AMC 10 Step by Step

2023 AMC 10A Problem 3

P1-10~2 minPrint

How many positive perfect squares less than 20232023 are divisible by 55 ?

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Key insight

A square is a multiple of 5 only when the number squared is a multiple of 5. Count 5, 10, 15, and so on while the square stays below 2023.

Solution

  1. First decide which squares are divisible by 55. Try a few: 52=255^2 = 25 and 102=10010^2 = 100 are, but 62=366^2 = 36 and 72=497^2 = 49 are not. Since 55 is prime, 55 can only divide n×nn \times n if it divides nn itself. So we want the squares of 5,10,15,20,…5, 10, 15, 20, \ldots
  2. Next find where the squares stop being less than 20232023. We know 402=160040^2 = 1600 and 452=202545^2 = 2025.
  3. 20252025 is just above 20232023, so 4545 is not allowed. The last one that works is 4040.
  4. List them: 5,10,15,20,25,30,35,405, 10, 15, 20, 25, 30, 35, 40. Their squares 25,100,225,400,625,900,1225,160025, 100, 225, 400, 625, 900, 1225, 1600 are all below 20232023 and all divisible by 55.
  5. That is 88 squares.

The answer is (A) 8\boxed{\textbf{(A)}\ 8}.

Why this works

A prime number divides a square only when it divides the number being squared, so the question about squares turns into a question about multiples of 55. The only tricky part is the cutoff: 452=202545^2 = 2025 misses 20232023 by just 22, and that single case decides between (A) and (B). Always check the last case by hand.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Because 55 is prime, 55 divides n2n^2 exactly when 55 divides nn. So we need to count positive multiples of 55 whose squares are below 20232023.

Since 442=1936<202344^2 = 1936 < 2023 and 452=2025>202345^2 = 2025 > 2023, the roots allowed are n=1,2,…,44n = 1, 2, \ldots, 44. The multiples of 55 in this range are 5,10,15,…,405, 10, 15, \ldots, 40, which is 88 numbers.

Their squares are 25,100,225,…,160025, 100, 225, \ldots, 1600, all below 20232023 and all divisible by 55.

The answer is (A) 8\boxed{\textbf{(A)}\ 8}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.