times a number can be found by doubling twice or by breaking the number apart by place value. 1. \(325 \times 4 = 1300\), because \(325 \times 2 = 650\) and \(650 \times 2 = 1300\). 2. \(550 \times 4 = 2200\), because \(500 \times 4 = 2000\) and \(50 \times 4 = 200\). 3. \(825 \times 4 = 3300\), because \(800 \times 4 = 3200\) and \(25 \times 4 = 100\). 4. \(1400 \times 4 = 5600\), because \(14 \times 4 = 56\) and the product is \(100\) times as large. 5. \(2350 \times 4 = 9400\), because \(2000 \times 4 = 8000\) and \(350 \times 4 = 1400\).

a) \(1300\) b) \(2200\) c) \(3300\) d) \(5600\) e) \(9400\)

- How can you find a whole when you know one half? - Which operation reverses halving? - Break each value into parts that are easy to double.

To find the whole number from its half, multiply by \(2\), or double the given value. 1. \(175 \times 2 = 350\). 2. \(430 \times 2 = 860\). 3. \(615 \times 2 = 1230\). 4. \(840 \times 2 = 1680\). 5. \(1350 \times 2 = 2700\).

a) \(350\) b) \(860\) c) \(1230\) d) \(1680\) e) \(2700\)

- Break the three-digit factor into hundreds, tens, and ones. - Use \(7 \times 3\) to help find \(7 \times 300\). - Which partial product is not needed for \(7 \times 304\)?

1. The partial products are \(7 \times 300 = 2100\), \(7 \times 60 = 420\), and \(7 \times 4 = 28\). 2. Add all three partial products: \(2100 + 420 + 28 = 2548\). Therefore, \(7 \times 364 = 2548\). 3. For \(7 \times 304\), use only the hundreds and ones partial products: \(2100 + 28 = 2128\).

1. Partial products: \(2100\), \(420\), \(28\) 2. \(7 \times 364 = 2548\) 3. \(7 \times 304 = 2128\)

- For part a, think of \(9\) as one less than \(10\). - For part b, find two equal addends whose sum is \(8\). - For each part, check that the decomposed factors combine to make the original factor.

1. Part a: Since \(9 = 10 - 1\), \(7 \times 9 = 7 \times 10 - 7 \times 1\). The missing number is \(1\). 2. Part b: Since \(8 = 4 + 4\), \(8 \times 6 = 4 \times 6 + 4 \times 6\). Both missing numbers are \(4\). 3. Part c: Since \(12 = 10 + 2\), \(4 \times 12 = 4 \times 10 + 4 \times 2\). The missing number is \(2\). 4. Part d: Since \(9 = 5 + 4\), \(9 \times 8 = 5 \times 8 + 4 \times 8\). The missing number is \(4\).

a) \(1\) b) \(4\) and \(4\) c) \(2\) d) \(4\)

- Decompose the two-digit factor into tens and ones. - Multiply the one-digit factor by each part. - Add the two partial products.

1. Decompose \(13\) as \(10 + 3\). Then \(7 \times 10 = 70\) and \(7 \times 3 = 21\). Add the partial products: \(70 + 21 = 91\). 2. Decompose \(24\) as \(20 + 4\). Then \(4 \times 20 = 80\) and \(4 \times 4 = 16\). Add: \(80 + 16 = 96\). 3. Decompose \(28\) as \(20 + 8\). Then \(3 \times 20 = 60\) and \(3 \times 8 = 24\). Add: \(60 + 24 = 84\). 4. Decompose \(15\) as \(10 + 5\). Then \(6 \times 10 = 60\) and \(6 \times 5 = 30\). Add: \(60 + 30 = 90\).

a) \(91\) b) \(96\) c) \(84\) d) \(90\)

- Decompose the multi-digit factor into hundreds, tens, and ones. - Multiply each place-value part by the one-digit factor. - Add the partial products.

1. For part a, \(100 \times 4 = 400\) and \(30 \times 4 = 120\). Add the partial products: \(400 + 120 = 520\). 2. For part b, \(200 \times 3 = 600\) and \(10 \times 3 = 30\). Add: \(600 + 30 = 630\). 3. For part c, \(100 \times 2 = 200\), \(20 \times 2 = 40\), and \(5 \times 2 = 10\). Add: \(200 + 40 + 10 = 250\). 4. For part d, \(100 \times 5 = 500\), \(10 \times 5 = 50\), and \(2 \times 5 = 10\). Add: \(500 + 50 + 10 = 560\). 5. For part e, \(200 \times 2 = 400\), \(40 \times 2 = 80\), and \(1 \times 2 = 2\). Add: \(400 + 80 + 2 = 482\).

a) \(520\) b) \(630\) c) \(250\) d) \(560\) e) \(482\)

- Multiply from the ones place toward the left. - Record a carry only when the column total is at least \(10\). - Compare the carry patterns after both products are complete.

1. a) Ones: \(9\times3=27\), so carry \(2\). Tens: \(1\times3+2=5\), so there is no carry. Hundreds: \(7\times3=21\), so carry \(2\). Thousands: \(4\times3+2=14\). The product is \(14{,}157\). The nonzero carries from right to left are \(2,2\). 2. b) Ones: \(6\times8=48\), so carry \(4\). Tens: \(2\times8+4=20\), so carry \(2\). Hundreds: \(8\times8+2=66\). The product is \(6608\). The nonzero carries from right to left are \(4,2\). 3. Part b has a carry after every nonfinal column; part a has no carry from the tens column to the hundreds column.

a) \(14{,}157\); nonzero carries from right to left: \(2,2\). b) \(6608\); nonzero carries from right to left: \(4,2\). Part b has a carry after every nonfinal column.

- Decompose \(124\) into hundreds, tens, and ones. - Multiply each part by \(6\). - Add the partial products.

1. Multiply \(124\) by \(6\): \(124 \times 6\). 2. Decompose \(124\) by place value: \(124 = 100 + 20 + 4\). 3. Multiply each part: \(100 \times 6 = 600\), \(20 \times 6 = 120\), and \(4 \times 6 = 24\). 4. Add the partial products: \(600 + 120 + 24 = 744\).

The baker needs to make \(744\) rolls.

- Decompose each two-digit number into tens and ones. - Multiply each part by the one-digit factor. - Add the partial products.

1. For \(58 \times 4\), decompose \(58\) as \(50 + 8\). Then \(50 \times 4 = 200\) and \(8 \times 4 = 32\). Add: \(200 + 32 = 232\). 2. For \(82 \times 9\), decompose \(82\) as \(80 + 2\). Then \(80 \times 9 = 720\) and \(2 \times 9 = 18\). Add: \(720 + 18 = 738\).

\(232\) and \(738\)

- Multiply the cost of one kit by the number of kits. - Break \(132\) into hundreds, tens, and ones. - Add the partial products.

1. Multiply the price by the number of kits: \(\$132 \times 3\). 2. Use place value: \(\$100 \times 3 = \$300\), \(\$30 \times 3 = \$90\), and \(\$2 \times 3 = \$6\). 3. Add the partial products: \(\$300 + \$90 + \$6 = \$396\).

The total cost is \(\$396\).

- Count how many times each addend appears. - Write each repeated addend as a multiplication expression. - Evaluate the products before adding them.

1. The addend \(14\) appears \(5\) times, and \(22\) appears \(4\) times. 2. Rewrite the expression as \(5 \times 14 + 4 \times 22\). 3. Evaluate the products: \(5 \times 14 = 70\) and \(4 \times 22 = 88\). 4. Add: \(70 + 88 = 158\).

\(5 \times 14 + 4 \times 22 = 158\)

- Equal groups can be represented with multiplication. - Identify the number in each group and the number of groups. - Break \(435\) into hundreds, tens, and ones.

1. Multiply the number of bags by the number of apples in each bag: \(435 \times 8\). 2. Use place value: \(400 \times 8=3200\), \(30 \times 8=240\), and \(5 \times 8=40\). 3. Add: \(3200+240+40=3480\).

They pack \(3480\) apples in all.

- Division is the inverse operation of multiplication. - Dividing your product by \(6\) should give the other factor. - Check each regrouping step in the multiplication algorithm.

1. Multiply by place value. Ones: \(2\times 6=12\), so write \(2\) and regroup \(1\) ten. Tens: \(5\times 6+1=31\), so write \(1\) and regroup \(3\) hundreds. Hundreds: \(1\times 6+3=9\). Thousands: \(4\times 6=24\). Therefore, \(4152\times 6=24{,}912\). 2. Use division to check the product: \(24{,}912\div 6=4152\). 3. The quotient equals the original factor, so the product is correct.

The product is \(24{,}912\). Check: \(24{,}912\div 6=4152\).

- Use division when a factor is missing. - Use place value and partial products for the multiplication. - Check missing factors by multiplication.

1. In a), \(130 \times 5 = 650\). 2. In b), \(720 \div 3 = 240\), so \(240 \times 3 = 720\). 3. In c), \(880 \div 4 = 220\), so \(4 \times 220 = 880\). 4. In d), \(250 \times 4 = 1000\).

a) \(650\) b) \(240\) c) \(220\) d) \(1000\)

- Multiply the number of trays by the number of seedlings in each tray. - Break \(240\) into hundreds and tens. - Add the partial products.

1. Multiply the number of trays by the number of seedlings in each tray: \(4 \times 240\). 2. Decompose \(240\): \(240 = 200 + 40\). 3. Find the partial products: \(4 \times 200 = 800\) and \(4 \times 40 = 160\). 4. Add: \(800 + 160 = 960\).

The gardener has \(960\) seedlings in all.

- Break \(213\) into hundreds, tens, and ones. - Multiply each part by \(3\). - Add the partial products.

1. Multiply the number in each carton by the number of cartons: \(213 \times 3\). 2. Decompose \(213\): \(213 = 200 + 10 + 3\). 3. Find the partial products: \(200 \times 3 = 600\), \(10 \times 3 = 30\), and \(3 \times 3 = 9\). 4. Add: \(600 + 30 + 9 = 639\).

The school receives \(639\) notebooks.

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

1. For a), decompose \(42\) as \(40 + 2\): \(40 \times 6 + 2 \times 6 = 240 + 12 = 252\). 2. For b), decompose \(83\) as \(80 + 3\): \(80 \times 4 + 3 \times 4 = 320 + 12 = 332\).

a) \(42 = 40 + 2\), so \(40 \times 6 + 2 \times 6 = 252\). b) \(83 = 80 + 3\), so \(80 \times 4 + 3 \times 4 = 332\).

- Break each three-digit factor into hundreds, tens, and ones. - Multiply each part by the one-digit factor. - Add the partial products. - Pay special attention to place values containing zero.

1. For part a, \(200 \times 3 = 600\), \(10 \times 3 = 30\), and \(3 \times 3 = 9\). Then \(600 + 30 + 9 = 639\). 2. For part b, \(100 \times 2 = 200\), \(20 \times 2 = 40\), and \(4 \times 2 = 8\). Then \(200 + 40 + 8 = 248\). 3. For part c, \(400 \times 2 = 800\), \(0 \times 2 = 0\), and \(2 \times 2 = 4\). Then \(800 + 0 + 4 = 804\). 4. For part d, \(300 \times 3 = 900\), \(10 \times 3 = 30\), and \(0 \times 3 = 0\). Then \(900 + 30 + 0 = 930\).

a) \(639\) b) \(248\) c) \(804\) d) \(930\)

- Break the two-digit factor into tens and ones. - Multiply each part by the one-digit factor. - Add the two partial products.

1. For part a, \(10 \times 6 = 60\) and \(5 \times 6 = 30\). Then \(60 + 30 = 90\). 2. For part b, \(20 \times 4 = 80\) and \(3 \times 4 = 12\). Then \(80 + 12 = 92\). 3. For part c, \(40 \times 8 = 320\) and \(2 \times 8 = 16\). Then \(320 + 16 = 336\). 4. For part d, \(70 \times 3 = 210\) and \(6 \times 3 = 18\). Then \(210 + 18 = 228\).

a) \(90\) b) \(92\) c) \(336\) d) \(228\)

- Break each three-digit factor into hundreds, tens, and ones. - Multiply each part by the one-digit factor. - Add all three partial products. - Pay attention to zeros in products such as \(50 \times 4\).

1. For part a, \(200 \times 3 = 600\), \(10 \times 3 = 30\), and \(4 \times 3 = 12\). Then \(600 + 30 + 12 = 642\). 2. For part b, \(100 \times 4 = 400\), \(50 \times 4 = 200\), and \(2 \times 4 = 8\). Then \(400 + 200 + 8 = 608\). 3. For part c, \(300 \times 2 = 600\), \(20 \times 2 = 40\), and \(6 \times 2 = 12\). Then \(600 + 40 + 12 = 652\). 4. For part d, \(100 \times 5 = 500\), \(10 \times 5 = 50\), and \(8 \times 5 = 40\). Then \(500 + 50 + 40 = 590\).

a) \(642\) b) \(608\) c) \(652\) d) \(590\)

- Break the two-digit factor into tens and ones. - Multiply each part, then add the partial products. - You may switch the order of the factors when that makes a fact easier.

1. For part a, \(10 \times 7 = 70\) and \(4 \times 7 = 28\). Then \(70 + 28 = 98\). 2. For part b, \(20 \times 4 = 80\) and \(3 \times 4 = 12\). Then \(80 + 12 = 92\). 3. For part c, \(30 \times 2 = 60\) and \(6 \times 2 = 12\). Then \(60 + 12 = 72\). 4. For part d, \(5 \times 10 = 50\) and \(5 \times 8 = 40\). Then \(50 + 40 = 90\). 5. For part e, \(9 \times 10 = 90\) and \(9 \times 2 = 18\). Then \(90 + 18 = 108\).

a) \(98\) b) \(92\) c) \(72\) d) \(90\) e) \(108\)

- Break each three-digit factor into hundreds and ones. - Multiply the hundreds part by the one-digit factor. - Multiply the ones part by the one-digit factor. - Add the two partial products.

1. For part a, \(100 \times 9 = 900\) and \(3 \times 9 = 27\). Then \(900 + 27 = 927\). 2. For part b, \(100 \times 7 = 700\) and \(5 \times 7 = 35\). Then \(700 + 35 = 735\). 3. For part c, \(100 \times 4 = 400\) and \(8 \times 4 = 32\). Then \(400 + 32 = 432\). 4. For part d, \(100 \times 8 = 800\) and \(6 \times 8 = 48\). Then \(800 + 48 = 848\). 5. For part e, \(100 \times 5 = 500\) and \(4 \times 5 = 20\). Then \(500 + 20 = 520\). 6. For part f, \(100 \times 3 = 300\) and \(7 \times 3 = 21\). Then \(300 + 21 = 321\).

a) \(927\) b) \(735\) c) \(432\) d) \(848\) e) \(520\) f) \(321\)

- Break each three-digit factor into hundreds, tens, and ones. - Find a separate partial product for each place-value part. - Add the partial products. - Watch for regrouping when you add the partial products.

1. For part a, \(200 \times 3 = 600\), \(30 \times 3 = 90\), and \(1 \times 3 = 3\). Then \(600 + 90 + 3 = 693\). 2. For part b, \(100 \times 7 = 700\), \(10 \times 7 = 70\), and \(2 \times 7 = 14\). Then \(700 + 70 + 14 = 784\). 3. For part c, \(100 \times 2 = 200\), \(40 \times 2 = 80\), and \(5 \times 2 = 10\). Then \(200 + 80 + 10 = 290\). 4. For part d, \(300 \times 2 = 600\), \(20 \times 2 = 40\), and \(4 \times 2 = 8\). Then \(600 + 40 + 8 = 648\). 5. For part e, \(100 \times 5 = 500\), \(10 \times 5 = 50\), and \(8 \times 5 = 40\). Then \(500 + 50 + 40 = 590\).

a) \(693\) b) \(784\) c) \(290\) d) \(648\) e) \(590\)

- Break the two-digit factor into tens and ones. - Multiply each part by the one-digit factor. - Add the partial products. - Use basic multiplication facts to check the partial products.

1. For part a, \(10 \times 7 = 70\) and \(3 \times 7 = 21\). Then \(70 + 21 = 91\). 2. For part b, \(10 \times 6 = 60\) and \(6 \times 6 = 36\). Then \(60 + 36 = 96\). 3. For part c, \(10 \times 8 = 80\) and \(4 \times 8 = 32\). Then \(80 + 32 = 112\). 4. For part d, \(10 \times 9 = 90\) and \(2 \times 9 = 18\). Then \(90 + 18 = 108\).

a) \(91\) b) \(96\) c) \(112\) d) \(108\)

- Break the two-digit factor into tens and ones. - Find the tens partial product. - Find the ones partial product. - Add the two results.

1. For part a, \(40 \times 4 = 160\) and \(2 \times 4 = 8\). Then \(160 + 8 = 168\). 2. For part b, \(30 \times 3 = 90\) and \(5 \times 3 = 15\). Then \(90 + 15 = 105\). 3. For part c, \(60 \times 5 = 300\) and \(1 \times 5 = 5\). Then \(300 + 5 = 305\). 4. For part d, \(20 \times 8 = 160\) and \(4 \times 8 = 32\). Then \(160 + 32 = 192\).

a) \(168\) b) \(105\) c) \(305\) d) \(192\)

- Use the basic fact \(7 \times 4\) and place value to find \(70 \times 4\). - Find the ones partial product. - Add the two partial products.

1. Multiply the tens: \(70 \times 4 = 280\). 2. Multiply the ones: \(6 \times 4 = 24\). 3. Add the partial products: \(280 + 24 = 304\).

\(70 \times 4 = 280\) \(6 \times 4 = 24\) \(280 + 24 = 304\) The product is \(304\).

- Break each three-digit factor into hundreds, tens, and ones. - Multiply each part by the one-digit factor. - Add the partial products.

1. For part a, \(100 \times 5 = 500\), \(10 \times 5 = 50\), and \(6 \times 5 = 30\). Then \(500 + 50 + 30 = 580\). 2. For part b, \(200 \times 4 = 800\), \(30 \times 4 = 120\), and \(4 \times 4 = 16\). Then \(800 + 120 + 16 = 936\). 3. For part c, \(100 \times 6 = 600\), \(40 \times 6 = 240\), and \(2 \times 6 = 12\). Then \(600 + 240 + 12 = 852\). 4. For part d, \(300 \times 3 = 900\), \(20 \times 3 = 60\), and \(7 \times 3 = 21\). Then \(900 + 60 + 21 = 981\).

a) \(580\) b) \(936\) c) \(852\) d) \(981\)

- Break each factor into hundreds, tens, and ones. - Find each partial product. - Remember that a zero in a place contributes a partial product of zero.

1. For part a, \(200 \times 3 = 600\), \(40 \times 3 = 120\), and \(3 \times 3 = 9\). Then \(600 + 120 + 9 = 729\). 2. For part b, \(100 \times 5 = 500\), \(60 \times 5 = 300\), and \(2 \times 5 = 10\). Then \(500 + 300 + 10 = 810\). 3. For part c, \(300 \times 2 = 600\), \(0 \times 2 = 0\), and \(4 \times 2 = 8\). Then \(600 + 0 + 8 = 608\).

a) \(729\) b) \(810\) c) \(608\)

- Break the two-digit factor into tens and ones. - Multiply each part by the one-digit factor. - Add the partial products.

1. For part a, \(10 \times 4 = 40\) and \(8 \times 4 = 32\). Then \(40 + 32 = 72\). 2. For part b, \(3 \times 20 = 60\) and \(3 \times 6 = 18\). Then \(60 + 18 = 78\). 3. For part c, \(10 \times 7 = 70\) and \(4 \times 7 = 28\). Then \(70 + 28 = 98\). 4. For part d, \(6 \times 10 = 60\) and \(6 \times 5 = 30\). Then \(60 + 30 = 90\). 5. For part e, \(20 \times 2 = 40\) and \(9 \times 2 = 18\). Then \(40 + 18 = 58\). 6. For part f, \(5 \times 10 = 50\) and \(5 \times 7 = 35\). Then \(50 + 35 = 85\).

a) \(72\) b) \(78\) c) \(98\) d) \(90\) e) \(58\) f) \(85\)

- Notice how each two-digit factor is broken into tens and ones. - Find both partial products. - Add the partial products to find the total.

1. For part a, \(30 \times 5 = 150\) and \(4 \times 5 = 20\). Then \(150 + 20 = 170\). 2. For part b, \(6 \times 20 = 120\) and \(6 \times 7 = 42\). Then \(120 + 42 = 162\). 3. For part c, \(4 \times 40 = 160\) and \(4 \times 8 = 32\). Then \(160 + 32 = 192\).

a) \(30 \times 5 = 150\), \(4 \times 5 = 20\), and \(34 \times 5 = 170\) b) \(6 \times 20 = 120\), \(6 \times 7 = 42\), and \(6 \times 27 = 162\) c) \(4 \times 40 = 160\), \(4 \times 8 = 32\), and \(4 \times 48 = 192\)

- Replace multiplication by \(9\) with multiplication by a nearby number. - Determine how much must be subtracted after multiplying by \(10\). - Apply the same strategy to each factor.

1. Use \(n \times 9 = n \times 10 - n\). 2. Part a: \(19 \times 9 = 190 - 19 = 171\). 3. Part b: \(34 \times 9 = 340 - 34 = 306\). 4. Part c: \(48 \times 9 = 480 - 48 = 432\). 5. Part d: \(72 \times 9 = 720 - 72 = 648\).

a) \(171\) b) \(306\) c) \(432\) d) \(648\)

- Look at the first two problems in each group. - Notice how their factors combine to make the factor in the third problem. - Add the related partial products.

1. For parts a and b, \(200 \times 4 = 800\) and \(30 \times 4 = 120\). 2. Since \(230 = 200 + 30\), \(230 \times 4 = 800 + 120 = 920\). 3. For parts d and e, \(300 \times 3 = 900\) and \(7 \times 3 = 21\). 4. Since \(307 = 300 + 7\), \(307 \times 3 = 900 + 21 = 921\).

a) \(800\) b) \(120\) c) \(920\) d) \(900\) e) \(21\) f) \(921\)

- Break each three-digit factor into place-value parts. - Multiply each part by the one-digit factor. - Add all partial products. - When a place has a zero digit, include its zero partial product correctly.

1. For part a, \(200 \times 4 = 800\), \(10 \times 4 = 40\), and \(2 \times 4 = 8\). Then \(800 + 40 + 8 = 848\). 2. For part b, \(100 \times 2 = 200\), \(30 \times 2 = 60\), and \(5 \times 2 = 10\). Then \(200 + 60 + 10 = 270\). 3. For part c, \(300 \times 3 = 900\), \(0 \times 3 = 0\), and \(6 \times 3 = 18\). Then \(900 + 0 + 18 = 918\). 4. For part d, \(100 \times 5 = 500\), \(50 \times 5 = 250\), and \(0 \times 5 = 0\). Then \(500 + 250 + 0 = 750\).

a) \(848\) b) \(270\) c) \(918\) d) \(750\)

- Break the two-digit factor into tens and ones. - Multiply each part by the one-digit factor. - Add the two partial products.

1. For part a, \(40 \times 4 = 160\) and \(3 \times 4 = 12\). Then \(160 + 12 = 172\). 2. For part b, \(60 \times 3 = 180\) and \(8 \times 3 = 24\). Then \(180 + 24 = 204\). 3. For part c, \(50 \times 6 = 300\) and \(2 \times 6 = 12\). Then \(300 + 12 = 312\). 4. For part d, \(90 \times 7 = 630\) and \(1 \times 7 = 7\). Then \(630 + 7 = 637\). 5. For part e, \(30 \times 8 = 240\) and \(5 \times 8 = 40\). Then \(240 + 40 = 280\).

a) \(172\) b) \(204\) c) \(312\) d) \(637\) e) \(280\)

- Break the two-digit factor into tens and ones. - Multiply each part by the one-digit factor. - Add the partial products.

1. For part a, \(10 \times 6 = 60\) and \(4 \times 6 = 24\). Then \(60 + 24 = 84\). 2. For part b, \(10 \times 4 = 40\) and \(7 \times 4 = 28\). Then \(40 + 28 = 68\). 3. For part c, \(20 \times 5 = 100\) and \(3 \times 5 = 15\). Then \(100 + 15 = 115\). 4. For part d, \(30 \times 8 = 240\) and \(2 \times 8 = 16\). Then \(240 + 16 = 256\).

a) \(84\) b) \(68\) c) \(115\) d) \(256\)

- Break each three-digit number into hundreds, tens, and ones. - Multiply each part by \(4\). - Add the partial products.

1. For \(125 \times 4\), the partial products are \(100 \times 4 = 400\), \(20 \times 4 = 80\), and \(5 \times 4 = 20\). Then \(400 + 80 + 20 = 500\). 2. For \(212 \times 4\), the partial products are \(200 \times 4 = 800\), \(10 \times 4 = 40\), and \(2 \times 4 = 8\). Then \(800 + 40 + 8 = 848\). 3. For \(180 \times 4\), the partial products are \(100 \times 4 = 400\) and \(80 \times 4 = 320\). Then \(400 + 320 = 720\).

a) \(500\) b) \(848\) c) \(720\)

- Rewrite \(9\) as \(10 - 1\). - Rewrite \(11\) as \(10 + 1\). - Multiply by \(10\), then adjust by one more group.

1. For a), use \(10 \times 7 - 1 \times 7 = 70 - 7 = 63\). 2. For b), use \(10 \times 6 + 1 \times 6 = 60 + 6 = 66\). 3. For c), use \(10 \times 12 - 1 \times 12 = 120 - 12 = 108\).

a) \(63\) b) \(66\) c) \(108\)

- What happens to a product when the factors switch places? - Break \(15\) into tens and ones. - Multiply each part by \(4\), then add the partial products.

1. By the commutative property, switching the factors does not change the product, so neither product is greater. 2. Decompose \(15\) into \(10 + 5\): \(15 \times 4 = 10 \times 4 + 5 \times 4 = 40 + 20 = 60\). 3. Therefore, both \(15 \times 4\) and \(4 \times 15\) equal \(60\).

a) The products are equal because the factors are switched. b) \(15 \times 4 = 10 \times 4 + 5 \times 4 = 60\), so both products are \(60\).

- First solve the related basic fact without the trailing zero. - Use place value for multiples of \(10\). - Break a three-digit factor into hundreds and tens when helpful. - Check the number of zeros in each factor and product.

1. a) \(40 \times 5 = 200\), using the fact \(4 \times 5 = 20\) and place value. 2. b) \(8 \times 30 = 240\), using the fact \(8 \times 3 = 24\) and place value. 3. c) \(70 \times 4 = 280\). 4. d) \(6 \times 90 = 540\). 5. e) \(120 \times 3 = 360\). 6. f) \(2 \times 450 = 900\). 7. g) \(150 \times 4 = 600\). 8. h) \(5 \times 110 = 550\).

a) \(200\) b) \(240\) c) \(280\) d) \(540\) e) \(360\) f) \(900\) g) \(600\) h) \(550\)

- Relate each equation to a basic multiplication fact. - Use place value to explain factors that are ten times as great. - Use the inverse operation for an unknown factor. - Decompose \(230\) into hundreds and tens.

1. a) \(4 \times 80 = 320\). 2. b) Use the related fact \(6 \times 7 = 42\), so \(60 \times 7 = 420\). 3. c) Use \(3 \times 9 = 27\), so \(30 \times 9 = 270\). 4. d) Decompose \(230\): \(200 \times 4 = 800\) and \(30 \times 4 = 120\). Then \(800 + 120 = 920\). 5. e) Use \(5 \times 9 = 45\), so \(5 \times 90 = 450\). 6. f) Use \(5 \times 7 = 35\), so \(50 \times 7 = 350\).

a) \(320\) b) \(7\) c) \(30\) d) \(920\) e) \(90\) f) \(50\)

- Break each three-digit factor into hundreds, tens, and ones. - Multiply each nonzero part by the one-digit factor. - Add the partial products. - Pay special attention to any zero place value.

1. For part a, \(100 \times 8 = 800\) and \(5 \times 8 = 40\). Then \(800 + 40 = 840\). 2. For part b, \(200 \times 4 = 800\) and \(3 \times 4 = 12\). Then \(800 + 12 = 812\). 3. For part c, \(100 \times 5 = 500\) and \(70 \times 5 = 350\). Then \(500 + 350 = 850\). 4. For part d, \(100 \times 7 = 700\) and \(20 \times 7 = 140\). Then \(700 + 140 = 840\).

a) \(840\) b) \(812\) c) \(850\) d) \(840\)

- Break the two-digit factor into tens and ones. - Multiply the tens part by the one-digit factor. - Multiply the ones part by the one-digit factor. - Add the partial products.

1. For part a, \(50 \times 4 = 200\) and \(7 \times 4 = 28\). Then \(200 + 28 = 228\). 2. For part b, \(6 \times 80 = 480\) and \(6 \times 3 = 18\). Then \(480 + 18 = 498\). 3. For part c, \(30 \times 8 = 240\) and \(9 \times 8 = 72\). Then \(240 + 72 = 312\). 4. For part d, \(7 \times 90 = 630\) and \(7 \times 2 = 14\). Then \(630 + 14 = 644\).

a) \(228\) b) \(498\) c) \(312\) d) \(644\)

- Compare the factors \(4\) and \(8\). - Use each product with \(4\) to help find the corresponding product with \(8\). - Check one pair of products first.

1. Multiplying by \(4\) gives \(60 \times 4 = 240\), \(80 \times 4 = 320\), \(110 \times 4 = 440\), and \(120 \times 4 = 480\). 2. Multiplying by \(8\) gives \(60 \times 8 = 480\), \(80 \times 8 = 640\), \(110 \times 8 = 880\), and \(120 \times 8 = 960\). 3. Since \(8\) is twice \(4\), each product with \(8\) is twice the corresponding product with \(4\).

Products with \(4\): \(240\), \(320\), \(440\), \(480\) Products with \(8\): \(480\), \(640\), \(880\), \(960\) Each product with \(8\) is twice the matching product with \(4\).

- Break each three-digit factor into place-value parts. - Multiply each part by the one-digit factor. - Add the partial products.

1. For part a, \(100 \times 6 = 600\) and \(40 \times 6 = 240\). Then \(600 + 240 = 840\). 2. For part b, \(100 \times 4 = 400\), \(20 \times 4 = 80\), and \(5 \times 4 = 20\). Then \(400 + 80 + 20 = 500\). 3. For part c, \(200 \times 3 = 600\) and \(30 \times 3 = 90\). Then \(600 + 90 = 690\). 4. For part d, \(100 \times 8 = 800\), \(10 \times 8 = 80\), and \(2 \times 8 = 16\). Then \(800 + 80 + 16 = 896\).

a) \(840\) b) \(500\) c) \(690\) d) \(896\)

- The product is the result of multiplication. - Break a factor into hundreds, tens, and ones if helpful. - Find and add the partial products.

1. \(145 \times 4 = 580\). 2. \(86 \times 7 = 602\). 3. \(213 \times 4 = 852\).

a) \(580\) b) \(602\) c) \(852\)

- Break the two-digit factor into tens and ones. - Multiply both parts by the one-digit factor. - Add the partial products. - Check for regrouping when adding the two results.

1. For part a, \(60 \times 4 = 240\) and \(8 \times 4 = 32\). Then \(240 + 32 = 272\). 2. For part b, \(30 \times 9 = 270\) and \(7 \times 9 = 63\). Then \(270 + 63 = 333\). 3. For part c, \(80 \times 6 = 480\) and \(2 \times 6 = 12\). Then \(480 + 12 = 492\). 4. For part d, \(5 \times 90 = 450\) and \(5 \times 4 = 20\). Then \(450 + 20 = 470\).

a) \(272\) b) \(333\) c) \(492\) d) \(470\)

- Break a multi-digit factor into hundreds and tens. - Use the distributive property to multiply each part. - For a multiple of \(10\), use a related basic fact and place value.

1. \(150 \times 3 = (100 \times 3) + (50 \times 3) = 300 + 150 = 450\). 2. \(4 \times 120 = (4 \times 100) + (4 \times 20) = 400 + 80 = 480\). 3. \(230 \times 2 = (200 \times 2) + (30 \times 2) = 400 + 60 = 460\). 4. \(70 \times 6 = (7 \times 6) \times 10 = 42 \times 10 = 420\). 5. \(9 \times 80 = (9 \times 8) \times 10 = 72 \times 10 = 720\).

a) \(450\) b) \(480\) c) \(460\) d) \(420\) e) \(720\)

- Identify the hundreds, tens, and ones in \(136\). - Multiply each place-value part by the one-digit factor. - Add the partial products.

1. For \(136 \times 5\), \(100 \times 5 = 500\), \(30 \times 5 = 150\), and \(6 \times 5 = 30\). Then \(500 + 150 + 30 = 680\). 2. For \(154 \times 4\), \(100 \times 4 = 400\), \(50 \times 4 = 200\), and \(4 \times 4 = 16\). Then \(400 + 200 + 16 = 616\).

1. \(100 \times 5 = 500\), \(30 \times 5 = 150\), \(6 \times 5 = 30\), and the product is \(680\). 2. \(154 \times 4 = 616\)

- Doubling means multiplying by \(2\). - Double the previous result to move one column to the right. - Look for the repeated-doubling pattern across each row. - Check the last column by multiplying the starting number by \(8\).

1. Starting with \(40\), double successively: \(40 \times 2 = 80\), \(80 \times 2 = 160\), and \(160 \times 2 = 320\). 2. Starting with \(60\), double successively: \(60 \times 2 = 120\), \(120 \times 2 = 240\), and \(240 \times 2 = 480\). 3. Starting with \(120\), double successively: \(120 \times 2 = 240\), \(240 \times 2 = 480\), and \(480 \times 2 = 960\).

For \(40\): \(80\), \(160\), \(320\) For \(60\): \(120\), \(240\), \(480\) For \(120\): \(240\), \(480\), \(960\)

- Multiply the number of nickels by \(5\) cents. - Break \(18\) into tens and ones. - Add the partial products.

1. Multiply the number of nickels by the value of each nickel: \(18 \times 5\). 2. Decompose \(18\): \(18 = 10 + 8\). 3. Find the partial products: \(10 \times 5 = 50\) and \(8 \times 5 = 40\). 4. Add: \(50 + 40 = 90\).

The coin bank contains \(90\) cents.

- Find the total number of rolls in the bags. - Decompose a factor by place value if helpful. - Compare the product with \(80\).

1. Find the total number of rolls: \(14 \times 6 = 84\). 2. Since \(84 > 80\), there are enough rolls.

Yes. The bakery has \(84\) rolls, which is enough for the order of \(80\).

- Break the two-digit factor into tens and ones. - Multiply each part by the one-digit factor. - Add the partial products. - Use basic multiplication facts to check your work.

1. For part a, \(40 \times 3 = 120\) and \(7 \times 3 = 21\). Then \(120 + 21 = 141\). 2. For part b, \(50 \times 6 = 300\) and \(2 \times 6 = 12\). Then \(300 + 12 = 312\). 3. For part c, \(70 \times 4 = 280\) and \(4 \times 4 = 16\). Then \(280 + 16 = 296\). 4. For part d, \(80 \times 5 = 400\) and \(3 \times 5 = 15\). Then \(400 + 15 = 415\). 5. For part e, \(90 \times 8 = 720\) and \(1 \times 8 = 8\). Then \(720 + 8 = 728\).

a) \(141\) b) \(312\) c) \(296\) d) \(415\) e) \(728\)

- Identify the hundreds, tens, and ones in \(423\). - Multiply each place-value part by \(6\). - Add the partial products carefully by place value.

1. Decompose \(423\) as \(400 + 20 + 3\). 2. Multiply each part by \(6\): \(400 \times 6 + 20 \times 6 + 3 \times 6\). 3. Add the partial products: \(2400 + 120 + 18 = 2538\).

\(423 \times 6 = (400 + 20 + 3) \times 6 = 400 \times 6 + 20 \times 6 + 3 \times 6 = 2400 + 120 + 18 = 2538\)

- Relate \(197\) to \(200\). - Multiply both parts by the one-digit factor. - Decompose \(305\) into hundreds and ones.

1. Lukas uses the distributive property with subtraction. 2. \(197 \times 3 = 200 \times 3 - 3 \times 3 = 600 - 9 = 591\). 3. Decompose \(305\) as \(300 + 5\): \(305 \times 6 = 300 \times 6 + 5 \times 6 = 1800 + 30 = 1830\).

a) Distributive property b) \(197 \times 3 = 200 \times 3 - 3 \times 3 = 591\) c) \(305 \times 6 = 300 \times 6 + 5 \times 6 = 1830\)

- Use division to find each missing bottom factor. - Multiply the two values in the second row for the top.

1. For the bottom-left value, solve \(x \times 3 = 6\), so \(x = 2\). 2. For the bottom-right value, solve \(3 \times y = 12\), so \(y = 4\). 3. The top is \(6 \times 12 = 72\).

Bottom row: \(2\), \(3\), \(4\) Second row: \(6\), \(12\) Top: \(72\)

- Read each section as rows times columns. - The \(12\) columns are split into \(10\) columns and \(2\) columns. - Add the two partial products.

1. Part a: The left section has \(6\) rows and \(10\) columns, so it represents \(6 \times 10 = 60\). The right section has \(6\) rows and \(2\) columns, so it represents \(6 \times 2 = 12\). 2. Part b: Add the partial products: \(60 + 12 = 72\). 3. Part c: The grid shows \(6 \times (10 + 2) = 6 \times 10 + 6 \times 2\).

a) \(6 \times 10\) and \(6 \times 2\) b) \(60 + 12 = 72\) c) \(6 \times (10 + 2) = 6 \times 10 + 6 \times 2\)

- Read the dimensions of the square array to determine how many tiles there are. - Decompose the number of tiles by place value before multiplying by \(3\).

1. The image is an \(8\times8\) array, so it contains \(64\) tiles. 2. Multiply \(64\times3\). 3. Decompose \(64\): \(60\times3+4\times3=180+12=192\) cents. 4. \(192\) cents is \(\$1.92\).

The mosaic costs \(192\) cents, or \(\$1.92\).

- Use the image to determine the whole rectangle's dimensions and the width of each colored section. - Write one multiplication expression for each colored section. - Combine the two partial products to recover the whole product.

1. The image shows \(7\) rows and \(12\) columns, split into \(10\) columns and \(2\) columns. 2. The partial products are \(7 \times 10 = 70\) and \(7 \times 2 = 14\). 3. Add the adjacent sections: \(70 + 14 = 84\). Therefore, \(7 \times 12 = 84\). 4. The partial products are added because the two sections together make the whole rectangle.

\(7 \times 12 = 7 \times 10 + 7 \times 2 = 70 + 14 = 84\). The two partial products are added because the sections together form the whole array.

- Read the multiplicand, multiplier, and result directly from the written calculation. - Trace where the marked carry comes from in the column immediately to its right. - Describe the carry in place-value language.

1. The written calculation shows \(243\times3=729\). 2. In the tens place, \(4\times3=12\) tens. Write \(2\) tens and regroup \(10\) tens as \(1\) hundred. 3. The small \(1\) records that regrouped hundred.

a) \(729\) b) It represents \(1\) hundred regrouped from \(12\) tens.

- Begin with the ones column. - A zero digit does not erase a regrouped amount from the previous column. - Follow the written calculation one place at a time.

1. The calculation is \(1204\times3\). 2. In the ones place, \(4\times3=12\), so write \(2\) ones and regroup \(1\) ten. 3. The tens digit is \(0\), but the regrouped ten still has to be included: \(0\times3+1=1\) ten. 4. Continuing gives \(1204\times3=3612\).

\(3612\). The \(1\) comes from regrouping \(12\) ones as \(1\) ten and \(2\) ones.

- Start with the rightmost multiplicand digit. - A carry is needed only if a column product is at least \(10\). - Use the four one-column products to build the hidden result.

1. Ones: \(4\times2=8\), so no carry is needed. 2. Tens: \(1\times2=2\), so no carry is needed. 3. Hundreds: \(2\times2=4\), so no carry is needed. 4. Thousands: \(3\times2=6\), so no carry is needed. 5. Therefore the product is \(6428\).

Ones: \(4\times2=8\), no carry. Tens: \(1\times2=2\), no carry. Hundreds: \(2\times2=4\), no carry. Thousands: \(3\times2=6\), no carry. Product: \(6428\)

- How can you combine \(10\), \(5\), and \(2\) to make \(17\), \(7\), or \(13\)? - You may add or subtract the known products. - For \(13\), can you first make a larger number and then subtract a part?

1. For a), write \(17\) as \(10 + 5 + 2\). Then \((10 + 5 + 2) \times 9 = 90 + 45 + 18 = 153\). 2. For b), write \(7\) as \(5 + 2\). Then \((5 + 2) \times 9 = 45 + 18 = 63\). 3. For c), write \(13\) as \(10 + 5 - 2\). Then \((10 + 5 - 2) \times 9 = 90 + 45 - 18 = 117\).

a) \(17 \times 9 = 90 + 45 + 18 = 153\) b) \(7 \times 9 = 45 + 18 = 63\) c) \(13 \times 9 = 90 + 45 - 18 = 117\)

- Products involving multiples of \(10\) or \(100\) may be efficient to find mentally. - For other products, break the multi-digit factor into hundreds, tens, and ones. - Add partial products carefully.

1. a) Use mental math: \(8 \times 4 = 32\), so \(8 \times 400 = 3200\). 2. b) Use place value: \(3 \times 200 = 600\) and \(3 \times 9 = 27\). Then \(600+27=627\). 3. c) Use mental math: \(7 \times 6 = 42\), so \(70 \times 6 = 420\). 4. d) Use place value: \(200 \times 7=1400\), \(40 \times 7=280\), and \(8 \times 7=56\). Then \(1400+280+56=1736\).

a) \(3200\) b) \(627\) c) \(420\) d) \(1736\)

- Break \(48\) into tens and ones for the exact product. - Round \(48\) to a nearby multiple of \(10\) for the estimate. - Make sure your exact answer is close to your estimate.

1. Estimate by rounding \(48\) to \(50\): \(6 \times 50 = 300\). 2. Find the exact product using place value: \(6 \times 40 = 240\) and \(6 \times 8 = 48\). 3. Add the partial products: \(240 + 48 = 288\).

Estimate: about \(300\) cans Exact answer: \(288\) cans

- Break \(135\) into hundreds, tens, and ones. - Multiply each part by \(7\). - Check that your answer is greater than \(7 \times 100\).

1. Break \(135\) into \(100 + 30 + 5\). 2. Find the partial products: \(7 \times 100 = 700\), \(7 \times 30 = 210\), and \(7 \times 5 = 35\). 3. Add the partial products: \(700 + 210 + 35 = 945\).

There are \(945\) marbles in the box.

- Break the two-digit factor into tens and ones for the exact product. - Choose a nearby or compatible number for the estimate. - Check that each exact product is reasonably close to its estimate.

1. For a), estimate with \(20 \times 7 = 140\). For the exact product, \(10 \times 7 = 70\) and \(8 \times 7 = 56\), so \(70 + 56 = 126\). 2. For b), use the compatible number \(25\) to estimate: \(25 \times 4 = 100\). For the exact product, \(20 \times 4 = 80\) and \(6 \times 4 = 24\), so \(80 + 24 = 104\). 3. For c), estimate with \(50 \times 3 = 150\). Since \(49\) is \(1\) less than \(50\), the exact product is \(150 - 3 = 147\).

a) Estimate: \(140\); exact product: \(126\) b) Estimate: \(100\); exact product: \(104\) c) Estimate: \(150\); exact product: \(147\)

- About how many seats would there be with \(10\) rows of \(20\) seats? - Break \(23\) into tens and ones. - Check that the exact answer is close to the estimate.

1. Estimate by using \(10 \times 20 = 200\). 2. For the exact product, break \(23\) into \(20 + 3\): \(9 \times 20 = 180\) and \(9 \times 3 = 27\). 3. Add the partial products: \(180 + 27 = 207\).

Estimate: about \(200\) seats Exact answer: \(207\) seats

- Is there a nearby number that makes a product easier to estimate? - Can you break a factor into tens and ones? - For \(198\), compare the product with the product using \(200\). - Check whether each exact product is reasonably close to its estimate.

1. For a), estimate with \(5 \times 200 = 1000\). For the exact product, subtract the extra \(5 \times 2\): \(1000 - 10 = 990\). 2. For b), estimate with \(20 \times 6 = 120\). For the exact product, \(10 \times 6 = 60\) and \(5 \times 6 = 30\), so \(60 + 30 = 90\). 3. For c), estimate with \(25 \times 10 = 250\). For the exact product, \(20 \times 8 = 160\) and \(5 \times 8 = 40\), so \(160 + 40 = 200\).

a) Estimate: about \(1000\); exact product: \(990\) b) Estimate: about \(120\); exact product: \(90\) c) Estimate: about \(250\); exact product: \(200\)

- Divide each target by one-digit numbers from \(2\) through \(9\). - Keep a factor pair when the quotient is a whole number. - The other factor does not have to be a multiple of \(10\).

1. Test one-digit factors by dividing the target product and checking for whole-number quotients. 2. For \(120\), examples include \(2 \times 60\), \(3 \times 40\), and \(4 \times 30\). 3. For \(180\), examples include \(2 \times 90\), \(3 \times 60\), and \(4 \times 45\). 4. For \(360\), examples include \(3 \times 120\), \(4 \times 90\), and \(8 \times 45\).

a) One possible set is \(2 \times 60\), \(3 \times 40\), \(4 \times 30\). b) One possible set is \(2 \times 90\), \(3 \times 60\), \(4 \times 45\). c) One possible set is \(3 \times 120\), \(4 \times 90\), \(8 \times 45\). Other valid factor pairs are also correct.

- Identify the hundreds, tens, and ones in each factor. - Reuse the given partial products instead of multiplying from the beginning. - Which given parts combine to make \(529\)?

1. For a), combine the hundreds and tens partial products: \(4000 + 160 = 4160\). 2. For b), combine the tens and ones partial products: \(160 + 72 = 232\). 3. For c), add all three partial products: \(4000 + 160 + 72 = 4232\). 4. For d), combine the hundreds and ones partial products: \(4000 + 72 = 4072\).

a) \(4160\) b) \(232\) c) \(4232\) d) \(4072\)

- Check each basic multiplication fact. - Correct the partial product before adding. - Compare \(635\) with \(630\) to identify which partial product is removed.

1. The first two partial products are correct. The error is \(4 \times 5=25\); the correct value is \(4 \times 5=20\). 2. Therefore, \(4 \times 635=2400+120+20=2540\). 3. For \(4 \times 630\), the ones partial product is not included: \(2400+120=2520\).

The incorrect partial product is \(4 \times 5=25\); it should be \(20\). \(4 \times 635=2540\). \(4 \times 630=2520\).

- Calculate both products before comparing them. - Decompose two-digit factors into tens and ones when helpful. - For part d, ask which one-digit factor makes a product of \(70\) with \(14\).

1. a) \(3\times27=81\) and \(4\times19=76\), so \(81>76\). 2. b) \(5\times16=80\) and \(2\times40=80\), so the products are equal. 3. c) \(7\times12=84\) and \(6\times15=90\), so \(84<90\). 4. d) Use multiplication facts: \(14\times5=70\), so the missing factor is \(5\).

a) \(>\) b) \(=\) c) \(<\) d) \(5\)

- Use the given decomposition of each multi-digit factor. - Find each smaller product first. - Add the partial products to find the final product.

1. For Problem 1, the partial products are \(100 \times 4 = 400\), \(40 \times 4 = 160\), and \(2 \times 4 = 8\). Add them: \(400 + 160 + 8 = 568\). 2. For Problem 2, the partial products are \(200 \times 3 = 600\), \(10 \times 3 = 30\), and \(5 \times 3 = 15\). Add them: \(600 + 30 + 15 = 645\).

Problem 1: \(142 \times 4 = 568\); partial products: \(400\), \(160\), and \(8\) Problem 2: \(215 \times 3 = 645\); partial products: \(600\), \(30\), and \(15\)

- Decompose \(138\) into hundreds, tens, and ones. - Multiply \(5\) by each part. - Add the three partial products.

1. Decompose \(138\) as \(100 + 30 + 8\). 2. Multiply each part by \(5\): \(5 \times 100 = 500\), \(5 \times 30 = 150\), and \(5 \times 8 = 40\). 3. Add the partial products: \(500 + 150 + 40 = 690\).

There are \(690\) drawing pads in the shipment.

- Find each group’s total separately. - Use partial products for each multiplication. - Compare the two totals.

1. For the Blue Group, decompose \(265\) as \(200 + 60 + 5\). Then \(3 \times 200 = 600\), \(3 \times 60 = 180\), and \(3 \times 5 = 15\). Add: \(600 + 180 + 15 = 795\). 2. For the Yellow Group, decompose \(198\) as \(100 + 90 + 8\). Then \(4 \times 100 = 400\), \(4 \times 90 = 360\), and \(4 \times 8 = 32\). Add: \(400 + 360 + 32 = 792\). 3. Compare the totals: \(795 > 792\).

The Blue Group collected more acorns: \(795\) acorns compared with \(792\) acorns.

- First find three times \(138\). - Decompose \(138\) into hundreds, tens, and ones. - Compare the product with \(400\).

1. Find the number of seats in the large auditorium by multiplying \(138\) by \(3\). 2. Use partial products: \(100 \times 3 = 300\), \(30 \times 3 = 90\), and \(8 \times 3 = 24\). 3. Add: \(300 + 90 + 24 = 414\). 4. Compare \(414\) with \(400\). Since \(414 > 400\), the large auditorium has enough seats.

a) The large auditorium has \(414\) seats. b) Yes. It has enough seats because \(414 > 400\).

- Identify the common factor in each expression. - Combine the other numbers to make a multiple of \(10\). - Multiply after rewriting the expression.

1. Part a: \((15 + 5) \times 3 = 20 \times 3 = 60\). 2. Part b: \((22 - 2) \times 4 = 20 \times 4 = 80\). 3. Part c: \(6 \times (17 - 7) = 6 \times 10 = 60\). 4. Part d: \((12 + 8) \times 5 = 20 \times 5 = 100\).

a) \(60\) b) \(80\) c) \(60\) d) \(100\)

- Calculate each product. - Find the positive difference between each product and \(3000\). - The smallest difference identifies the closest product.

1. Calculate the products: A is \(428 \times 7=2996\), B is \(596 \times 5=2980\), and C is \(382 \times 8=3056\). 2. Find each distance from \(3000\): \(3000-2996=4\), \(3000-2980=20\), and \(3056-3000=56\). 3. Since \(4<20<56\), product A is closest to \(3000\).

A: \(428 \times 7=2996\) is closest to \(3000\).

- Use inverse multiplication or division by a one-digit factor to recover a missing value. - Estimate the missing value before calculating. - Substitute your digit and verify the multiplication.

1. In part a, \(744\div 3=248\), so the missing digit is \(2\). 2. In part b, test the one-digit factor using multiplication: \(125\times 6=750\), so the missing factor is \(6\). 3. In part c, \(856\div 4=214\), so the missing digit is \(1\).

a) \(2\) b) \(6\) c) \(1\)

- Calculate both products before comparing them. - To find the difference, subtract the smaller value from the greater value.

1. Calculate the first value: \(185 \times 4=740\). 2. Calculate the second value: \(92 \times 8=736\). 3. Compare: \(740>736\). 4. Find the difference: \(740-736=4\).

Four times \(185\) is greater. The difference is \(4\).

- First find the total number of books. - Then multiply the number of books by the price of one book. - Check each multiplication using place value.

1. The shipment contains \(6 \times 115 = 690\) books. 2. The total sales amount is \(690 \times \$9 = \$6210\).

The bookstore will earn \(\$6210\).

- Calculate the left product first. - Then calculate the right product. - Compare the results. - Decompose each two-digit factor into tens and ones.

1. Decompose \(84\): \(7 \times 80 = 560\) and \(7 \times 4 = 28\). Thus, \(7 \times 84 = 588\). 2. Decompose \(98\): \(6 \times 90 = 540\) and \(6 \times 8 = 48\). Thus, \(6 \times 98 = 588\). 3. Since \(588 = 588\), insert \(=\).

\(=\)

- Break each two-digit factor into tens and ones. - Calculate each product separately. - Compare all four results.

1. Expression A: \(3 \times 24 = 3 \times 20 + 3 \times 4 = 60 + 12 = 72\). 2. Expression B: \(4 \times 18 = 4 \times 10 + 4 \times 8 = 40 + 32 = 72\). 3. Expression C: \(6 \times 12 = 6 \times 10 + 6 \times 2 = 60 + 12 = 72\). 4. Expression D: \(8 \times 9 = 72\). 5. All four expressions equal \(72\), so Lina is correct.

Yes. Lina is correct because \(3 \times 24 = 4 \times 18 = 6 \times 12 = 8 \times 9 = 72\).

- Decompose two-digit factors into tens and ones. - Find each product. - Compare all five results.

1. A: \(4 \times 10 = 40\) and \(4 \times 6 = 24\), so \(4 \times 16 = 64\). 2. B: \(2 \times 30 = 60\) and \(2 \times 2 = 4\), so \(2 \times 32 = 64\). 3. C: \(5 \times 10 = 50\) and \(5 \times 3 = 15\), so \(5 \times 13 = 65\). 4. D: \(8 \times 8 = 64\). 5. E: \(3 \times 20 = 60\) and \(3 \times 1 = 3\), so \(3 \times 21 = 63\). 6. Problems A, B, and D all have a product of \(64\).

A) \(4 \times 16 = 64\) B) \(2 \times 32 = 64\) C) \(5 \times 13 = 65\) D) \(8 \times 8 = 64\) E) \(3 \times 21 = 63\) Problems A, B, and D have the same product, \(64\).

- Calculate the product on each side before comparing. - Write down the partial products. - Decompose a number such as \(17\) as \(10 + 7\).

1. For part a, \(4 \times 17 = 40 + 28 = 68\), and \(3 \times 23 = 60 + 9 = 69\). Since \(68 < 69\), insert \(<\). 2. For part b, \(6 \times 14 = 60 + 24 = 84\), and \(7 \times 12 = 70 + 14 = 84\). Since \(84 = 84\), insert \(=\). 3. For part c, \(5 \times 18 = 50 + 40 = 90\), and \(4 \times 22 = 80 + 8 = 88\). Since \(90 > 88\), insert \(>\). 4. For part d, \(2 \times 49 = 80 + 18 = 98\), and \(3 \times 33 = 90 + 9 = 99\). Since \(98 < 99\), insert \(<\).

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

- Find how many sheets one student receives. - You can also find the red and green totals separately. - Both methods should give the same total.

1. Combine the sheets for one student, then multiply: \(5 + 3 = 8\), and \(24 \times 8 = 192\). 2. Find each color separately, then add: \(24 \times 5 = 120\), \(24 \times 3 = 72\), and \(120 + 72 = 192\).

One method is \(24 \times (5 + 3) = 24 \times 8 = 192\). Another method is \(24 \times 5 + 24 \times 3 = 120 + 72 = 192\). The class receives \(192\) sheets altogether.

- Decompose each multi-digit factor into hundreds, tens, and ones. - Add the partial products. - Compare the three final products.

1. For part a, \(200 \times 4 = 800\), \(10 \times 4 = 40\), and \(4 \times 4 = 16\). Add: \(800 + 40 + 16 = 856\). 2. For part b, \(300 \times 3 = 900\), \(20 \times 3 = 60\), and \(1 \times 3 = 3\). Add: \(900 + 60 + 3 = 963\). 3. For part c, \(400 \times 2 = 800\) and \(50 \times 2 = 100\). Add: \(800 + 100 = 900\). 4. Compare: \(963 > 900 > 856\). The greatest product is \(963\).

a) \(856\) b) \(963\) c) \(900\) Part b has the greatest product: \(321 \times 3 = 963\).

- Break \(123\) into hundreds, tens, and ones. - Use the given partial products to find the missing factors. - Add the three partial products.

1. The hundreds part must be \(100\), because \(100 \times 4 = 400\). 2. Multiply the tens: \(20 \times 4 = 80\). 3. The ones part must be \(3\), because \(3 \times 4 = 12\). 4. Add the partial products: \(400 + 80 + 12 = 492\).

\(123 \times 4 = 492\) \(100 \times 4 = 400\) \(20 \times 4 = 80\) \(3 \times 4 = 12\)

- Find each product before comparing. - You can break each two-digit factor into tens and ones. - Compare the two results in each part.

1. For part a, \(12 \times 8 = 96\) and \(16 \times 6 = 96\), so the products are equal. 2. For part b, \(24 \times 3 = 72\) and \(18 \times 4 = 72\), so the products are equal. 3. For part c, \(15 \times 7 = 105\) and \(13 \times 8 = 104\), so \(105 > 104\).

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

- Notice how each two-digit number is broken into tens and ones. - Find the partial products, then add them. - In part b, combine the two partial factors to find the original factor.

1. For part a, multiply the tens: \(40 \times 6 = 240\). 2. Multiply the ones: \(7 \times 6 = 42\). 3. Add the partial products: \(240 + 42 = 282\), so \(47 \times 6 = 282\). 4. For part b, the partial factors are \(30\) and \(6\), so the first factor is \(30 + 6 = 36\). 5. The equation is \(36 \times 8 = 288\).

a) \(47 \times 6 = 282\) \(40 \times 6 = 240\) \(7 \times 6 = 42\) \(240 + 42 = 282\) b) \(36 \times 8 = 288\)

- Calculate both products in each comparison. - Compare the products from the greatest place value to the least. - Estimate first to check whether each exact result is reasonable.

1. For a), \(178 \times 4=712\) and \(142 \times 5=710\), so \(712>710\). 2. For b), \(235 \times 3=705\) and \(118 \times 6=708\), so \(705<708\). 3. For c), \(156 \times 7=1092\) and \(182 \times 6=1092\), so the products are equal.

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

- Think about what \(4 \times 215\) means as repeated addition. - Compare the digits in \(215\) and \(251\) carefully. - Break \(215\) into hundreds, tens, and ones.

1. The expression \(4 \times 215\) means four groups of \(215\), so Method A represents the multiplication. 2. Use partial products: \(4 \times 200 = 800\), \(4 \times 10 = 40\), and \(4 \times 5 = 20\). 3. Add: \(800 + 40 + 20 = 860\).

Method A represents the multiplication. The product is \(860\).

- Write \(124\) once for each of the \(7\) boxes. - Find \(7 \times 100\), \(7 \times 20\), and \(7 \times 4\). - Add the partial products.

1. Seven boxes of \(124\) bulbs can be written as \(124 + 124 + 124 + 124 + 124 + 124 + 124\). 2. Find the partial products: \(7 \times 100 = 700\), \(7 \times 20 = 140\), and \(7 \times 4 = 28\). 3. Add the partial products: \(700 + 140 + 28 = 868\).

a) \(124 + 124 + 124 + 124 + 124 + 124 + 124\) b) There are \(868\) tulip bulbs in all.

- Break the three-digit number into hundreds, tens, and ones. - Multiply each part by \(4\). - Add the three partial products.

1. Break apart \(218\) as \(200 + 10 + 8\). 2. Find the partial products: \(200 \times 4 = 800\), \(10 \times 4 = 40\), and \(8 \times 4 = 32\). 3. Add: \(800 + 40 + 32 = 872\).

There are \(872\) bags of flour on the shelves in all.

- Count the school days from Monday through Friday. - Decide how many times the daily distance must be counted. - Break \(124\) into hundreds, tens, and ones, then multiply and add. - Write the final answer in miles.

1. There are \(5\) school days from Monday through Friday. 2. Break apart \(124\) as \(100 + 20 + 4\). 3. Find the partial products: \(100 \times 5 = 500\), \(20 \times 5 = 100\), and \(4 \times 5 = 20\). 4. Add: \(500 + 100 + 20 = 620\).

The bus travels \(620\) miles during the school week.

- Organize or mark matching addends before counting them. - Write one product for each different addend. - Check that every original addend is represented exactly once.

1. The addend \(9\) appears \(4\) times, \(13\) appears \(3\) times, and \(11\) appears \(2\) times. 2. Rewrite the expression as \(4 \times 9 + 3 \times 13 + 2 \times 11\). 3. Evaluate: \(4 \times 9 = 36\), \(3 \times 13 = 39\), and \(2 \times 11 = 22\). 4. Add the products: \(36 + 39 + 22 = 97\).

\(4 \times 9 + 3 \times 13 + 2 \times 11 = 97\)

- Dividing a product by one factor should give the other factor. - A related division equation reverses the multiplication. - A remainder or a different quotient shows that the stated product is incorrect.

1. Part a: \(236\times 4=944\). The related check is \(944\div 4=236\). 2. Part b: \(1116\div 7=159\text{ R }3\), not \(158\), so Emma’s product is incorrect. 3. The correct product is \(158\times 7=1106\). 4. Check: \(1106\div 7=158\).

a) \(236\times 4=944\); check: \(944\div 4=236\) b) \(1116\div 7=159\text{ R }3\), so the stated product is incorrect. The correct product is \(1106\).

- Find the product on each side. - Use place value and a related basic fact when a factor is a multiple of \(10\). - Compare the two products.

1. For part a, \(160 \times 4 = 640\) and \(210 \times 3 = 630\), so \(640 > 630\). 2. For part b, \(150 \times 6 = 900\) and \(300 \times 3 = 900\), so the products are equal. 3. For part c, \(120 \times 8 = 960\) and \(240 \times 4 = 960\), so the products are equal.

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

- Break \(245\) into hundreds, tens, and ones. - Multiply each part by \(3\). - Add the partial products.

1. Break apart \(245\) as \(200 + 40 + 5\). 2. Find the partial products: \(200 \times 3 = 600\), \(40 \times 3 = 120\), and \(5 \times 3 = 15\). 3. Add: \(600 + 120 + 15 = 735\).

The new bookcases can hold \(735\) books in all.

- Identify which place-value part of \(148\) has not been used. - Multiply that part by \(6\). - Add all three partial products.

1. The ones part of \(148\) has not been multiplied, so the missing partial product is \(8 \times 6 = 48\). 2. Add all the partial products: \(600 + 240 + 48 = 888\).

The missing partial product is \(8 \times 6 = 48\). Lara has \(888\) beads in all.

- Break \(136\) into hundreds, tens, and ones. - Multiply each part by \(4\). - Add the partial products.

1. Break apart \(136\) as \(100 + 30 + 6\). 2. Find the partial products: \(4 \times 100 = 400\), \(4 \times 30 = 120\), and \(4 \times 6 = 24\). 3. Add: \(400 + 120 + 24 = 544\).

The teacher buys \(544\) notebooks in all.

- Break each larger factor into place-value parts. - Use basic multiplication facts for each part. - Record each partial product. - Add the partial products for each multiplication problem.

1. For \(138 \times 4\), find the partial products: \(100 \times 4 = 400\), \(30 \times 4 = 120\), and \(8 \times 4 = 32\). 2. Add: \(400 + 120 + 32 = 552\). 3. For \(92 \times 8\), find the partial products: \(90 \times 8 = 720\) and \(2 \times 8 = 16\). 4. Add: \(720 + 16 = 736\).

\(138 \times 4 = 552\) \(92 \times 8 = 736\)

- Identify the next multiple of \(10\). - Multiply using the multiple of \(10\), then subtract the extra part. - Apply the one-digit factor to both terms.

1. For a), write \(59\) as \(60 - 1\): \((60 - 1) \times 3 = 60 \times 3 - 1 \times 3 = 180 - 3 = 177\). 2. For b), write \(38\) as \(40 - 2\): \((40 - 2) \times 7 = 40 \times 7 - 2 \times 7 = 280 - 14 = 266\).

a) \(177\) b) \(266\)

- Decompose a factor using \(100\) or \(200\). - Decide whether a sum or a difference is more convenient. - Multiply each part, then combine the partial products.

1. For a), decompose \(104\) as \(100 + 4\): \(100 \times 8 + 4 \times 8 = 800 + 32 = 832\). 2. For b), decompose \(198\) as \(200 - 2\): \(9 \times 200 - 9 \times 2 = 1800 - 18 = 1782\).

a) \(104 = 100 + 4\), so \(100 \times 8 + 4 \times 8 = 832\). b) \(198 = 200 - 2\), so \(9 \times 200 - 9 \times 2 = 1782\).

- First find the number of pencils in the hundreds packs and the tens packs. - Add those amounts to find the number in one large box. - Multiply the result from part a by \(4\) for part b.

1. For part a, \(2 \times 100 = 200\) and \(4 \times 10 = 40\). Then \(200 + 40 = 240\). 2. For part b, multiply \(240\) by \(4\). The partial products are \(200 \times 4 = 800\) and \(40 \times 4 = 160\). 3. Add: \(800 + 160 = 960\).

a) One large box contains \(240\) pencils. b) Four large boxes contain \(960\) pencils.

- Find the product on the left side of each comparison. - Find the product on the right side. - Look for relationships such as doubling one factor while halving the other. - Compare the final products.

1. For part a, \(120 \times 4 = 480\) and \(240 \times 2 = 480\), so the products are equal. 2. For part b, \(303 \times 3 = 909\) and \(450 \times 2 = 900\), so \(909 > 900\). 3. For part c, \(112 \times 4 = 448\) and \(221 \times 2 = 442\), so \(448 > 442\). 4. For part d, \(205 \times 4 = 820\) and \(410 \times 2 = 820\), so the products are equal.

a) \(120 \times 4 = 240 \times 2\) b) \(303 \times 3 > 450 \times 2\) c) \(112 \times 4 > 221 \times 2\) d) \(205 \times 4 = 410 \times 2\)

- Find the product on each side. - Use partial products when needed. - Compare the two results in each part.

1. For part a, \(14 \times 8 = 112\) and \(16 \times 7 = 112\), so the products are equal. 2. For part b, \(22 \times 6 = 132\) and \(33 \times 4 = 132\), so the products are equal. 3. For part c, \(45 \times 4 = 180\) and \(28 \times 6 = 168\), so \(180 > 168\). 4. For part d, \(19 \times 9 = 171\) and \(25 \times 7 = 175\), so \(171 < 175\).

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

- Find the product on the left side. - Find the product on the right side. - Look for factor relationships that may help, especially in part b. - Compare the final products.

1. For part a, \(124 \times 3 = 372\) and \(185 \times 2 = 370\), so \(372 > 370\). 2. For part b, \(215 \times 4 = 860\) and \(430 \times 2 = 860\), so the products are equal. 3. For part c, \(132 \times 6 = 792\) and \(265 \times 3 = 795\), so \(792 < 795\).

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

- Break \(112\) into hundreds, tens, and ones. - Multiply each part by \(8\). - Add the partial products.

1. Break apart \(112\) as \(100 + 10 + 2\). 2. Find the partial products: \(100 \times 8 = 800\), \(10 \times 8 = 80\), and \(2 \times 8 = 16\). 3. Add: \(800 + 80 + 16 = 896\).

The garden center delivers \(896\) seedlings in all.

- Count how many times \(214\) appears in the sum. - Break \(214\) into hundreds, tens, and ones for the multiplication. - Compare the two final results.

1. Add step by step: \(214 + 214 = 428\), \(428 + 214 = 642\), and \(642 + 214 = 856\). 2. Since \(214\) is added four times, the related multiplication equation is \(4 \times 214\). 3. Use partial products: \(4 \times 200 = 800\), \(4 \times 10 = 40\), and \(4 \times 4 = 16\). Then \(800 + 40 + 16 = 856\). 4. Both methods give the same result.

a) \(856\) b) \(4 \times 214 = 856\) c) The results are equal.

- Find the value of the repeated addition and the multiplication separately. - Compare the two values. - Think about what the factor \(5\) tells you about the number of equal addends.

1. Add the five equal addends: \(136 + 136 = 272\), \(272 + 136 = 408\), \(408 + 136 = 544\), and \(544 + 136 = 680\). 2. Use partial products for \(5 \times 136\): \(5 \times 100 = 500\), \(5 \times 30 = 150\), and \(5 \times 6 = 30\). Then \(500 + 150 + 30 = 680\). 3. The statement is true because multiplication represents repeated addition of equal groups.

The statement is true. Both expressions equal \(680\). The multiplication represents adding \(136\) five times.

- Notice how each two-digit factor is broken into tens and ones. - Use the same one-digit factor in both partial products. - Add the partial products to find the total.

1. For part a, break \(17\) into \(10 + 7\). Then \(10 \times 4 = 40\), \(7 \times 4 = 28\), and \(40 + 28 = 68\). 2. For part b, break \(25\) into \(20 + 5\). Then \(20 \times 3 = 60\), \(5 \times 3 = 15\), and \(60 + 15 = 75\). 3. For part c, break \(13\) into \(10 + 3\). Then \(10 \times 6 = 60\), \(3 \times 6 = 18\), and \(60 + 18 = 78\).

a) \(17 \times 4 = (10 \times 4) + (7 \times 4) = 40 + 28 = 68\) b) \(25 \times 3 = (20 \times 3) + (5 \times 3) = 60 + 15 = 75\) c) \(13 \times 6 = 10 \times 6 + 3 \times 6 = 60 + 18 = 78\)

- Find the product on the left side of each comparison. - Find the product on the right side. - Compare the two results. - In part c, check the ones digits carefully.

1. For part a, \(128 \times 4 = 512\) and \(256 \times 2 = 512\), so the products are equal. 2. For part b, \(115 \times 7 = 805\) and \(132 \times 6 = 792\), so \(805 > 792\). 3. For part c, \(219 \times 3 = 657\) and \(164 \times 4 = 656\), so \(657 > 656\).

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

- 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 factor shown in each multiplication problem. - Break each three-digit number into place-value parts if helpful. - Add the partial products carefully.

1. For a), \(145 \times 3 = 435\). 2. For b), \(232 \times 3 = 696\). 3. For c), \(118 \times 7 = 826\). 4. For d), \(106 \times 7 = 742\).

a) \(435\) b) \(696\) c) \(826\) d) \(742\)

- Estimate by comparing the hundreds before calculating. - Find both products in each part. - Break each three-digit factor into hundreds and ones if helpful.

1. For part a, \(105 \times 4 = 420\) and \(104 \times 5 = 520\), so \(420 < 520\). 2. For part b, \(108 \times 3 = 324\) and \(103 \times 8 = 824\), so \(324 < 824\). 3. For part c, \(102 \times 9 = 918\) and \(109 \times 2 = 218\), so \(918 > 218\).

a) \(105 \times 4 < 104 \times 5\) b) \(108 \times 3 < 103 \times 8\) c) \(102 \times 9 > 109 \times 2\)

- Identify how each two-digit factor is broken into tens and ones. - Use the given partial product to work backward when a factor is missing. - Add the partial products to find the total.

1. For part a, \(50 \times 5 = 250\) and \(7 \times 5 = 35\). Then \(250 + 35 = 285\). 2. For part b, the missing partial factors are \(80\) and \(4\), because \(80 \times 4 = 320\) and \(4 \times 4 = 16\). Then \(320 + 16 = 336\). 3. For part c, both missing factors are \(9\), because \(20 \times 9 = 180\) and \(9 \times 9 = 81\). Then \(180 + 81 = 261\).

a) \(57 \times 5 = 285\), \(50 \times 5 = 250\), and \(7 \times 5 = 35\) b) \(84 \times 4 = 336\), \(80 \times 4 = 320\), and \(4 \times 4 = 16\) c) \(29 \times 9 = 261\), \(20 \times 9 = 180\), and \(9 \times 9 = 81\)

- Find the product on the left and the product on the right. - Compare the two values. - Choose the symbol that makes each statement true.

1. For part a, \(18 \times 7 = 126\) and \(15 \times 8 = 120\), so \(126 > 120\). 2. For part b, \(24 \times 4 = 96\) and \(12 \times 8 = 96\), so the products are equal. 3. For part c, \(13 \times 9 = 117\) and \(16 \times 7 = 112\), so \(117 > 112\).

a) \(18 \times 7 > 15 \times 8\) b) \(24 \times 4 = 12 \times 8\) c) \(13 \times 9 > 16 \times 7\)

- Find each product separately. - Break the two-digit factors into tens and ones if helpful. - Compare the final products.

1. For part a, \(46 \times 4 = 184\) and \(62 \times 3 = 186\), so \(184 < 186\). 2. For part b, \(38 \times 5 = 190\) and \(94 \times 2 = 188\), so \(190 > 188\).

a) \(46 \times 4 < 62 \times 3\) b) \(38 \times 5 > 94 \times 2\)

- Identify the tens and ones parts of each two-digit factor. - Multiply both parts by the one-digit factor. - Add the partial products.

1. For part a, \(50 \times 6 = 300\) and \(4 \times 6 = 24\). Then \(300 + 24 = 324\). 2. For part b, \(80 \times 3 = 240\) and \(7 \times 3 = 21\). Then \(240 + 21 = 261\). 3. For part c, \(60 \times 4 = 240\) and \(9 \times 4 = 36\). Then \(240 + 36 = 276\).

a) \(50 \times 6 = 300\), \(4 \times 6 = 24\), and \(54 \times 6 = 324\) b) \(80 \times 3 = 240\), \(7 \times 3 = 21\), and \(87 \times 3 = 261\) c) \(60 \times 4 = 240\), \(9 \times 4 = 36\), and \(69 \times 4 = 276\)

- In part a), multiply each number by \(6\). - In part b), ask which factor makes \(120 \times \square = 600\). - Use a related basic fact, such as \(12 \times \square = 60\), if helpful. - Use the same factor for the other products in part b).

1. For a), \(110 \times 6 = 660\), \(145 \times 6 = 870\), and \(160 \times 6 = 960\). 2. For b), the missing factor is \(5\), because \(120 \times 5 = 600\). 3. Using the same factor, \(140 \times 5 = 700\) and \(180 \times 5 = 900\).

a) \(660\), \(870\), \(960\) b) Factor: \(5\); products: \(700\), \(900\)

- 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

- Decompose the two-digit factor into tens and ones. - Pay attention to whether the decomposition uses addition or subtraction. - Multiply each part, then combine the partial products.

1. For a), \(14 \times 6 = 10 \times 6 + 4 \times 6 = 60 + 24 = 84\). 2. For b), \(18 \times 4 = 20 \times 4 - 2 \times 4 = 80 - 8 = 72\). 3. For c), \(23 \times 3 = 20 \times 3 + 3 \times 3 = 60 + 9 = 69\).

a) \(10 \times 6 + 4 \times 6 = 60 + 24 = 84\) b) \(20 \times 4 - 2 \times 4 = 80 - 8 = 72\) c) \(20 \times 3 + 3 \times 3 = 60 + 9 = 69\)

- Count how many times \(156\) appears in the sum. - Break \(156\) into hundreds, tens, and ones. - Find and add the partial products.

1. Add the four equal addends: \(156 + 156 + 156 + 156 = 624\). 2. Since \(156\) is added four times, the multiplication expression is \(4 \times 156\). 3. Find the partial products: \(4 \times 100 = 400\), \(4 \times 50 = 200\), and \(4 \times 6 = 24\). 4. Add: \(400 + 200 + 24 = 624\).

a) \(624\) b) \(4 \times 156\) c) \(400 + 200 + 24 = 624\)

- The factor \(8\) tells how many equal addends to write. - Break \(125\) into hundreds, tens, and ones. - Add the three partial products carefully.

1. The repeated addition is \(125 + 125 + 125 + 125 + 125 + 125 + 125 + 125\). 2. The partial products are \(8 \times 100 = 800\), \(8 \times 20 = 160\), and \(8 \times 5 = 40\). 3. Add: \(800 + 160 + 40 = 1000\).

a) \(125 + 125 + 125 + 125 + 125 + 125 + 125 + 125\) b) \(800 + 160 + 40 = 1000\)

- The factor \(6\) tells how many equal addends to write. - Break \(132\) into hundreds, tens, and ones. - Multiply each part by \(6\), then add.

1. The repeated addition is \(132 + 132 + 132 + 132 + 132 + 132\). 2. Break apart \(132\) as \(100 + 30 + 2\). 3. Find the partial products: \(6 \times 100 = 600\), \(6 \times 30 = 180\), and \(6 \times 2 = 12\). 4. Add: \(600 + 180 + 12 = 792\).

a) \(132 + 132 + 132 + 132 + 132 + 132\) b) \(6 \times 132 = 792\)

- Break each three-digit factor into hundreds and tens. - Find and add the partial products. - Compare the three final products.

1. For A, \(100 \times 6 = 600\) and \(40 \times 6 = 240\). Then \(600 + 240 = 840\). 2. For B, \(100 \times 4 = 400\) and \(90 \times 4 = 360\). Then \(400 + 360 = 760\). 3. For C, \(200 \times 3 = 600\) and \(80 \times 3 = 240\). Then \(600 + 240 = 840\). 4. Expressions A and C have the same value.

A and C have the same value, \(840\).

- Check each line of Leo’s work separately. - Verify both basic multiplication facts. - Add the corrected partial products.

1. The first partial product, \(40 \times 8 = 320\), is correct. 2. The second partial product is incorrect: \(7 \times 8 = 56\), not \(48\). 3. Add the correct partial products: \(320 + 56 = 376\).

Leo’s error is the second partial product. \(7 \times 8 = 56\), not \(48\). The correct product is \(376\).

- Break \(156\) into hundreds, tens, and ones. - Multiply each part by \(4\), then add. - Subtract the product from \(1000\) to find the difference.

1. Break apart \(156\) as \(100 + 50 + 6\). 2. Find the partial products: \(100 \times 4 = 400\), \(50 \times 4 = 200\), and \(6 \times 4 = 24\). 3. Add: \(400 + 200 + 24 = 624\). 4. Find the difference: \(1000 - 624 = 376\).

The product is \(624\). It is \(376\) less than \(1000\).

- Find both products in each part. - Use partial products to make the multiplication easier. - Compare the final values.

1. For part a, \(135 \times 3 = 405\) and \(112 \times 4 = 448\), so \(405 < 448\). 2. For part b, \(160 \times 5 = 800\) and \(198 \times 4 = 792\), so \(800 > 792\).

a) \(135 \times 3 < 112 \times 4\) b) \(160 \times 5 > 198 \times 4\)

- Estimate first, then calculate both products. - Use partial products if needed. - Compare the final values.

1. For part a, \(104 \times 7 = 728\) and \(107 \times 4 = 428\), so \(728 > 428\). 2. For part b, \(120 \times 8 = 960\) and \(160 \times 6 = 960\), so the products are equal. 3. For part c, \(150 \times 4 = 600\) and \(130 \times 5 = 650\), so \(600 < 650\).

a) \(104 \times 7 > 107 \times 4\) b) \(120 \times 8 = 160 \times 6\) c) \(150 \times 4 < 130 \times 5\)

- Find both products using partial products. - Compare the two results. - Subtract the smaller product from the larger product.

1. For Problem A, \(40 \times 6 = 240\) and \(7 \times 6 = 42\). Then \(240 + 42 = 282\). 2. For Problem B, \(30 \times 9 = 270\) and \(2 \times 9 = 18\). Then \(270 + 18 = 288\). 3. Since \(288 > 282\), Problem B has the greater product. 4. The difference is \(288 - 282 = 6\).

Problem B has the greater product: \(288\) compared with \(282\). The difference is \(6\).

- Find the value on each side of the box. - Use place-value parts to simplify the multiplication. - Compare the final values.

1. For part a, \(120 \times 4 = 480\) and \(110 \times 5 = 550\), so \(480 < 550\). 2. For part b, \(7 \times 60 = 420\), so \(420 > 400\). 3. For part c, \(250 \times 2 = 500\) and \(100 \times 5 = 500\), so the values are equal. 4. For part d, \(9 \times 30 = 270\), so \(270 < 280\). 5. For part e, \(80 \times 4 = 320\) and \(3 \times 110 = 330\), so \(320 < 330\).

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

- Find each product separately. - Use partial products for the larger factors. - Compare the three final values.

1. For part a, \(100 \times 4 = 400\), \(20 \times 4 = 80\), and \(4 \times 4 = 16\). Then \(400 + 80 + 16 = 496\). 2. For part b, \(200 \times 2 = 400\), \(40 \times 2 = 80\), and \(8 \times 2 = 16\). Then \(400 + 80 + 16 = 496\). 3. For part c, \(60 \times 8 = 480\) and \(2 \times 8 = 16\). Then \(480 + 16 = 496\). 4. All three products are equal.

a) \(496\) b) \(496\) c) \(496\) All three products are equal.

- Find the cost of the first \(40\) tickets. - Find the cost of the remaining \(5\) tickets. - Add the two costs.

1. Find the cost of \(40\) tickets: \(40 \times 8 = 320\), so the cost is \(\$320\). 2. Find the cost of \(5\) tickets: \(5 \times 8 = 40\), so the cost is \(\$40\). 3. Add the two partial costs: \(\$320 + \$40 = \$360\). 4. Because \(40 + 5 = 45\), the costs of those two groups combine to give the cost of all \(45\) tickets.

1. \(40\) tickets cost \(\$320\). 2. \(5\) tickets cost \(\$40\). 3. All \(45\) tickets cost \(\$360\). Add the first two costs because \(40 + 5 = 45\).

- Break each three-digit factor into place-value parts. - Add the partial products carefully. - For part e, find both products before comparing.

1. For part a, \(100 \times 6 = 600\), \(40 \times 6 = 240\), and \(5 \times 6 = 30\). Then \(600 + 240 + 30 = 870\). 2. For part b, \(200 \times 3 = 600\) and \(60 \times 3 = 180\). Then \(600 + 180 = 780\). 3. For part c, \(300 \times 3 = 900\), \(10 \times 3 = 30\), and \(2 \times 3 = 6\). Then \(900 + 30 + 6 = 936\). 4. For part d, \(100 \times 7 = 700\) and \(8 \times 7 = 56\). Then \(700 + 56 = 756\). 5. For part e, \(160 \times 4 = 640\) and \(210 \times 3 = 630\), so \(640 > 630\).

a) \(870\) b) \(780\) c) \(936\) d) \(756\) e) \(160 \times 4\) is greater.

- Find all six products first. - Look for factor pairs related by doubling one factor and halving the other. - Match equal results.

1. The products are \(40 \times 8 = 320\), \(140 \times 3 = 420\), \(500 \times 2 = 1000\), \(160 \times 2 = 320\), \(70 \times 6 = 420\), and \(250 \times 4 = 1000\). 2. The equal pairs are \(40 \times 8\) and \(160 \times 2\); \(140 \times 3\) and \(70 \times 6\); and \(500 \times 2\) and \(250 \times 4\).

\(40 \times 8 = 320\) and \(160 \times 2 = 320\) \(140 \times 3 = 420\) and \(70 \times 6 = 420\) \(500 \times 2 = 1000\) and \(250 \times 4 = 1000\)

- Find the product on each side. - Record the intermediate results if helpful. - Choose the comparison symbol that makes each statement true.

1. For part a, \(120 \times 4 = 480\) and \(60 \times 8 = 480\), so the products are equal. 2. For part b, \(150 \times 3 = 450\) and \(110 \times 4 = 440\), so \(450 > 440\). 3. For part c, \(210 \times 2 = 420\) and \(80 \times 5 = 400\), so \(420 > 400\). 4. For part d, \(130 \times 3 = 390\) and \(100 \times 4 = 400\), so \(390 < 400\).

a) \(120 \times 4 = 60 \times 8\) b) \(150 \times 3 > 110 \times 4\) c) \(210 \times 2 > 80 \times 5\) d) \(130 \times 3 < 100 \times 4\)

- Find each product separately. - Compare the five results. - Group expressions with matching products.

1. The products are \(120 \times 4 = 480\), \(160 \times 3 = 480\), \(240 \times 2 = 480\), \(110 \times 4 = 440\), and \(220 \times 2 = 440\). 2. Parts a, b, and c have the same value. Parts d and e have the same value.

a), b), and c) each equal \(480\); d) and e) each equal \(440\).

- Break the larger factor into hundreds, tens, and ones. - Multiply each place-value part by the one-digit factor. - Add the partial products. - In part b, account for the zero in the tens place.

1. For part a, \(100 \times 3 = 300\), \(30 \times 3 = 90\), and \(2 \times 3 = 6\). Then \(300 + 90 + 6 = 396\). 2. For part b, \(200 \times 4 = 800\), \(0 \times 4 = 0\), and \(5 \times 4 = 20\). Then \(800 + 0 + 20 = 820\). 3. For part c, \(3 \times 200 = 600\) and \(3 \times 10 = 30\). Then \(600 + 30 = 630\).

a) \(300\), \(90\), \(6\); product: \(396\) b) \(800\), \(0\), \(20\); product: \(820\) c) \(600\), \(30\); product: \(630\)

- Identify the tens and ones parts of each two-digit factor. - Determine which missing factor produces each stated partial product. - For \(280\), think about which number multiplied by \(7\) gives that product. - Add the partial products.

1. For part a, \(60 \times 4 = 240\) and \(8 \times 4 = 32\). Then \(240 + 32 = 272\). 2. For part b, \(43\) is broken into \(40 + 3\). Then \(40 \times 7 = 280\), \(3 \times 7 = 21\), and \(280 + 21 = 301\).

a) \(68 \times 4 = 272\), \(60 \times 4 = 240\), \(8 \times 4 = 32\), and the product is \(272\). b) \(43 \times 7 = 301\), \(40 \times 7 = 280\), \(3 \times 7 = 21\), and the product is \(301\).

- Break each two-digit factor into tens and ones. - Add the partial products for each problem. - Compare parts a and b, then parts c and d. - Check how both the factor and the product change.

1. For part a, \(10 \times 4 = 40\) and \(3 \times 4 = 12\). Then \(40 + 12 = 52\). 2. For part b, \(20 \times 4 = 80\) and \(6 \times 4 = 24\). Then \(80 + 24 = 104\). 3. For part c, \(10 \times 3 = 30\) and \(5 \times 3 = 15\). Then \(30 + 15 = 45\). 4. For part d, \(30 \times 3 = 90\). 5. Since \(26\) is twice \(13\), \(104\) is twice \(52\). Since \(30\) is twice \(15\), \(90\) is twice \(45\). When one factor is doubled and the other factor stays the same, the product doubles.

a) \(52\) b) \(104\) c) \(45\) d) \(90\) When the first factor is doubled, the product doubles.

- Begin with a connected group that has only one unknown value. - Use division when a product and one factor are known.

1. The left brick in the second row is \(2 \times 1 = 2\). 2. For the missing bottom value, solve \(1 \times x = 3\), so \(x = 3\). 3. The right brick in the second row is \(3 \times 2 = 6\). 4. The third row is \(2 \times 3 = 6\) and \(3 \times 6 = 18\). 5. The top is \(6 \times 18 = 108\).

Bottom row: \(2\), \(1\), \(3\), \(2\) Second row: \(2\), \(3\), \(6\) Third row: \(6\), \(18\) Top: \(108\)

- Evaluate each pair of partial products separately. - Compare the two ways of decomposing the factor \(12\). - A valid distributive decomposition must keep the same total factor.

1. Student A: \(6\times10+6\times2=60+12=72\). 2. Student B: \(6\times6+6\times6=36+36=72\). 3. Student A may be more efficient because multiplying by \(10\) is especially simple. Student B is also valid because \(6+6=12\). 4. Both represent \(6\times12\) because the factor \(12\) is decomposed into addends whose sum is still \(12\).

a) Both decompositions equal \(72\). b) Sample answer: Student A is more efficient because \(6\times10\) is easy to find. c) Both are valid because \(10+2=12\) and \(6+6=12\).

- Use the picture to determine the width of each colored section. - Write a multiplication expression for each section before adding. - Check that each partial product counts every row in that section.

1. The large section has \(10\) columns, so it represents \(7 \times 10 = 70\). 2. The smaller section has \(2\) columns, so it represents \(7 \times 2 = 14\), not \(2\). 3. Add the partial products: \(70 + 14 = 84\). Therefore, \(7 \times 12 = 84\). 4. One check is to reverse the factors and view the same array as \(12 \times 7 = 84\).

a) The student used \(2\) instead of \(7 \times 2\) for the smaller partial product. b) \(7 \times 12 = 7 \times 10 + 7 \times 2 = 70 + 14 = 84\) c) One possible check is \(12 \times 7 = 84\).

- Read the number of rows, books per row, and shaded books per row from the diagram. - You can find how many remain in each row first. - Or find the original total and subtract all borrowed books.

1. The diagram shows \(5\) rows of \(12\) books, with \(4\) shaded books in each row. 2. Find the number left in each row, then multiply: \(5\times(12-4)=5\times8=40\). 3. Or subtract all borrowed books from the original total: \(5\times12-5\times4=60-20=40\).

\(40\) books remain. Two methods are \(5\times(12-4)=40\) and \(5\times12-5\times4=40\).

- Compare both side lengths in the image with \(10\). - Use a nearby \(\times 10\) product to calculate the exact value. - Compare the exact result with the estimate.

1. The array has \(9\) rows and \(11\) columns. Since both dimensions are close to \(10\), estimate with \(10 \times 10 = 100\). 2. The exact number is \(9 \times 11 = 99\). Using the distributive property, \(9 \times (10 + 1) = 90 + 9 = 99\). 3. One dimension is \(1\) less than \(10\), and the other is \(1\) greater than \(10\), so the exact product is only \(1\) less than \(100\).

The estimate is about \(100\) dots. The exact number is \(99\) dots.

- Use the visible carry into the tens column. - The tens-column total determines both the tens result digit and the carry to hundreds.

1. The shown carry from the ones column is \(5\), because \(8\times7=56\). 2. In the tens column, \(6\times7+5=47\). 3. Write \(7\) in the tens place and carry \(4\) to the hundreds place. Therefore, the hidden carry is \(4\).

The hidden carry is \(4\). The tens-column total is \(6\times7+5=47\).

- Start in the ones column to determine whether anything is carried into tens. - The tens-column total must end in the shown product digit \(6\) and also create the hidden carry.

1. Ones: \(2\times4=8\), so there is no carry into the tens place. 2. The tens result digit is \(6\), so the hidden tens digit must satisfy \(4\times d=16\). 3. Thus \(d=4\), and the tens-column total \(16\) sends a carry of \(1\) to the hundreds place.

The missing multiplicand digit is \(4\), the hidden carry is \(1\), and the tens-column equation is \(4\times4=16\).

- Check the calculation from right to left rather than recomputing the whole product at once. - A carry shown above a column must be included in that column's multiplication. - Once you find the first inconsistent digit, continue correctly from there.

1. Ones: \(8\times4=32\), so writing \(2\) and carrying \(3\) is correct. 2. Tens should use that carry: \(3\times4+3=15\). The tens digit should be \(5\), not \(2\). 3. Continue with regrouping: \(5\times4+1=21\), then \(1\times4+2=6\). 4. The correct product is \(6152\).

The first error is in the tens column. The correct product is \(6152\).

- Check the carry created by the ones column first. - Carries are written over the column that receives them.

1. The rightmost digit of the multiplicand is \(6\). 2. Ones: \(6\times5=30\), so write \(0\) in the ones place and carry \(3\) to the tens place. 3. The displayed carry \(2\) above the tens place is therefore incorrect; it must be \(3\).

The carry above the tens place is wrong. It should be \(3\), because \(6\times5=30\).

- Check the ones column first because its carry is used in the tens column. - The tens-column total determines the carry written above the hundreds column.

1. Ones: \(7\times6=42\). Write \(2\) and carry \(4\) to the tens column. 2. Tens: \(8\times6+4=52\). Write \(2\) and carry \(5\) to the hundreds column. 3. Therefore, the carry above the hundreds column should be \(5\), not \(4\). The correct carries are \(5\) above the hundreds column and \(4\) above the tens column.

The incorrect carry is the \(4\) above the hundreds column. Correct carries: hundreds column \(5\); tens column \(4\). Column equations: \(7\times6=42\) and \(8\times6+4=52\).

- Begin with the ones column; its carry is needed in the tens column. - Use the first carry when computing the second one.

1. Ones: \(6\times4=24\), so carry \(2\) into the tens column. 2. Tens: \(5\times4+2=22\), so carry \(2\) into the hundreds column. 3. Those are the two hidden carry digits.

Carry into tens: \(2\), from \(6\times4=24\). Carry into hundreds: \(2\), from \(5\times4+2=22\).

- The ones column determines the hidden carry before you solve for the hidden tens digit. - Use both the carry and the shown tens result digit in one column equation.

1. Ones: \(8\times7=56\), so the hidden carry into the tens column is \(5\). 2. The tens result digit is \(7\). The hidden digit \(d\) must satisfy \(7d+5=47\). 3. Therefore, \(d=6\), and the complete tens-column equation is \(6\times7+5=47\).

Missing multiplicand digit: \(6\). Hidden carry into tens: \(5\). Tens-column equation: \(6\times7+5=47\).