Sample problem #1 (beginner): 65 ÷ 5. Place the 5 outside the division bar, and the 65 inside it. It should look like 5厂65, but with the 65 underneath the horizontal line. Sample problem #2 (intermediate): 136 ÷ 3. Place the 3 outside the division bar, and the 136 inside it. It should look like 3厂136, but with the 136 underneath the horizontal line.
Sample problem #1 (beginner): 65 ÷ 5. Place the 5 outside the division bar, and the 65 inside it. It should look like 5厂65, but with the 65 underneath the horizontal line. Sample problem #2 (intermediate): 136 ÷ 3. Place the 3 outside the division bar, and the 136 inside it. It should look like 3厂136, but with the 136 underneath the horizontal line.
In sample problem #1 (5厂65), 5 is the divisor and 6 is the first digit of the dividend (65). 5 goes into 6 one time, so place a 1 on the top of the divisor bar, aligned above the 6. In sample problem #2 (3厂136), 3 (the divisor) does not go into 1 (the first digit of the dividend) and result in a whole number. In this case, write a 0 above the division bar, aligned above the 1.
In sample problem #1 (5厂65), multiply the number above the bar (1) by the divisor (5), which results in 1 x 5 = 5, and place the answer (5) just below the 6 in 65. In sample problem #2 (3厂136), there is a zero above the division bar, so when you multiply this by 3 (the divisor), your result is zero. Write a zero on a new line just below the 1 in 136.
In sample problem #1 (5厂65), subtract the 5 (the multiplication result in the new row) from the 6 right above it (the first digit of the dividend): 6 - 5 = 1. Place the result (1) in another new row right below the 5. In sample problem #2 (3厂136), subtract 0 (the multiplication result in the new row) from the 1 right above it (the first digit in the dividend). Place the result (1) in another new row right below the 0.
In sample problem #1 (5厂65), drop the 5 from 65 down so that it’s beside the 1 that you got from subtracting 5 from 6. This gives you 15 in this row. In sample problem #2 (3厂136), carry down the 3 from 136 and place it beside the 1, giving you 13.
To continue 5厂65, divide 5 (the dividend) into the new number (15), and write the result (3, since 15 ÷ 5 = 3) to the right of the 1 above the division bar. Then, multiply this 3 above the bar by 5 (the dividend) and write the result (15, since 3 x 5 = 15) below the 15 under the division bar. Finally, subtract 15 from 15 and write 0 in a new bottom row. Sample problem #1 is now complete, since there are no more digits in the divisor to carry down. Your answer (13) is above the division bar.
For 3厂136: Determine how many times 3 goes into 13, and write the answer (4) to the right of the 0 above the division bar. Then, multiply 4 by 3 and write the answer (12) below the 13. Finally, subtract 12 from 13 and write the answer (1) below the 12.
For 3厂136: Continue the process for another round. Drop down the 6 from 136, making 16 in the bottom row. Divide 3 into 16, and write the result (5) above the division line. Multiply 5 by 3, and write the result (15) in a new bottom row. Subtract 15 from 16, and write the result (1) in a new bottom row. Because there are no more digits to carry down in the dividend, you’re done with the problem and the 1 on the bottom line is the remainder (the amount left over). Write it above the division bar with an “r. ” in front of it, so that your final answer reads “45 r. 1”.
In order to do short division, your divisor can’t have more than one digit. Sample problem: 518 ÷ 4. In this case, the 4 will be outside the division bar, and the 518 inside it.
In the sample problem, 4 (the divisor) goes into 5 (the first digit of the dividend) 1 time, with a remainder of 1 (5 ÷ 4 = 1 r. 1). Place the quotient, 1, above the long division bar. Place a small, superscript 1 beside the 5, to remind yourself that you had a remainder of 1. The 518 under the bar should now look like this: 5118.
In the sample problem, the number formed by the remainder and the second number of the dividend is 11. The divisor, 4, goes into 11 twice, leaving a remainder of 3 (11 ÷ 4 = 2 r. 3). Write the 2 above the division line (giving you 12) and the 3 as a superscript number beside the 1 in 518. The original dividend, 518, should now look like this: 51138.
In the sample problem, the next (and final) dividend number is 38—the remainder 3 from the previous step, and the number 8 as the last term of the dividend. The divisor, 4, goes into 38 nine times with a remainder of 2 (38 ÷ 4 = 9 r. 2), because 4 x 9 = 36, which is 2 short of 38. Write this final remainder (2) above the division bar to complete your answer. Therefore, your final answer above the division bar is 129 r. 2.
Your problem might be, for example, 3/4 ÷ 5/8. For convenience, use horizontal instead of diagonal lines to separate the numerator (top number) and denominator (bottom number) of each fraction.
In the sample problem, reverse 5/8 so the 8 is on top and the 5 is on the bottom.
For example: 3/4 x 8/5.
In this case, the numerators are 3 and 8, and 3 x 8 = 24.
The denominators are 4 and 5 in the sample problem, and 4 x 5 = 20.
In the sample problem, then, 3/4 x 8/5 = 24/20.
In the case of 24/20, 4 is the largest number that goes evenly into both 24 and 20. You can confirm this by writing out all of the factors of both numbers and picking out the largest number that is a factor of both: 24: 1, 2, 3, 4, 6, 8, 12, 24 20: 1, 2, 4, 5, 10, 20 Since 4 is the greatest common factor of 24 and 20, divide both numbers by 4 to reduce the fraction. 24/4 = 6 20/4 = 5 24/20 = 6/5. Therefore, 3/4 ÷ 5/8 = 6/5
In the sample problem, 5 goes into 6 one time with a remainder of 1. Therefore, the new whole number is 1, the new numerator is 1, and the denominator remains 5. As a result, 6/5 = 1 1/5.
As a beginner, start with a sample problem in which both numbers with exponents already have the same base—for instance, 38 ÷ 35.
In the sample problem: 8 - 5 = 3.
Therefore: 38 ÷ 35 = 33.
For the example 65. 5 ÷ 0. 5, 0. 5 goes outside the division bar, and 65. 5 goes inside it.
In the sample problem, you only need to move the decimal point over one spot for both the divisor and dividend. So, 0. 5 becomes 5, and 65. 5 becomes 655. If, however, the sample problem used 0. 5 and 65. 55, you’d need to move the decimal point 2 places in 65. 55, making it 6555. As a result, you’d also have to move the decimal point in 0. 5 2 places. To do this, you’d add a zero to the end and make it 50.
In the sample problem, the decimal in 655 would appear after the last 5 (as 655. 0). So, write the decimal point above the division line right above where that decimal point in 655 would appear.
Divide 5 into the hundredths digit, 6. You get 1 with a remainder of 1. Place 1 in the hundredths place on top of the long division bar, and subtract 5 from 6 below the number six. Your remainder, 1, is left over. Carry the first five in 655 down to create the number 15. Divide 5 into 15 to get 3. Place the three above the long division bar, next to the 1. Carry down the last 5. Divide 5 into 5 to get 1, and place the 1 on top of the long division bar. There is no remainder, since 5 goes into 5 evenly. The answer is the number above the long division bar (131), so 655 ÷ 5 = 131. If you pull out a calculator, you’ll see that this is also the answer to the original division problem, 65. 5 ÷ 0. 5.
