Remainders in Elementary Number Problems

Learning outcome

Express division with a valid remainder and simplify elementary remainder problems involving sums, differences and products.

Concepts and reasons

Here the dividend N is a nonnegative (zero or positive) integer and divisor d is a positive integer. Write N = d × q + r: q is the whole-number quotient and r is the remainder, with 0 ≤ r < d.

The remainder is what remains after removing the maximum number of complete groups of size d. It cannot equal or exceed d: that would allow another complete group. A remainder of 0 means exact divisibility. When N < d, the quotient is 0 and remainder is N.

For the same divisor, replace numbers by their remainders when finding the remainder of a sum, difference or product. Then reduce the result again into the allowed range.

Why does replacement work? Write A = d × a + r and B = d × b + s. Adding or subtracting leaves r + s or r - s apart from multiples of d. Multiplication leaves r × s apart from multiples of d. Those complete multiples contribute no remainder.

A negative intermediate difference is not the final remainder under our convention. Add d to put it into the allowed range.

Worked examples

Example 1. Divide 157 by 12. Since 12 × 13 = 156, write 157 = 12 × 13 + 1. The remainder is 1, which lies between 0 and 11.

Example 2. A and B leave remainders 5 and 4 when divided by 7. Their sum leaves the remainder of 5 + 4 = 9, namely 2. Their product leaves the remainder of 5 × 4 = 20, namely 6.

Example 3. Suppose A ≥ B and their remainders upon division by 7 are 2 and 5. For A - B, the provisional difference is -3. Add 7: the required remainder is 4, not -3.

Common traps

A remainder is not a quotient or a decimal part. Divide oversized results again to find the remainder. Do not combine remainders obtained using different divisors, and never divide by zero.

Practice questions

1. Find the remainder when 246 is divided by 11.
2. N leaves remainder 5 upon division by 8. Find the remainder of 3N + 7 upon division by 8.
3. N leaves remainder 3 upon division by 5. Find the remainder of (N + 1)(N + 2) upon division by 5.

Answers and explanations

1. Since 11 × 22 = 242, write 246 = 11 × 22 + 4. The remainder is 4, smaller than 11.
2. Replace N by 5: 3 × 5 + 7 = 22. Since 22 = 8 × 2 + 6, the remainder is 6.
3. N + 1 leaves 4, while N + 2 leaves 0. Their product therefore leaves 4 × 0 = 0.