Free math worksheets

- Compare \(615{,}000\) with the known product \(615\). - Determine the place-value factor relating the two dividends. - Apply that same factor to the known quotient \(3\).

1. Since \(615{,}000=615\times1000\), replace \(615\) with \(205\times3\). 2. Then \(615{,}000=(205\times3)\times1000=205\times3000\). 3. Therefore, \(615{,}000\div205=3000\).

\(3000\)

- Start with the leftmost dividend digit and compare it with the divisor. - After each subtraction, bring down the next digit. - Check the quotient by multiplying it by \(6\).

1. Divide from left to right, using one dividend digit at each stage when possible. 2. Each quotient digit is written above the place being divided. 3. The written work gives the quotient \(156\). 4. Check: \(156\times6=936\).

\(156\)

- Begin with the smallest leading part of the dividend that is at least as large as the divisor. - At each place, divide, multiply, subtract, and bring down the next digit. - Multiply the final quotient by the divisor to check the dividend.

1. Divide \(84\) by \(36\): write \(2\), subtract \(72\), and bring down \(6\) to make \(126\). 2. Divide \(126\) by \(36\): write \(3\), subtract \(108\), and bring down \(7\) to make \(187\). 3. Divide \(187\) by \(36\): write \(5\), subtract \(180\), and bring down \(2\) to make \(72\). 4. Divide \(72\) by \(36\): write \(2\). The quotient is \(2352\).

\(2352\)

- Notice where the first partial dividend ends in the written layout. - Each time a new dividend digit is brought down, the next quotient digit moves one place to the right. - Use multiplication to confirm the final quotient.

1. The first usable partial dividend is \(43\), so the first quotient digit is placed over the hundreds place of the dividend. 2. After subtracting and bringing down the next digit, the next quotient digit is placed over the tens place. 3. The final brought-down digit determines the ones-place quotient digit. 4. The quotient is \(312\). Check: \(312\times14=4368\).

\(312\). Each quotient digit is aligned with the dividend place used at that stage of the standard algorithm.

- Follow the quotient one place at a time; do not skip a dividend place. - Ask what happens when the current partial dividend is smaller than the divisor. - A zero may be needed as a place holder even when later quotient digits are nonzero.

1. The first quotient digit leaves a remainder that forms the next partial dividend \(12\) when the next digit is brought down. 2. Because \(12\) is less than \(24\), there are zero groups of \(24\) in that place, so a \(0\) must be written in the quotient to hold the place. 3. Continue the standard algorithm through the remaining digits to obtain \(3052\). 4. Check: \(3052\times24=73{,}248\).

\(3052\). The \(0\) is required because the partial dividend \(12\) is smaller than \(24\), so that quotient place contains zero groups.

- Keep dividing until every digit of the dividend has been brought down. - The final leftover must be smaller than the divisor. - Check with divisor times quotient plus remainder.

1. Apply the standard algorithm until all dividend digits have been used. 2. The quotient is \(385\), and the amount left after the final subtraction is \(18\). 3. The remainder is valid because \(18<37\). 4. Check: \(37\times385+18=14{,}263\).

\(385\) remainder \(18\)

- In this context, the quotient counts complete groups of \(48\). - Interpret the remainder as items that do not make another full group. - Verify that full boxes plus leftovers account for every bolt.

1. Divide \(18{,}725\) by \(48\) with the standard algorithm. 2. The quotient \(390\) counts full boxes, and the remainder \(5\) counts bolts that do not fill another box. 3. Check: \(48\times390+5=18{,}725\).

\(390\) full boxes and \(5\) bolts left over

- Use long division one place at a time. - After each multiplication, subtract to find the partial remainder before bringing down the next digit. - A final remainder must be smaller than the divisor. - Verify each result with divisor \(\times\) quotient \(+\) remainder.

1. For a), \(23\) goes into \(124\) five times: \(5\times23=115\), leaving \(9\). Bring down \(7\) to get \(97\). Then \(23\) goes into \(97\) four times: \(4\times23=92\), leaving \(5\). Thus \(1247\div23=54\text{ R }5\). 2. Check: \(23\times54+5=1242+5=1247\). 3. For b), \(37\) goes into \(238\) six times: \(6\times37=222\), leaving \(16\). Bring down \(5\) to get \(165\). Then \(37\) goes into \(165\) four times: \(4\times37=148\), leaving \(17\). Thus \(2385\div37=64\text{ R }17\). 4. Check: \(37\times64+17=2368+17=2385\).

a) \(54\text{ R }5\) b) \(64\text{ R }17\) c) \(1247=23\times54+5\) and \(2385=37\times64+17\)

- Reverse the structure of a division with remainder. - First rebuild the part divided into complete groups. - Add the leftover amount only after multiplying.

1. Use \(\text{dividend}=\text{divisor}\times\text{quotient}+\text{remainder}\). 2. Multiply: \(72\times1408=101{,}376\). 3. Add the remainder: \(101{,}376+35=101{,}411\).

\(101{,}411\)

- First find the number of complete groups and the leftover amount. - Decide whether the leftover passengers can be ignored in this context. - A nonzero remainder may require one more group than the whole-number quotient shows.

1. Divide \(58\) by \(48\): write \(1\), subtract \(48\), and bring down \(9\) to make \(109\). 2. Divide \(109\) by \(48\): write \(2\), subtract \(96\), and bring down \(4\) to make \(134\). 3. Divide \(134\) by \(48\): write \(2\), subtract \(96\), and bring down \(3\) to make \(383\). 4. Divide \(383\) by \(48\): write \(7\), leaving remainder \(47\). Thus, \(58{,}943\div48=1227\) remainder \(47\). 5. The remainder represents \(47\) passengers who need one additional shuttle, so \(1228\) shuttles are needed.

\(1228\) shuttles

- Complete each division independently before comparing. - With a fixed positive dividend, think about how divisor size affects quotient size. - Subtract the smaller quotient from the larger after both are verified.

1. For \(96{,}768\div32\), divide \(96\) by \(32\): write \(3\) and subtract \(96\), leaving \(0\). 2. Bring down the next digit, \(7\). Since \(7<32\), write \(0\) in the quotient. 3. Bring down the next digit, \(6\), to make \(76\). Write \(2\), subtract \(64\), and leave remainder \(12\). 4. Bring down \(8\) to make \(128\). Write \(4\). Thus, \(96{,}768\div32=3024\). 5. For \(96{,}768\div48\), divide \(96\) by \(48\): write \(2\) and subtract \(96\), leaving \(0\). 6. Bring down \(7\). Since \(7<48\), write \(0\) in the quotient. 7. Bring down \(6\) to make \(76\). Write \(1\), subtract \(48\), and leave remainder \(28\). 8. Bring down \(8\) to make \(288\). Write \(6\). Thus, \(96{,}768\div48=2016\). 9. The first quotient is greater, and \(3024-2016=1008\).

\(3024\) and \(2016\); the first quotient is greater by \(1008\).

- Use the related multiplication equation to identify the missing factor. - Divide the dividend by the known quotient. - Substitute the divisor back into the original division statement.

1. Rewrite the relationship as \(3990d=287{,}280\). 2. Divide: \(d=287{,}280\div3990=72\). 3. Check: \(287{,}280\div72=3990\).

\(d=72\)

- Choose a nearby dividend that is easy to divide by \(64\). - Use the estimate to predict the quotient's size and number of digits. - Complete the exact division and verify by multiplication.

1. Use the compatible value \(512{,}000\): \(512{,}000\div64=8000\), so the exact quotient should be a little more than \(8000\). 2. Divide \(526\) by \(64\): write \(8\), subtract \(512\), and bring down \(1\) to make \(141\). 3. Divide \(141\) by \(64\): write \(2\), subtract \(128\), and bring down \(4\) to make \(134\). 4. Divide \(134\) by \(64\): write \(2\), subtract \(128\), and bring down \(4\) to make \(64\). 5. Divide \(64\) by \(64\): write \(1\). The exact quotient is \(8221\). 6. The result is slightly greater than \(8000\), as predicted. Check: \(8221\times64=526{,}144\).

Estimate: about \(8000\); exact quotient: \(8221\)

- Continue the division until every dividend digit has been used. - Keep the leftover only if it is smaller than the divisor. - Check with divisor times quotient plus remainder.

1. Divide \(682\) by \(93\): write \(7\), subtract \(651\), and bring down \(9\) to make \(319\). 2. Divide \(319\) by \(93\): write \(3\), subtract \(279\), and bring down \(1\) to make \(401\). 3. Divide \(401\) by \(93\): write \(4\), subtract \(372\), and bring down \(7\) to make \(297\). 4. Divide \(297\) by \(93\): write \(3\), leaving remainder \(18\). 5. Check: \(93\times7343+18=682{,}917\).

\(7343\) remainder \(18\)

- Look for a common ending zero that can be removed from both numbers. - Make the same change to dividend and divisor so the quotient stays equal. - Use the standard algorithm on the simpler equivalent expression.

1. Divide both the dividend and divisor by \(10\): \(987{,}840\div240=98{,}784\div24\). 2. The quotient is unchanged because both numbers were divided by the same nonzero factor. 3. Divide \(98\) by \(24\): write \(4\), subtract \(96\), and bring down \(7\) to make \(27\). 4. Divide \(27\) by \(24\): write \(1\), subtract \(24\), and bring down \(8\) to make \(38\). 5. Divide \(38\) by \(24\): write \(1\), subtract \(24\), and bring down \(4\) to make \(144\). 6. Divide \(144\) by \(24\): write \(6\). The quotient is \(4116\).

\(4116\)

- Keep every dividend place aligned throughout long division. - A zero in the dividend may create a zero quotient digit or become part of a later partial dividend. - Use multiplication to verify the final place values.

1. Divide \(200\) by \(96\): write \(2\), subtract \(192\), and bring down \(0\) to make \(80\). 2. Since \(80<96\), write \(0\) in the quotient and bring down the next \(0\) to make \(800\). 3. Divide \(800\) by \(96\): write \(8\), subtract \(768\), and bring down \(6\) to make \(326\). 4. Divide \(326\) by \(96\): write \(3\), subtract \(288\), and bring down \(4\) to make \(384\). 5. Divide \(384\) by \(96\): write \(4\). The quotient is \(20{,}834\). 6. The zeros must remain aligned: one creates the zero quotient digit, and the next becomes part of the partial dividend \(800\). Check: \(20{,}834\times96=2{,}000{,}064\).

\(20{,}834\)

- Round both numbers to compatible values for a benchmark. - Estimate each quotient digit before multiplying and subtracting. - Use the exact inverse multiplication to complete the check.

1. Use compatible values: \(3{,}200{,}000\div320=10{,}000\), so the exact quotient should be near \(10{,}000\). 2. Divide \(345\) by \(321\): write \(1\), subtract \(321\), and bring down \(6\) to make \(246\). 3. Since \(246<321\), write \(0\) in the quotient and bring down \(8\) to make \(2468\). 4. Divide \(2468\) by \(321\): write \(7\), subtract \(2247\), and bring down \(4\) to make \(2214\). 5. Divide \(2214\) by \(321\): write \(6\), subtract \(1926\), and bring down \(9\) to make \(2889\). 6. Divide \(2889\) by \(321\): write \(9\). The exact quotient is \(10{,}769\). 7. The quotient is close to the estimate. Verify: \(10{,}769\times321=3{,}456{,}849\).

Estimate: about \(10{,}000\) Exact quotient: \(10{,}769\)

- Focus on the partial dividend directly below the missing quotient place. - Choose the largest single digit whose product with \(28\) fits in that partial dividend. - Keep every later quotient place aligned, including any zero place holder.

1. The first quotient digit is \(2\), leaving a remainder that forms the next partial dividend \(85\). 2. The next quotient digit must be the greatest whole number whose product with \(28\) does not exceed \(85\). 3. That digit is \(3\). The later \(0\) is still required when the next partial dividend is smaller than \(28\). 4. The completed quotient is \(2304\). Check: \(2304\times28=64{,}512\).

\(3\); the quotient is \(2304\).

- A subtraction row in long division is the divisor multiplied by the quotient digit for that step. - Use the quotient digit directly above the partial dividend \(229\). - Check that the completed subtraction row is not greater than the partial dividend.

1. The quotient digit above this step is \(6\), so the subtraction row must equal \(36\times6\). 2. Compute \(36\times6=216\). 3. Therefore, the missing digit is \(1\). 4. The completed written division gives \(58{,}968\div36=1638\).

\(1\)

- In each long-division step, the amount subtracted cannot exceed the current partial dividend. - Reconsider the first quotient digit before continuing any later steps. - After correcting the first step, complete the algorithm and check by multiplication.

1. The first quotient digit cannot be \(4\) because \(4\times23=92\), which is greater than the partial dividend \(74\). 2. The greatest whole-number multiple of \(23\) that fits in \(74\) is obtained with quotient digit \(3\). 3. Continuing the standard algorithm with valid quotient digits gives \(324\). 4. Check: \(324\times23=7452\).

The first quotient digit is too large because \(92>74\). The correct quotient is \(324\).

- Look for a nearby dividend that is an easy multiple of \(48\). - Use the estimate to predict the size of the exact quotient before dividing. - Compare the exact result with the estimate, then verify by multiplication.

1. Use the compatible value \(264{,}000\): \(264{,}000\div48=5500\), so the exact quotient should be close to \(5500\). 2. Apply the standard algorithm to \(263{,}520\div48\). 3. The exact quotient is \(5490\). 4. The exact quotient is only \(10\) less than the estimate, which is reasonable because the exact dividend is slightly less than \(264{,}000\). 5. Check: \(5490\times48=263{,}520\).

Estimate: about \(5500\) Exact quotient: \(5490\)

- For each quotient place, estimate how many groups of \(384\) fit in the current partial dividend. - Multiply \(384\) by that quotient digit to fill the hidden subtrahend, subtract, and then bring down the next dividend digit. - Use the multiplication check only after you have completed the hidden long-division rows.

1. The first usable partial dividend is \(1234\). Since \(3\times384=1152\) and \(4\times384>1234\), the first quotient digit is \(3\). Subtracting gives \(82\), then bringing down \(5\) gives \(825\). 2. Since \(2\times384=768\), the next quotient digit is \(2\). Subtracting gives \(57\), then bringing down \(6\) gives \(576\). 3. Since \(1\times384=384\), the next quotient digit is \(1\). Subtracting gives \(192\), then bringing down \(0\) gives \(1920\). 4. Since \(5\times384=1920\), the final quotient digit is \(5\), leaving remainder \(0\). 5. Therefore, the quotient is \(3215\). Check: \(3215\times384=1{,}234{,}560\).

Quotient: \(3215\) Subtrahends: \(1152,768,384,1920\) Later partial dividends: \(825,576,1920\) Remainder: \(0\)