Remainder: The Leftover After Division
Key Remainder Properties
| Property | Rule |
|---|---|
| Sum of remainders | $\text{rem}(a+b) = \text{rem}(\text{rem}(a) + \text{rem}(b))$ |
| Product of remainders | $\text{rem}(a \times b) = \text{rem}(\text{rem}(a) \times \text{rem}(b))$ |
| Remainder of $n$ divided by $n$ | Always 0 |
| If $r = 0$ | $n$ is divisible by $d$ (no remainder) |
Cyclical Patterns & Units Digits
The units digit of any number is the remainder when you divide by 10. Powers of integers cycle in their units digits — use this for "units digit of large powers."
Remainder in DS Problems
10 Remainder Traps
1. Remainder is always $< d$
The remainder when dividing by 5 can only be 0,1,2,3,4 — never 5 or above.
2. $n = qd + r$ form is universally useful
Express "n leaves remainder 3 when divided by 7" as $n = 7k+3$ for algebraic work.
3. Remainder of 0 means divisible
If $n ÷ d$ has remainder 0, then $d | n$ (d divides n).
4. Units digit cycles: find cycle length first
Powers of 4 cycle with period 2: {4,6}. Powers of 7 cycle with period 4: {7,9,3,1}.
5. Remainder of product = product of remainders (mod d)
$\text{rem}(a \times b, d) = \text{rem}(\text{rem}(a,d) \times \text{rem}(b,d), d)$. Useful for large powers.
6. Don't assume integer division
$17/5 = 3.4$, not 3. Quotient is 3 (floor) and remainder is 2.
7. Adding remainders may exceed d
If $\text{rem}(a)=4$ and $\text{rem}(b)=4$ (both mod 5), their sum remainder = $(4+4) \mod 5 = 3$, not 8.
8. Remainder problems in DS: test multiple values
Two numbers can have the same remainder for different reasons — don't conclude uniqueness too quickly.
9. Large exponents: use remainder of exponent ÷ cycle length
To find $3^{47} \mod 10$: cycle length 4, $47 \mod 4 = 3$ → 3rd element of {3,9,7,1} = 7.
10. Remainder when dividing by 1
Any number divided by 1 has remainder 0. Trivially true but sometimes appears in DS.
10 GMAT Practice Questions
What is the remainder when 137 is divided by 7?
(D) 4. $137 = 7 \times 19 + 4$. Remainder = 4.
When a positive integer $n$ is divided by 6, the remainder is 4. What is the remainder when $n + 10$ is divided by 6?
(B) 2. $n = 6k+4$ for some integer $k$. $n+10 = 6k+14 = 6(k+2)+2$. Remainder = 2.
What is the units digit of $7^{25}$?
(C) 7. Units digits of powers of 7 cycle: {7,9,3,1} (period 4). $25 \mod 4 = 1$. 1st position → 7.
When positive integer $p$ is divided by 8, the remainder is 5. What is the remainder when $3p$ is divided by 8?
(D) 7. $p \equiv 5 \pmod{8}$. $3p \equiv 15 \pmod{8}$. $15 = 8+7$ → remainder = 7.
If $n$ leaves a remainder of 3 when divided by 9, what is the remainder when $n^2$ is divided by 9?
(D) 6. Wait — $n \equiv 3 \pmod{9}$. $n^2 \equiv 9 \equiv 0 \pmod{9}$. Hmm: $3^2=9$, and $9 \mod 9 = 0$. So remainder = 0. (A) 0 is correct.
What is the units digit of $2^{100}$?
(C) 6. Units digits of powers of 2 cycle: {2,4,8,6} (period 4). $100 \mod 4 = 0$ → 4th element = 6.
When integer $n$ is divided by 5, what is the remainder?
(1) When $n$ is divided by 10, the remainder is 7.
(2) When $n+3$ is divided by 5, the remainder is 0.
(D) Each alone sufficient. (1): $n = 10k+7$ → $n \mod 5$: $10k$ leaves 0, 7 leaves 2 → remainder = 2. Sufficient. (2): $n+3 \equiv 0 \pmod{5}$ → $n \equiv -3 \equiv 2 \pmod{5}$. Remainder = 2. Sufficient. Both → (D).
The sum of three consecutive integers leaves a remainder of 1 when divided by 3. The integers could be:
(E) Impossible. Three consecutive integers: $n + (n+1) + (n+2) = 3n+3 = 3(n+1)$. This is always divisible by 3, leaving remainder 0. It can NEVER leave a remainder of 1.
What is the remainder when $17^{12} \times 13^7$ is divided by 5?
(A) 0. Wait — $17 \mod 5 = 2$. $2^{12} = 4096$; $4096 \mod 5$: cycle of 2 mod 5 is {2,4,3,1} period 4. $12 \mod 4 = 0$ → 4th element = 1. $13 \mod 5 = 3$. Cycle of 3: {3,4,2,1} period 4. $7 \mod 4 = 3$ → 3rd element = 2. Product remainder: $1 \times 2 = 2 \mod 5 = 2$. (C) 2 is correct.
If positive integer $k$ has a remainder of 2 when divided by 6, which of the following could be a value of $k$?
(D) 20. $k \equiv 2 \pmod{6}$: valid values are 2,8,14,20,26,... $14 \mod 6 = 2$ ✓. Wait, 14 is also valid! $14 = 6(2)+2$, remainder 2. So (A) 14 is also correct. But the answer choices list 14 as (A). Both (A) 14 and (D) 20 work. The first valid answer in the list: (A) 14.
Lesson Summary — Key Takeaways
n = qd + r, where 0 ≤ r < d
The fundamental remainder equation. Express any remainder problem in this form.
Rem(a×b) = rem(rem(a) × rem(b))
Work with remainders throughout. No need to compute the full product.
Units digit cycles: period ≤ 4
Powers of 2,3,7,8 cycle with period 4. Powers of 4,9 cycle with period 2. Powers of 5,6 are constant.
DS: express as kd + r
"Remainder r when divided by d" → $n = kd+r$. Use this form algebraically.