Remainders • Topic 1 of 3

Remainder Theorem

The single habit that solves most remainder problems: never carry the full number — replace it by its remainder immediately. Because remainders respect addition and multiplication, you can reduce each factor mod n before combining. So the remainder of a product is the product of the remainders (then reduce again), and the remainder of a power a^k mod n is (a mod n)^k mod n. The CAT skill is spotting a cycle: powers of any base repeat their remainders periodically, so find the cycle length and reduce the exponent mod that length. For example 2^1,2^2,… mod 7 cycle through 2,4,1 with period 3, so 2^100 mod 7 = 2^(100 mod 3) = 2^1 = 2. Always shrink the base first, then hunt for the smallest power that gives 1 or −1.

✅ Solved examples

1. Find the remainder when 43 × 57 × 61 is divided by 8.
Reduce each: 43 ≡ 3, 57 ≡ 1, 61 ≡ 5 (mod 8). Product of remainders = 3 × 1 × 5 = 15 ≡ 7 (mod 8). Remainder = 7.
2. Find the remainder when 2^100 is divided by 7.
Powers of 2 mod 7 cycle 2,4,1 with period 3. 100 mod 3 = 1, so 2^100 ≡ 2^1 = 2 (mod 7). Remainder = 2.
3. Find the remainder when 3^50 is divided by 5.
Powers of 3 mod 5 cycle 3,4,2,1 with period 4. 50 mod 4 = 2, so 3^50 ≡ 3^2 = 9 ≡ 4 (mod 5). Remainder = 4.
4. Find the remainder when 15^23 × 23^15 is divided by 7.
15 ≡ 1, 23 ≡ 2 (mod 7). So expression ≡ 1^23 × 2^15 = 2^15. 2 has cycle 2,4,1 (period 3); 15 mod 3 = 0 ⇒ 2^15 ≡ last term of cycle = 1 (mod 7). Remainder = 1.

✏️ Practice — try these, take hints as needed

1. Remainder when 51 × 52 × 53 is divided by 5.
Reduce each mod 5.
51 ≡ 1, 52 ≡ 2, 53 ≡ 3.
Multiply 1 × 2 × 3 = 6.
1
2. Remainder when 4^61 is divided by 6.
Find the cycle of 4 mod 6.
4^1 ≡ 4, 4^2 ≡ 4, …
It is constant at 4 for all k ≥ 1.
4
3. Remainder when 7^84 is divided by 9.
7 ≡ −2 (mod 9), or find the cycle.
7^1=7, 7^2=4, 7^3=1 (period 3).
84 mod 3 = 0 ⇒ last term of cycle.
1
4. Remainder when 2^256 is divided by 17.
Find the smallest power of 2 that is 1 mod 17.
2^4 = 16 ≡ −1, so 2^8 ≡ 1 (period 8).
256 mod 8 = 0.
1
5. Remainder when 12^17 × 17^12 is divided by 5.
12 ≡ 2, 17 ≡ 2 (mod 5).
Expression ≡ 2^17 × 2^12 = 2^29.
2 cycles 2,4,3,1 (period 4); 29 mod 4 = 1.
2

📝 Topic test — 8 questions

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