Competition · AMC preparation · step 4 of 4

AMC 10 · 2020B · #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
📘 View easy version →
Problem
Count the positive perfect squares that are less than 2020 and are also even multiples of 3.

Pick an answer.

(A)
7
(B)
8
(C)
9
(D)
10
(E)
12

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

Rewrite as multiples of 6

"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

Show n is a multiple of 6

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

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

Compare each to 2020

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

Match against the choices

7 matches choice (A).

7 → (A)
Answer
7
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.

  • Rewrite as multiples of 6
  • Show n is a multiple of 6
  • List the squares in order
  • Compare each to 2020
  • Match against the choices

A parent dashboard for the family lives at sensimlab.com.

More like this

Same problem type. Tags show what each one shares with this problem.