AMC 8 · 2005 · #5

Grade 4 arithmetic

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

optimization-countingdivisibility-rulesmultiplesmulti-digit-arithmetic optimization-countingsystematic-enumeration ↑ Prerequisites: multi-digit-arithmeticdivisibility-rules

📏 Medium solution 💡 3 insights

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Problem

Soda is sold in packs of 6, 12, and 24 cans. What is the fewest packs you can buy that gives exactly 90 cans?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

Tool #7 (Identify Subproblems) splits "minimum packs for 90 cans" into a short chain: peel off as many 24-packs as fit, then peel off 12-packs from what's left, then finish with 6-packs. Each sub-problem is one division-with-remainder, and stacking the largest packs first keeps the pack count as small as possible. Tool #3 (Eliminate Possibilities) then verifies the minimum: the smaller answer choice (A) 4 packs is impossible, so the candidate from Tool #7 really is the minimum.

1STEP 1

Start biggest: 90 ÷ 24 = 3 packs (72 cans), leaving 18 cans.

90 ÷ 24 = 3 R 18 → 3 packs of 24 = 72 cans, 18 cans left

2STEP 2

Next size: 18 ÷ 12 = 1 pack (12 cans), leaving 6 cans.

18 ÷ 12 = 1 R 6 → 1 pack of 12 = 12 cans, 6 cans left

3STEP 3

Last 6 cans = one 6-pack, 0 left — total is exactly 90.

6 ÷ 6 = 1 R 0 → 1 pack of 6 = 6 cans, 0 cans left

4STEP 4

Add the pack counts: 3 + 1 + 1 = 5 packs.

Total packs = 3 + 1 + 1 = 5

5STEP 5

Check (A) 4 fails: every pack is a multiple of 6, and no mix of 4 packs hits 90, so 5 is truly the minimum.

5 packs → (B)

Answer

Add the chosen packs back: 3 × 24 + 1 × 12 + 1 × 6 = 72 + 12 + 6 = 90. The total matches exactly, and the pack count is 5, so choice (B) is consistent. Choices (C) 6, (D) 8, and (E) 15 are all larger and would only be the answer if smaller counts didn't work; we already hit 90 with 5 packs, so they can't be minimal. Choice (A) 4 was ruled out above. Another quick check: 15 packs of 6 also gives 90, which is why (E) appears, but that's the maximum, not the minimum.

💡Key takeaway

When you want the fewest packs that total an exact amount, start with the biggest pack and peel off as many as fit, then move to the next size. That "biggest first" plan turns this AMC 8 problem into three short Grade 4 division-with-remainder steps.

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