Remainders

The number 17 is not divisible by 5. When you divide 17 by 5, using long division, you get a remainder: a number left over. In this case, the remainder is 2, as shown here:

You can also write that 17 is 2 more than 15, or 2 more than a multiple of 5. In other words, you can write 17 = 15 + 2 = 3 × 5 + 2. Every number that leaves a remainder of 2 after it is divided by 5 can be written this way: as a multiple of 5, plus 2.

On simpler remainder problems, it is often easiest to pick numbers. Simply add the desired remainder to a multiple of the divisor. For instance, if you need a number that leaves a remainder of 4 after division by 7, first pick a multiple of 7, such as 14. Then add 4 to get 18, which satisfies the requirement (18 = 7 × 2 + 4).

A remainder is defined as the integer portion of the dividend (or numerator) that is not evenly divisible by the divisor (or denominator). Here is an example written in fractional notation:

The quotient is the resulting integer portion that can be divided out (in this case, the quotient is 5). Note that the dividend, divisor, quotient, and remainder will always be integers. Sometimes, the quotient may be zero. For instance, when 3 is divided by 5, the remainder is 3 but the quotient is 0 (because 0 is the biggest multiple of 5 that can be divided out of 3).

Algebraically, this relationship can be written as the Remainder Formula:

This framework is often easiest to use on GRE problems when you multiply through by the divisor N:

Again, remember that x, Q, N, and R must all be integers. It should also be noted that R must be equal to or greater than 0, but less than N (the divisor).

Range of Possible Remainders

When you divide an integer by 7, the remainder could be 0, 1, 2, 3, 4, 5, or 6. Notice that you cannot have a negative remainder or a remainder larger than 7, and that you have exactly 7 possible remainders. You can see these remainders repeating themselves on the Remainder Ruler:

This pattern can be generalized. When you divide an integer by a positive integer N, the possible remainders range from 0 to (N − 1). There are thus N possible remainders. Negative remainders are not possible, nor are remainders equal to or larger than N.

If a ÷ b yields a remainder of 3, c ÷ d yields a remainder of 4, and a, b, c, and d are all integers, what is the smallest possible value for b + d?

The remainder must be smaller than the divisor, therefore, 3 must be smaller than b. Because b must be an integer, then b is at least 4. Similarly, 4 must be smaller than d, and d must be an integer, so d must be at least 5. Therefore, the smallest possible value for b + d is 4 + 5 = 9.

Remainder of 0

If x divided by y yields a remainder of 0 (commonly referred to as “no remainder”), then x is divisible by y. Conversely, if x is divisible by y, then x divided by y yields a remainder of 0 (or “no remainder”).

Similarly, if x divided by y yields a remainder greater than 0, then x is not divisible by y, and vice versa.

Arithmetic with Remainders

Two useful tips for arithmetic with remainders, if you have the same divisor throughout:

1. You can add and subtract remainders directly, as long as you correct excess or negative remainders. “Excess remainders” are remainders larger than or equal to the divisor. To correct excess or negative remainders, just add or subtract the divisor. For instance, if x leaves a remainder of 4 after division by 7, and y leaves a remainder of 2 after division by 7, then x + y leaves a remainder of 4 + 2 = 6 after division by 7. You do not need to pick numbers or write algebraic expressions for x and y. Simply write R4 + R2 = R6.
2. If x leaves a remainder of 4 after division by 7 and z leaves a remainder of 5 after division by 7, then adding the remainders together yields 9. This number is too high, however. The remainder must be non-negative and less than 7. You can take an additional 7 out of the remainder, because 7 is the excess portion. The correct remainder is thus R4 + R5 = R9 = R2 (subtracting a 7 out).
3. With the same x and z, subtraction of the remainders gives −1, which is also an unacceptable remainder (it must be non-negative). In this case, add an extra 7 to see that x − z leaves a remainder of 6 after division by 7. Using R’s, you can write R4 − R5 = R(−1)= R6 (adding a 7 in).
4. You can multiply remainders, as long as you correct excess remainders at the end.
5. Again, if x has a remainder of 4 upon division by 7 and z has a remainder of 5 upon division by 7, then 4 × 5 gives 20. Two additional 7’s can be taken out of this remainder, so x × z will have remainder 6 upon division by 7. In other words, (R4)(R5) = R20 = R6 (taking out two 7’s). You can prove this by again picking x = 25 and z = 12 (try the algebraic method on your own!):
   

   

Check Your Skills

25. What is the remainder when 13 is divided by 6?

26. What’s the first double-digit number that results in a remainder of 4 when divided by 5?

27. If x has a remainder of 4 when divided by 9 and y has a remainder of 3 when divided by 9, what’s the remainder when x + y is divided by 9?

28. Using the example from #27, what’s the remainder when xy is divided by 9?