2007 AIME I Problem 1

Attempt Problem 1 of the 2007 AIME I below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2007 AIME I solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

1.

How many positive perfect squares less than 10610^6 are multiples of 24?24?
Answer: 83
Concepts:perfect squareprime factorizationmultiple
Small hint:
A perfect square is a multiple of 24=23⋅324 = 2^3 \cdot 3 exactly when its square root is a multiple of 1212
Big hint:
Count the multiples of 1212 whose squares are less than 106,10^6, i.e. the multiples of 1212 less than 10001000
Solution:
Since 24=23⋅3,24 = 2^3 \cdot 3, a square N2N^2 is a multiple of 2424 exactly when NN is a multiple of 12:12: the factor 232^3 forces NN to contain 22,2^2, the factor 33 forces NN to contain 3,3, and conversely (12m)2=144m2(12m)^2 = 144m^2 is always a multiple of 24.24. Also N2<106N^2 \lt 10^6 exactly when N<1000.N \lt 1000.
The multiples of 1212 less than 10001000 are 12,24,…,996,12, 24, \ldots, 996, and there are 99612=83\frac{996}{12} = 83 of them.
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