2.3 Division Algorithm & Remainder Arithmetic

The division algorithm is the official remainder definition in GMAT Math Review 3.1. For any integer dividend N and any positive integer divisor d, there exist unique integers q (quotient) and r (remainder) such that

N = d × q + r, where 0 ≤ r < d.

The remainder is never negative and never equal to the divisor. When you divide by 7, the only possible remainders are 0, 1, 2, 3, 4, 5, and 6. Remainder 0 means d divides N.

GMAT Quantitative Reasoning almost always uses positive integers for both N and d. If a rare item uses a negative dividend, still force 0 ≤ r < d. Example: −17 divided by 5 is not “remainder −2.” Rewrite −17 = 5 × (−4) + 3 so the remainder is 3. Programming languages that return a negative modulo are not the GMAT convention.

If 0 < N < d, then q = 0 and r = N. Example: 4 divided by 9 is 0 remainder 4.

Remainder arithmetic

Remainders add, multiply, and exponentiate after you reduce each piece modulo d.

Sums

The remainder of (A + B) divided by d equals the remainder of (rem(A) + rem(B)) divided by d.

Example: remainder of 1,003 + 2,005 when divided by 7. 1,001 = 7 × 143, so 1,003 leaves 2. 2,002 = 7 × 286, so 2,005 leaves 3. 2 + 3 = 5, already less than 7, so the remainder is 5. If the sum of remainders meets or exceeds d, subtract d (or divide again) until the result sits in 0 through d − 1.

Products

The remainder of (A × B) divided by d equals the remainder of (rem(A) × rem(B)) divided by d.

Example: remainder of 73 × 89 when divided by 8. 72 = 8 × 9, so 73 leaves 1. 88 = 8 × 11, so 89 leaves 1. Product remainder 1.

If rem(A) × rem(B) is larger than d, reduce again. 11 × 11 divided by 8: each 11 leaves 3, and 3 × 3 = 9 = 8 + 1, remainder 1.

Powers

The remainder of A^k divided by d equals the remainder of [rem(A)]^k divided by d. Compute a short cycle of successive powers of that reduced base.

Example: remainder of 3^100 when divided by 7.

• 3^1 leaves 3
• 3^2 = 9 leaves 2
• 3^3 = 27 leaves 6, which is also −1 (mod 7)
• 3^6 = (3^3)^2 leaves (−1)^2 = 1

Then 3^100 = (3^6)^16 × 3^4 leaves 1^16 × 81. 81 = 7 × 11 + 4, remainder 4.

Remainder tables beat memory

For a fixed divisor, the integers that leave remainder r are an arithmetic sequence with difference d: r, r+d, r+2d, …. Every integer of the form d k + r is exactly that set.

A compact GMAT move: last-digit work is remainder-mod-10 work. Last-two-digit work is remainder-mod-100 work, which is why the divisibility-by-4 test looks only at the last two digits.

Units digits

The units digit of an integer is its remainder when divided by 10. Units digits of powers repeat in short cycles, never longer than 4.

Method: isolate the units digit of the base; divide the exponent by the cycle length (usually 4); the remainder picks the position in the cycle. Remainder 0 means you are at the last position of the cycle, not at “units digit 0.”

Worked: units digit of 7^85. Cycle of 7 is 7, 9, 3, 1. 85 divided by 4 leaves remainder 1, so the units digit is 7 (first position).

Worked: units digit of 2^34. Cycle 2, 4, 8, 6. 34 divided by 4 leaves remainder 2, so the units digit is 4.

Worked: units digit of 7^85 + 3^47 − 2^34. 3^47: cycle 3, 9, 7, 1 and 47 divided by 4 leaves 3, so units digit 7. Combine 7 + 7 − 4 = 10, units digit 0. If a subtraction of units digits goes negative, add 10 (borrow) before reading the units digit.

Two remainder conditions at once

If N leaves remainder 3 when divided by 7 and remainder 2 when divided by 5, list numbers of the form 7a + 3 and test the second condition:

3, 10, 17, 24, 31, …

17 divided by 5 leaves 2. First match: 17. The combined period is LCM(7, 5) = 35, so N = 35k + 17.

Equal-deficit shortcut

If each remainder is the same amount short of its divisor — remainder 4 when divided by 6, remainder 7 when divided by 9, remainder 13 when divided by 15 — each deficit is 2. Then N + 2 is a common multiple of 6, 9, and 15, so N + 2 is a multiple of LCM(6, 9, 15) = 90, and N = 90k − 2. The smallest N > 10 is 88.

Same idea: numbers that leave remainder 1 when divided by 2, 3, and 4 are 1 less than a multiple of LCM(2, 3, 4) = 12, namely 11, 23, 35, ….

Counting integers with a fixed remainder

The integers from 1 through 100 that leave remainder 3 when divided by 7 are 3, 10, 17, …, 94. That is an arithmetic sequence with first term 3, last term 94, common difference 7. Number of terms = ((94 − 3)/7) + 1 = 14.

Equivalently, N = 7k + 3 with 1 ≤ N ≤ 100 gives k from 0 through 13.

Trap: “how many multiples of 7 between 1 and 100” is remainder 0, first 7 last 98, count 14 — not 15, and not automatically the same count as remainder 3. Inclusive versus exclusive endpoints change the first or last term; always write the actual sequence before counting.

Another counting form: the number of integers from A through B that leave remainder r when divided by d equals the number of terms in {N : A ≤ N ≤ B and N = d k + r}. Solve for k and count integers k.