Problem 21 · 2018 AMC 8
Hard
Number Theory
divisibilitycasework
How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?
Show answer
Answer: E — 5 integers.
Show hints
Hint 1 of 2
Three separate remainder conditions look like a mess — until you notice each remainder is exactly 4 short of its divisor (2 = 6−4, 5 = 9−4, 7 = 11−4). That shared "4 short" is the whole problem.
Still stuck? Show hint 2 →
Hint 2 of 2
The technique: if x is 4 short of a multiple of 6, of 9, and of 11, then x + 4 is a common multiple of all three — so it's a multiple of their lcm. Three conditions collapse into one.
Show solution
Approach: spot the common shift, then use lcm
- Rewrite each condition as "4 short": remainder 2 mod 6 means x + 4 is divisible by 6; remainder 5 mod 9 means x + 4 divisible by 9; remainder 7 mod 11 means x + 4 divisible by 11.
- So x + 4 is a multiple of lcm(6, 9, 11) = 2 · 32 · 11 = 198.
- Three-digit x means 100 ≤ x ≤ 999, so 104 ≤ x + 4 ≤ 1003. The multiples of 198 there are 198, 396, 594, 792, 990 — 5 values.
- You'll see it again: when every remainder is the same fixed amount below its divisor, shift the variable by that amount and the simultaneous system becomes a single lcm divisibility — the cleanest route past full Chinese Remainder Theorem casework.
Mark:
· log in to save