AMC 10 · 2020 · #7

Grade 5 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration

multiplesperfect-squaresdivisibility-rulesbound-inequality-then-enumerate easier-related-problembound-inequality-then-enumeratesystematic-enumeration ↑ Prerequisites: perfect-squaresdivisibility-rules

📏 Short solution 💡 2 insights

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Problem

Count the positive perfect squares that are less than 2020 and are also even multiples of 3.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

Tool #9 (Easier) replaces "even multiple of 3" with the simpler equivalent "multiple of 6". Tool #2 then lists the candidates 6², 12², 18², … in order and stops when they exceed 2020. Tool #5 reveals these are exactly the squares (6k)² = 36k², so we just need to count how many k work. Tool #3 confirms against the answer choices.

1STEP 1

"Even multiple of 3" means a multiple of both 2 and 3 — that is just a multiple of 6.

even ∩ multiple of 3 = multiple of 6

2STEP 2

Since 2 and 3 are prime, n² being a multiple of 6 forces n = 6k — so we need 36k² under 2020.

n = 6k, k ∈ {1, 2, 3, …}, n² = 36k² ＜ 2020

3STEP 3

List the squares in order until one passes 2020. With n = 6k, compute 36k² for k = 1, 2, 3, …

k=1: 36, k=2: 144, k=3: 324, k=4: 576, k=5: 900, k=6: 1296, k=7: 1764, k=8: 2304

4STEP 4

The first seven squares stay under 2020; the eighth (2304) overshoots — so k runs 1 to 7, giving 7 values.

1764 ＜ 2020 ＜ 2304 → k ∈ {1, 2, …, 7}

5STEP 5

7 matches choice (A).

7 → (A)

Answer

Cross-check with a bound argument: 36k² ＜ 2020 ⇔ k² ＜ 56.1 ⇔ k ≤ 7 (since 7² = 49 ＜ 56 but 8² = 64 ＞ 56). So k ranges over 1, 2, …, 7 — exactly 7 values. ✓ The listed squares 36, 144, 324, 576, 900, 1296, 1764 are all even (end in even digits), all multiples of 3 (digit sums 9, 9, 9, 18, 9, 18, 18), and all under 2020. ✓

💡Key takeaway

This AMC 10 problem only needs Grade 5 multiplication and factor reasoning you already know — "even multiple of 3" just means "multiple of 6", so list the squares (6 · 1)², (6 · 2)², … until one passes 2020. The first seven (36, 144, 324, 576, 900, 1296, 1764) fit, the eighth (2304) doesn't — answer 7.

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