AMC 8 · 2018 · #21

Grade 6 number-theory

Archetype Modular Cycle Frequency Count └ Sub-archetype Floor Count Across Residue Bins

modular-arithmeticlcmdivisibility-rulespattern-recognition pattern-recognitionmodular-arithmeticsystematic-enumeration ↑ Prerequisites: modular-arithmeticlcm

📏 Medium solution 💡 3 insights

Problem

We want to count the positive three-digit integers N (so 100 ≤ N ≤ 999) that leave 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 — all three conditions at once.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Look for a Pattern

The three remainder conditions look messy until you compare each divisor with its remainder: 6 - 2 = 4, 9 - 5 = 4, 11 - 7 = 4. That repeating 4 is a Tool #5 pattern: in every case, N is exactly 4 short of a multiple of the divisor. So N + 4 is a common multiple of 6, 9, and 11 — meaning N + 4 must be a multiple of lcm(6, 9, 11) = 198. Once we have N = 198k - 4, Tool #2 (Systematic List) takes over: list the values for k = 1, 2, 3, … in order and stop as soon as we leave the three-digit range, then count.

1STEP 1

Each divisor minus its remainder gives the same gap: 6-2 = 9-5 = 11-7 = 4, so N + 4 is a multiple of 6, 9, and 11.

6 - 2 = 9 - 5 = 11 - 7 = 4 → N + 4 is a multiple of 6, 9, and 11

2STEP 2

Combine the three into one requirement: lcm(6, 9, 11) = 2 × 3² × 11 = 198, so N + 4 must be a multiple of 198.

lcm(6, 9, 11) = 2 × 3² × 11 = 198

3STEP 3

So N + 4 = 198k for a positive integer k, giving the general form N = 198k - 4.

N + 4 = 198k ⟹ N = 198k - 4 (k = 1, 2, 3, …)

4STEP 4

List N = 198k - 4 and keep the three-digit values: 194, 392, 590, 788, 986; k = 6 gives 1184, past 999.

k=1: & N = 198 - 4 = 194 ✓ ; k=2: & N = 396 - 4 = 392 ✓ ; k=3: & N = 594 - 4 = 590 ✓ ; k=4: & N = 792 - 4 = 788 ✓ ; k=5: & N = 990 - 4 = 986 ✓ ; k=6: & N = 1188 - 4 = 1184 × (too big)

5STEP 5

Count the list — 194, 392, 590, 788, 986 — that's five integers, so the answer is (E).

{194, 392, 590, 788, 986} → 5 integers → (E)

Answer

Spot-check the smallest candidate N = 194: 194 ÷ 6 = 32 remainder 2 (✓), 194 ÷ 9 = 21 remainder 5 (✓), 194 ÷ 11 = 17 remainder 7 (✓). The solutions are spaced exactly 198 apart, so the count is the number of multiples of 198 in [104, 1003], which is ⌊

\frac{1003}{198}

⌋ - ⌈

\frac{104}{198}

⌉ + 1 = 5 - 1 + 1 = 5. That matches the answer (E).

💡Key takeaway

This AMC 8 problem only needs Grade 6 least common multiple plus a clever "divisor minus remainder" pattern you already know!

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