- Look only at the hundreds before moving to the next column. - How many hundreds does each of three shares get? - Name the place value of that first quotient digit.

1. The first part is \(6\) hundreds divided into \(3\) equal shares. 2. Each share gets \(2\) hundreds, so the first quotient digit is \(2\) in the hundreds place.

\(2\) in the hundreds place.

- Divide from left to right, one place at a time. - Carry each remainder to the next place. - Write a zero in the quotient when the divisor does not fit into a place after a nonzero quotient digit has already been written.

1. For a), \(7 \div 6=1\) with remainder \(1\). Bring down the \(3\) to make \(13\). Then \(13 \div 6=2\) with remainder \(1\). Bring down the \(2\) to make \(12\), and \(12 \div 6=2\). Therefore, \(732 \div 6=122\). 2. For b), \(9 \div 4=2\) with remainder \(1\). Bring down the \(1\) to make \(11\). Then \(11 \div 4=2\) with remainder \(3\). Bring down the \(2\) to make \(32\), and \(32 \div 4=8\). Therefore, \(912 \div 4=228\). 3. For c), \(5 \div 5=1\). Bring down the \(2\). Since \(5\) goes into \(2\) zero times, write \(0\) in the tens place of the quotient. Bring down the final \(5\) to make \(25\), and \(25 \div 5=5\). Therefore, \(525 \div 5=105\).

a) \(122\) b) \(228\) c) \(105\)

- A remainder is the amount left when the divisor does not divide the dividend evenly. - The remainder must be less than the divisor. - Include a zero in the quotient when needed to preserve place value.

1. For a), \(8 \div 8=1\). Bring down the \(4\). Since \(8\) goes into \(4\) zero times, write \(0\) in the tens place of the quotient. Bring down the \(9\) to make \(49\). Then \(49 \div 8=6\) with remainder \(1\). Therefore, \(849 \div 8=106\text{ R }1\). Check: \(8 \times 106+1=849\). 2. For b), begin with \(67\) because \(6<7\). Then \(67 \div 7=9\) with remainder \(4\). Bring down the final \(4\) to make \(44\). Then \(44 \div 7=6\) with remainder \(2\). Therefore, \(674 \div 7=96\text{ R }2\). Check: \(7 \times 96+2=674\).

a) \(106\text{ R }1\) b) \(96\text{ R }2\)

- If the first digit is less than the divisor, begin with the first two digits. - Divide from left to right. - Write zero in a quotient place when needed to preserve place value.

1. For a), begin with \(15\) because \(1<3\). Then \(15 \div 3=5\). Bring down the \(3\), and \(3 \div 3=1\). Bring down the \(6\), and \(6 \div 3=2\). Therefore, \(1536 \div 3=512\). 2. For b), \(24 \div 8=3\). Bring down the \(0\). Since \(8\) goes into \(0\) zero times, write \(0\) in the tens place of the quotient. Bring down the \(8\), and \(8 \div 8=1\). Therefore, \(2408 \div 8=301\).

a) \(512\) b) \(301\)

- Divide from left to right. - Write zero in a quotient place when needed. - Multiply the quotient by the divisor to check.

1. For a), \(8 \div 6=1\) with remainder \(2\). Bring down the \(1\) to make \(21\), and \(21 \div 6=3\) with remainder \(3\). Bring down the \(6\) to make \(36\), and \(36 \div 6=6\). Therefore, \(816 \div 6=136\). Check: \(136 \times 6=816\). 2. For b), \(8 \div 4=2\). Bring down the \(1\). Since \(4\) goes into \(1\) zero times, write \(0\) in the tens place of the quotient. Bring down the \(6\) to make \(16\), and \(16 \div 4=4\). Therefore, \(816 \div 4=204\). Check: \(204 \times 4=816\). 3. For c), \(8 \div 8=1\). Bring down the \(1\). Since \(8\) goes into \(1\) zero times, write \(0\) in the tens place of the quotient. Bring down the \(6\) to make \(16\), and \(16 \div 8=2\). Therefore, \(816 \div 8=102\). Check: \(102 \times 8=816\).

a) \(136\); check: \(136 \times 6=816\) b) \(204\); check: \(204 \times 4=816\) c) \(102\); check: \(102 \times 8=816\)

- Estimate the quotient first. - Divide from left to right and carry each remainder to the next place. - Check your answer with multiplication.

1. Divide the total amount by the number of teams: \(7536 \div 6\). 2. Using the standard division algorithm gives \(1256\). 3. Check: \(1256 \times 6=7536\).

Each team receives \(\$1256\).

- Use multiplication to check a quotient. - When there is a remainder, add it after multiplying the quotient and divisor. - Keep the division steps aligned by place value.

1. For a), begin with \(54\) because \(5<8\). Then \(54 \div 8=6\) with remainder \(6\). Bring down the \(3\) to make \(63\), and \(63 \div 8=7\) with remainder \(7\). Bring down the \(2\) to make \(72\), and \(72 \div 8=9\). Therefore, \(5432 \div 8=679\). Check: \(679 \times 8=5432\). 2. For b), \(7 \div 4=1\) with remainder \(3\). Bring down the \(8\) to make \(38\), and \(38 \div 4=9\) with remainder \(2\). Bring down the \(1\) to make \(21\), and \(21 \div 4=5\) with remainder \(1\). Bring down the \(5\) to make \(15\), and \(15 \div 4=3\) with remainder \(3\). Therefore, \(7815 \div 4=1953\text{ R }3\). Check: \(1953 \times 4+3=7815\).

a) \(679\); check: \(679 \times 8=5432\) b) \(1953\text{ R }3\); check: \(1953 \times 4+3=7815\)

- Divide one place at a time from left to right. - Carry each remainder to the next digit. - The final remainder must be less than the divisor.

1. For a), begin with \(39\) because \(3<7\). Then \(39 \div 7=5\) with remainder \(4\). Bring down the \(4\) to make \(44\). Then \(44 \div 7=6\) with remainder \(2\). Bring down the \(9\) to make \(29\). Then \(29 \div 7=4\) with remainder \(1\). Therefore, \(3949 \div 7=564\text{ R }1\). 2. For b), begin with \(25\) because \(2<4\). Then \(25 \div 4=6\) with remainder \(1\). Bring down the \(0\) to make \(10\). Then \(10 \div 4=2\) with remainder \(2\). Bring down the \(6\) to make \(26\). Then \(26 \div 4=6\) with remainder \(2\). Therefore, \(2506 \div 4=626\text{ R }2\). 3. For c), begin with \(61\) because \(6<8\). Then \(61 \div 8=7\) with remainder \(5\). Bring down the \(2\) to make \(52\). Then \(52 \div 8=6\) with remainder \(4\). Bring down the \(3\) to make \(43\). Then \(43 \div 8=5\) with remainder \(3\). Therefore, \(6123 \div 8=765\text{ R }3\).

a) \(564\text{ R }1\) b) \(626\text{ R }2\) c) \(765\text{ R }3\)

- Write zero in a quotient place when the divisor does not fit into that place. - Carry each remainder to the next digit. - The final remainder must be less than the divisor.

1. For a), \(8 \div 4=2\). Bring down the \(1\). Since \(4\) goes into \(1\) zero times, write \(0\) in the tens place of the quotient. Bring down the \(6\) to make \(16\), and \(16 \div 4=4\). Therefore, \(816 \div 4=204\). 2. For b), \(7 \div 3=2\) with remainder \(1\). Bring down the \(2\) to make \(12\), and \(12 \div 3=4\). Bring down the \(5\), and \(5 \div 3=1\) with remainder \(2\). Therefore, \(725 \div 3=241\text{ R }2\). 3. For c), \(9 \div 5=1\) with remainder \(4\). Bring down the \(0\) to make \(40\), and \(40 \div 5=8\). Bring down the \(5\), and \(5 \div 5=1\). Therefore, \(905 \div 5=181\). 4. For d), \(6 \div 6=1\). Bring down the \(3\). Since \(6\) goes into \(3\) zero times, write \(0\) in the tens place of the quotient. Bring down the \(8\) to make \(38\), and \(38 \div 6=6\) with remainder \(2\). Therefore, \(638 \div 6=106\text{ R }2\).

a) \(204\) b) \(241\text{ R }2\) c) \(181\) d) \(106\text{ R }2\)

- Divide from left to right. - Write zero in a quotient place when the divisor does not fit into that place. - Check each quotient with multiplication.

1. For a), begin with \(45\) because \(4<5\). Then \(45 \div 5=9\). Bring down the \(0\), and write \(0\) in the tens place of the quotient. Bring down the \(5\), and \(5 \div 5=1\). Therefore, \(4505 \div 5=901\). 2. For b), \(12 \div 6=2\). Bring down the \(4\). Since \(6\) goes into \(4\) zero times, write \(0\) in the tens place of the quotient. Bring down the \(8\) to make \(48\), and \(48 \div 6=8\). Therefore, \(1248 \div 6=208\). 3. For c), begin with \(56\) because \(5<7\). Then \(56 \div 7=8\). Bring down the \(8\), and \(8 \div 7=1\) with remainder \(1\). Bring down the \(4\) to make \(14\), and \(14 \div 7=2\). Therefore, \(5684 \div 7=812\). 4. For d), \(8 \div 8=1\). Bring down the \(1\). Since \(8\) goes into \(1\) zero times, write \(0\) in the hundreds place of the quotient. Bring down the \(6\) to make \(16\), and \(16 \div 8=2\). Bring down the final \(8\), and \(8 \div 8=1\). Therefore, \(8168 \div 8=1021\).

a) \(901\) b) \(208\) c) \(812\) d) \(1021\)

- Divide from left to right, one place at a time. - Carry each remainder to the next digit. - Check using multiplication and the remainder.

1. Begin with \(7\). Then \(7 \div 4=1\) with remainder \(3\). Bring down the \(4\) to make \(34\), and \(34 \div 4=8\) with remainder \(2\). Bring down the \(5\) to make \(25\), and \(25 \div 4=6\) with remainder \(1\). Bring down the \(2\) to make \(12\), and \(12 \div 4=3\). Therefore, \(7452 \div 4=1863\). 2. Begin with \(9\). Then \(9 \div 7=1\) with remainder \(2\). Bring down the \(6\) to make \(26\), and \(26 \div 7=3\) with remainder \(5\). Bring down the \(0\) to make \(50\), and \(50 \div 7=7\) with remainder \(1\). Bring down the \(5\) to make \(15\), and \(15 \div 7=2\) with remainder \(1\). Therefore, \(9605 \div 7=1372\text{ R }1\). 3. Begin with \(4\). Then \(4 \div 3=1\) with remainder \(1\). Bring down the \(8\) to make \(18\), and \(18 \div 3=6\). Bring down the \(1\). Since \(3\) goes into \(1\) zero times, write \(0\) in the tens place of the quotient. Bring down the \(5\) to make \(15\), and \(15 \div 3=5\). Therefore, \(4815 \div 3=1605\). 4. Begin with \(5\). Since \(6\) does not fit into \(5\), use \(52\). Then \(52 \div 6=8\) with remainder \(4\). Bring down the \(3\) to make \(43\), and \(43 \div 6=7\) with remainder \(1\). Bring down the \(6\) to make \(16\), and \(16 \div 6=2\) with remainder \(4\). Therefore, \(5236 \div 6=872\text{ R }4\).

1. \(1863\) 2. \(1372\text{ R }1\) 3. \(1605\) 4. \(872\text{ R }4\)

- Divide from left to right. - Write zero in a quotient place when needed. - Check whether a final remainder is left and verify with multiplication.

1. For a), begin with \(43\) because \(4<7\). Then \(43 \div 7=6\) with remainder \(1\). Bring down the \(6\) to make \(16\), and \(16 \div 7=2\) with remainder \(2\). Bring down the \(8\) to make \(28\), and \(28 \div 7=4\). Therefore, \(4368 \div 7=624\). 2. For b), begin with \(29\) because \(2<4\). Then \(29 \div 4=7\) with remainder \(1\). Bring down the \(5\) to make \(15\), and \(15 \div 4=3\) with remainder \(3\). Bring down the \(2\) to make \(32\), and \(32 \div 4=8\). Therefore, \(2952 \div 4=738\). 3. For c), begin with \(50\) because \(5<8\). Then \(50 \div 8=6\) with remainder \(2\). Bring down the \(4\) to make \(24\), and \(24 \div 8=3\). Bring down the \(3\). Since \(8\) goes into \(3\) zero times, write \(0\) in the ones place of the quotient, with remainder \(3\). Therefore, \(5043 \div 8=630\text{ R }3\).

a) \(624\) b) \(738\) c) \(630\text{ R }3\)

- Write zero in a quotient place when the divisor fits zero times there. - A final remainder must be less than the divisor. - Check division by multiplying the quotient and divisor, then adding any remainder.

1. For a), begin with \(48\) because \(4<8\). Then \(48 \div 8=6\). Bring down the \(1\). Since \(8\) goes into \(1\) zero times, write \(0\) in the tens place of the quotient. Bring down the \(6\) to make \(16\), and \(16 \div 8=2\). Therefore, \(4816 \div 8=602\). 2. For b), begin with \(36\) because \(3<4\). Then \(36 \div 4=9\). Bring down the \(0\), and write \(0\) in the tens place of the quotient. Bring down the \(8\), and \(8 \div 4=2\). Therefore, \(3608 \div 4=902\). 3. For c), begin with \(18\) because \(1<9\). Then \(18 \div 9=2\). Bring down the \(4\). Since \(9\) goes into \(4\) zero times, write \(0\) in the tens place of the quotient. Bring down the \(8\) to make \(48\), and \(48 \div 9=5\) with remainder \(3\). Therefore, \(1848 \div 9=205\text{ R }3\). 4. For d), \(7 \div 7=1\). Bring down the \(2\). Since \(7\) goes into \(2\) zero times, write \(0\) in the hundreds place of the quotient. Bring down the \(1\) to make \(21\), and \(21 \div 7=3\). Bring down the final \(0\), and write \(0\) in the ones place of the quotient. Therefore, \(7210 \div 7=1030\). 5. Check for d): \(1030 \times 7=7210\).

a) \(602\) b) \(902\) c) \(205\text{ R }3\) d) \(1030\); check: \(1030 \times 7=7210\)

- Follow the quotient digits from left to right. - Each subtrahend is the divisor multiplied by the quotient digit for that step. - Match the middle quotient digit to the subtraction under the partial dividend \(6\).

1. The calculation is \(864\div4=216\). 2. After \(8\div4=2\), the next partial dividend is \(6\). 3. The divisor \(4\) fits into \(6\) once, so the quotient digit is \(1\) and the subtrahend is \(1\times4=4\). 4. The final step uses \(24\div4=6\), giving quotient \(216\).

a) \(216\) b) Because the quotient digit in that place is \(1\), so \(1\times4=4\).

- Track the remainder from the first subtraction before bringing down the next digit. - Compare nearby multiples of \(5\) with the partial dividend \(23\). - Continue the same process for the final digit.

1. The calculation is \(735\div5\). 2. After \(7-5=2\), bring down the \(3\) to make \(23\). 3. The divisor \(5\) fits into \(23\) four times because \(4\times5=20\) and \(5\times5=25\) is too large. 4. Continuing with \(35\div5=7\) gives quotient \(147\).

The quotient digit \(4\) comes from \(23\div5\), and the quotient is \(147\).

- Match each quotient digit with the subtraction step directly below its partial dividend. - Each subtrahend is a multiple of the divisor. - Read the multiplication facts in the same left-to-right order as the quotient digits.

1. The calculation is \(936\div3=312\). 2. The first quotient digit \(3\) uses \(3\times3=9\). 3. The next quotient digit \(1\) uses \(1\times3=3\). 4. The final quotient digit \(2\) uses \(2\times3=6\).

The facts are \(3\times3=9\), \(1\times3=3\), and \(2\times3=6\). The quotient is \(312\).

- Read the first subtraction in the displayed calculation. - What digit is brought down next? - Explain the transition before reading the final quotient.

1. The first step is \(13-12=1\), leaving a remainder of \(1\). 2. Bring down the next digit, \(2\), next to the remaining \(1\) ten. The next number to divide is \(12\). 3. The shown quotient is \(221\).

a) \(1\) b) \(12\), formed by bringing down \(2\) next to the remaining \(1\). c) \(221\)

- Use each partial dividend shown in the written work. - The difference after subtracting the middle row becomes part of the next partial dividend. - Check each quotient digit against the divisor.

1. \(8\div4=2\). Subtract \(8\) and bring down the next digit, \(6\). 2. \(6\div4=1\). The middle subtraction entry is \(4\), leaving \(2\) tens. 3. Bring down the \(4\) ones to make \(24\). Divide \(24\div4=6\) and subtract \(24\), leaving remainder \(0\). 4. The completed quotient is \(216\).

Quotient: \(216\). Missing subtraction entries: \(4\) and \(24\). Remainder: \(0\). In the middle step, \(6-4=2\) tens remain. Bring down the \(4\) ones to make \(24\) ones.

- Divide the total number of apples by the number in each bag. - The quotient represents full bags. - The remainder represents apples left over.

1. Divide: \(4589 \div 7\). 2. The standard division algorithm gives \(655\text{ R }4\). 3. Therefore, \(655\) full bags can be packed, with \(4\) apples left. Check: \(655 \times 7+4=4589\).

\(655\) full bags can be packed, and \(4\) apples will be left over.

- Choose a nearby number that is easy to divide by the divisor. - Then use the standard division algorithm for the exact quotient. - Compare each exact result with its estimate.

1. For a), use the compatible number \(1680\): \(1638 \div 7\approx1680 \div 7=240\). For the exact quotient, begin with \(16\). Then \(16 \div 7=2\) with remainder \(2\). Bring down the \(3\) to make \(23\), and \(23 \div 7=3\) with remainder \(2\). Bring down the \(8\) to make \(28\), and \(28 \div 7=4\). Therefore, \(1638 \div 7=234\). 2. For b), use the compatible number \(2250\): \(2304 \div 9\approx2250 \div 9=250\). For the exact quotient, begin with \(23\). Then \(23 \div 9=2\) with remainder \(5\). Bring down the \(0\) to make \(50\), and \(50 \div 9=5\) with remainder \(5\). Bring down the \(4\) to make \(54\), and \(54 \div 9=6\). Therefore, \(2304 \div 9=256\).

a) Estimate: \(240\); exact quotient: \(234\) b) Estimate: \(250\); exact quotient: \(256\)

- Align each division carefully in the standard algorithm. - Compare consecutive dividends. - Divide the change in the dividend by \(4\). - Look for that change in the quotients.

1. Part a: \(24\div 4=6\), so write \(6\) in the hundreds place. Since \(3<4\), write \(0\) in the tens place. Bring down \(6\) to make \(36\), and \(36\div 4=9\). Thus, \(2436\div 4=609\). 2. Part b: \(28\div 4=7\). Since \(3<4\), write \(0\), then use \(36\div 4=9\). Thus, \(2836\div 4=709\). 3. Part c: \(32\div 4=8\). Since \(3<4\), write \(0\), then use \(36\div 4=9\). Thus, \(3236\div 4=809\). 4. Part d: \(36\div 4=9\). Since \(3<4\), write \(0\), then use \(36\div 4=9\). Thus, \(3636\div 4=909\). 5. The dividend increases by \(400\) each time. Since \(400\div 4=100\), the quotient increases by \(100\) each time.

a) \(609\) b) \(709\) c) \(809\) d) \(909\) Each increase of \(400\) in the dividend increases the quotient by \(100\).

- Use the standard division algorithm for each quotient. - Compare the divisors from one part to the next. - Compare each quotient with the previous quotient. - Use multiplication to check the last quotient.

1. Part a: \(5\div 2=2\) remainder \(1\); \(11\div 2=5\) remainder \(1\); \(18\div 2=9\); and \(4\div 2=2\). Therefore, \(5184\div 2=2592\). 2. Part b: \(5\div 4=1\) remainder \(1\); \(11\div 4=2\) remainder \(3\); \(38\div 4=9\) remainder \(2\); and \(24\div 4=6\). Therefore, \(5184\div 4=1296\). 3. Part c: Since \(5<8\), use \(51\). Then \(51\div 8=6\) remainder \(3\); \(38\div 8=4\) remainder \(6\); and \(64\div 8=8\). Therefore, \(5184\div 8=648\). 4. Check part c: \(648\times 8=5184\). 5. The dividend stays fixed while the divisor doubles each time, so the quotient is halved each time.

a) \(2592\) b) \(1296\) c) \(648\) Check: \(648\times 8=5184\). When the divisor doubles and the dividend stays fixed, the quotient is halved.

- Complete both divisions through the final remainder. - Compare only the remainders. - A remainder must always be less than its divisor.

1. For a), begin with \(14\) because \(1<5\). Then \(14 \div 5=2\) with remainder \(4\). Bring down the \(2\) to make \(42\). Then \(42 \div 5=8\) with remainder \(2\). Bring down the \(7\) to make \(27\). Then \(27 \div 5=5\) with remainder \(2\). Therefore, \(1427 \div 5=285\text{ R }2\). 2. For b), \(9 \div 9=1\). Bring down the \(8\). Since \(9\) goes into \(8\) zero times, write \(0\) in the hundreds place of the quotient. Bring down the \(3\) to make \(83\). Then \(83 \div 9=9\) with remainder \(2\). Bring down the \(5\) to make \(25\). Then \(25 \div 9=2\) with remainder \(7\). Therefore, \(9835 \div 9=1092\text{ R }7\). 3. Since \(7>2\), part b) has the greater remainder.

a) \(285\text{ R }2\) b) \(1092\text{ R }7\) Part b) has the greater remainder.

- Follow the place values from left to right. - Ask how many times the divisor fits into the amount at each stage. - Preserve the quotient place with a zero when the answer is zero times.

1. For a), \(8 \div 4=2\). Bring down the \(1\). Since \(4\) goes into \(1\) zero times, write \(0\) in the tens place of the quotient. Bring down the \(6\) to make \(16\), and \(16 \div 4=4\). Therefore, \(816 \div 4=204\). 2. For b), \(9 \div 3=3\). Bring down the \(2\). Since \(3\) goes into \(2\) zero times, write \(0\) in the tens place of the quotient. Bring down the \(1\) to make \(21\), and \(21 \div 3=7\). Therefore, \(921 \div 3=307\). 3. In each division, the digit brought down for the tens place is less than the divisor. The divisor fits zero times in that place, so a zero must be written in the tens place of the quotient before continuing.

a) \(204\) b) \(307\) c) The zero is required because \(1<4\) and \(2<3\), so the divisor fits zero times in the tens place.

- Check Niran's quotient by multiplying it by \(6\) and comparing with the dividend. - In the standard algorithm, keep the quotient digits aligned with the dividend's place values. - When a partial dividend is too small to make a nonzero quotient digit, consider what must be recorded to preserve place value.

1. Part a: Since \(4<6\), use the first two digits. \(48\div 6=8\), so write \(8\) in the hundreds place. Bring down \(3\). Because \(3<6\), write \(0\) in the tens place. Bring down \(6\), and \(36\div 6=6\). Therefore, \(4836\div 6=806\). 2. Niran probably omitted the zero in the tens place and wrote \(86\) instead of \(806\). 3. Part c: \(7\div 7=1\), so write \(1\) in the thousands place. Bring down \(0\) and write \(0\) in the hundreds place. Bring down \(5\) and write \(0\) in the tens place because \(5<7\). Bring down \(6\), and \(56\div 7=8\). Therefore, \(7056\div 7=1008\).

a) \(806\) b) Niran omitted the zero in the tens place. c) \(1008\)

- Estimate how many digits the quotient should have. - In the standard algorithm, write a zero in the quotient when a partial dividend is smaller than the divisor. - Multiply the quotient by \(5\) to check your result.

1. Round \(2015\) to \(2000\). Then \(2000\div 5=400\). Since \(43\) is much smaller than \(400\), Max’s answer cannot be correct. 2. Begin with \(20\) because \(2<5\). Then \(20\div 5=4\). Bring down the \(1\). Since \(5\) goes into \(1\) zero times, write \(0\) in the tens place of the quotient. Bring down the \(5\) to make \(15\), and \(15\div 5=3\). Therefore, \(2015\div 5=403\). 3. Check: \(403\times 5=2015\).

1. One estimate is \(2000\div 5=400\), so \(43\) is much too small. 2. The correct quotient is \(403\). 3. Check: \(403\times 5=2015\).

- Divide the product by the known factor to find the missing factor. - Use multiplication to check a division result. - Keep track of each remainder as you divide from left to right.

1. Divide the product by the known factor: \(8136\div 8\). 2. Since \(8\div 8=1\), write \(1\). Bring down the \(1\). Since \(8\) goes into \(1\) zero times, write \(0\) in the hundreds place of the quotient. Bring down the \(3\) to make \(13\), and \(13\div 8=1\) with remainder \(5\). Bring down the \(6\) to make \(56\), and \(56\div 8=7\). Therefore, \(8136\div 8=1017\). 3. Check by multiplication: \(1017\times 8=8136\). The product matches the given value.

a) The unknown number is \(1017\). b) Check: \(1017\times 8=8136\).

- Complete part a) with the standard division algorithm. - Compare \(2688\) with \(1344\). - Decide how the same scale factor affects the quotient when the divisor is unchanged.

1. Part a: Since \(1<6\), use \(13\). Then \(13\div 6=2\) remainder \(1\). Bring down \(4\): \(14\div 6=2\) remainder \(2\). Bring down the final \(4\): \(24\div 6=4\). Therefore, \(1344\div 6=224\). 2. The dividend \(2688\) is twice \(1344\). 3. With the divisor unchanged, doubling the dividend doubles the quotient, so \(224\times 2=448\).

a) \(224\) b) \(448\). The dividend is doubled while the divisor stays the same, so the quotient is doubled.

- For each estimate, choose a nearby dividend that is easy to divide by the divisor. - In the standard algorithm, keep every place represented in the quotient, even when a partial dividend is too small to contain the divisor. - Check each exact quotient by multiplication.

1. Estimate \(4800\div8=600\). Using the standard algorithm, \(4656\div8=582\). 2. Estimate \(6000\div6=1000\). Using the standard algorithm, \(6216\div6=1036\). The \(0\) in the hundreds place of the quotient is needed because after dividing \(6\) thousands, the next partial dividend is \(2\), which is less than \(6\).

1) Estimate: \(600\); exact quotient: \(582\) 2) Estimate: \(1000\); exact quotient: \(1036\)

- Keep every place of the dividend represented in the quotient. - Ask how many groups of \(4\) are in the middle partial dividend. - A zero quotient digit can be a necessary placeholder.

1. The calculation is \(804\div4\). 2. \(8\div4=2\), so the first quotient digit is \(2\). 3. The next partial dividend is \(0\). The divisor \(4\) fits into \(0\) zero times, so a \(0\) must be written in the tens place of the quotient. 4. Finally, \(4\div4=1\), giving quotient \(201\).

The zero is needed because the tens-place partial dividend is \(0\). The quotient is \(201\).

- Compare the first dividend digit with the divisor. - If the divisor does not fit into the first digit, use the first two digits as the initial partial dividend. - Follow the alignment shown in the written algorithm.

1. The calculation is \(3126\div6\). 2. The first digit \(3\) is smaller than \(6\), so division cannot begin with \(3\) alone. 3. Use the first two digits: \(31\div6=5\) with \(1\) left, so the first quotient digit aligns with the second dividend digit. 4. The later steps are \(12\div6=2\) and \(6\div6=1\), giving quotient \(521\).

Division begins with \(31\) because \(3<6\). The quotient is \(521\).

- Focus on the step whose partial dividend is \(12\). - The quotient digit and subtrahend in one long-division step are linked by multiplication.

1. The second partial dividend is \(12\). 2. Three groups of \(4\) fit in \(12\), so the hidden quotient digit is \(3\). 3. The subtrahend for that step is \(3\times4=12\).

The quotient digit is \(3\), the subtrahend is \(12\), and the connecting fact is \(3\times4=12\).

- Match the hidden subtrahend with the quotient digit directly above its step. - Multiply that quotient digit by the divisor. - Check that the subtraction leaves the remainder needed for the next partial dividend.

1. The calculation is \(1568\div7=224\). 2. In the middle step, the quotient digit is \(2\), so the subtrahend must be \(2\times7=14\). 3. Therefore, the hidden digit in \(1\square\) is \(4\). 4. The remaining steps confirm the quotient \(224\) with no remainder.

The missing digit is \(4\).

- Track the place represented by every dividend digit. - What quotient digit corresponds to a partial dividend smaller than the divisor? - Omitting that zero would change the place values of the later quotient digits.

1. The calculation is \(2835\div7\). 2. \(28\div7=4\), so the hundreds quotient digit is \(4\). 3. The next partial dividend is \(3\), which contains \(0\) groups of \(7\). A \(0\) must be written in the tens place of the quotient. 4. Bringing down \(5\) gives \(35\), and \(35\div7=5\). 5. The quotient is \(405\).

The zero is a necessary tens-place placeholder because \(3<7\). The quotient is \(405\).

- Follow the dividend digits in each panel from left to right. - Compare where the zero partial dividend or too-small partial dividend occurs. - Keep quotient place value aligned with the corresponding dividend place.

1. In a), \(1632\div8=204\). After \(16\div8=2\), the next partial dividend is \(3\), which is smaller than \(8\), so the tens quotient digit is \(0\). 2. In b), \(1680\div8=210\). The final partial dividend is \(0\), so the quotient ends in \(0\). 3. Thus \(204\) has an internal zero, while \(210\) has a zero in the ones place.

a) \(204\) has the internal zero. b) It is needed because the tens-place partial dividend is \(3<8\). c) The quotients are \(204\) and \(210\).

- Look only at the final partial dividend and subtrahend first. - The remainder is what is left after the last subtraction. - Check that the remainder is smaller than \(6\).

1. The calculation is \(745\div6\). 2. The final partial dividend is \(25\), and the final subtrahend is \(24\). 3. \(25-24=1\), so the hidden remainder is \(1\). 4. The complete result is \(745\div6=124\) remainder \(1\).

The remainder is \(1\).

- What remains after the first subtraction? - Look at what is shown after the next digit is brought down. - Work backward from the next subtraction to the missing number.

1. The first subtraction is \(22-20=2\). Bring down the next digit. The next number to divide must begin with \(2\). 2. The next subtraction takes away \(24\). The following step shows \(16\), formed by bringing down the last digit \(6\). So the subtraction before it must leave \(1\). 3. The missing number in the middle step is \(24+1=25\). Its final digit, \(5\), is the missing digit in the original number. 4. Check: \(25-24=1\), then bring down \(6\) to get \(16\), and \(16-16=0\). The completed dividend is \(2256\).

Missing dividend digit: \(5\) Missing number in the written work: \(25\)

- In each step, the subtrahend equals the divisor times the quotient digit for that place. - Use the first quotient digit and first subtrahend together. - Check the recovered divisor against the later subtraction steps.

1. The quotient is \(216\), so the first quotient digit is \(2\). 2. The first subtrahend is \(14\). It must equal \(2\times\text{divisor}\). 3. Since \(2\times7=14\), the hidden divisor is \(7\). 4. The full check is \(1512\div7=216\).

The divisor is \(7\).

- Check each number subtracted against the quotient digit directly above it. - After correcting the first wrong step, continue from the corrected remainder. - Check the result using multiplication and the remainder.

1. The first shown quotient digit is \(3\), so the first subtraction uses \(3\times6=18\), which is correct. 2. The next quotient digit is \(5\), so the next number subtracted should be \(5\times6=30\), not the displayed \(24\). 3. Correcting that step gives \(33-30=3\). Bring down \(4\) to make \(34\). 4. Now \(34\div6=5\) remainder \(4\). The complete division is \(2134\div6=355\) remainder \(4\).

The second subtraction is wrong: subtract \(30\), not \(24\). Correct quotient: \(355\), remainder: \(4\)

- Follow the steps one digit at a time after each subtraction. - Check which dividend digit has not yet been used. - Correct the first misplaced step before finishing the division.

1. The first step is correct: \(32-32=0\). 2. Bring down only the next unused digit. The next number to divide should be \(6\), not \(65\). 3. Divide \(6\) by \(4\): write \(1\) in the quotient, subtract \(4\), and keep a remainder of \(2\). 4. Bring down the last digit \(5\) to make \(25\). Divide \(25\) by \(4\) to get \(6\) with remainder \(1\). 5. The correct result is \(3265\div4=816\) remainder \(1\).

First wrong number: \(65\); it should be \(6\). Quotient: \(816\); remainder: \(1\)