AMC 8 · 2024 · #5

Grade 4 number-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In

divisibility-rulesfactorsmultiples caseworksystematic-enumeration ↑ Prerequisites: multi-digit-arithmeticfactors

📏 Long solution 💡 3 insights

Problem

Aaliyah rolls two standard 6-sided dice (each die shows a number from 1 to 6). The product of the two numbers happens to be a multiple of 6. Among the listed integers 5, 6, 7, 8, 9, we must find the one that CANNOT possibly be the sum of those two dice.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

There are only 5 candidate sums and at most a handful of dice pairs for each, so we can simply LIST every pair (d₁, d₂) with 1 ≤ d₁ ≤ d₂ ≤ 6 that adds to a given sum (Tool #2). For each sum we ELIMINATE the candidate the moment we find one pair whose product is a multiple of 6 — that proves the sum is reachable (Tool #3). The whole task splits into the SUBPROBLEM 'is d₁ d₂ divisible by 6?', which we break further into 'is it divisible by 2?' and 'is it divisible by 3?' (Tool #7). No algebra is needed.

1STEP 1

Since 6 = 2 × 3, a product is a multiple of 6 only when the two dice supply at least one even value and at least one multiple of 3.

6 = 2 × 3 → d₁ × d₂ ≡ 0 (mod 6) ⇔ (at least one even) AND (at least one multiple of 3)

2STEP 2

Pairs adding to 5: (1,4)→4 and (2,3)→6. Since 2 × 3 = 6 is a multiple of 6, sum 5 works, so eliminate (A).

1+4=5, 1× 4 = 4 (no) ; 2+3=5, 2× 3 = 6 ✓

3STEP 3

Pairs adding to 6: (1,5)→5, (2,4)→8, (3,3)→9. None is a multiple of 6, so sum 6 fails every pair.

1× 5 = 5 (no) ; 2× 4 = 8 (no) ; 3× 3 = 9 (no)

4STEP 4

Pairs adding to 7: (1,6)→6, (2,5)→10, (3,4)→12. Both 6 and 12 are multiples of 6, so sum 7 works — eliminate (C).

1× 6 = 6 ✓ ; 2× 5 = 10 (no) ; 3× 4 = 12 ✓

5STEP 5

Pairs adding to 8: (2,6)→12, (3,5)→15, (4,4)→16. Here 2 × 6 = 12 is a multiple of 6, so sum 8 works — eliminate (D).

2× 6 = 12 ✓ ; 3× 5 = 15 (no) ; 4× 4 = 16 (no)

6STEP 6

Pairs adding to 9: (3,6)→18, (4,5)→20. Since 3 × 6 = 18 is a multiple of 6, sum 9 works — eliminate (E).

3× 6 = 18 ✓ ; 4× 5 = 20 (no)

7STEP 7

A, C, D, E each yielded an explicit multiple-of-6 pair; only (B) survives, so 6 cannot be the sum.

Eliminated: 5, 7, 8, 9 → Impossible sum = 6 → (B)

Answer

Why is sum 6 the special case? A multiple of 6 needs BOTH a factor of 2 AND a factor of 3. The dice values that supply a factor of 3 are only 3 and 6. If one die is 6, the other must be 6 - 6 = 0 (impossible). If one die is 3, the other must be 6 - 3 = 3, but then the product is 3 × 3 = 9, which has no factor of 2. So no pair adding to 6 can carry both factors — exactly matching what our casework showed. The answer (B) 6 is consistent with the structural reason.

💡Key takeaway

This AMC 8 problem only needs Grade 4 multiples-and-factors thinking you already know!

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