1996 AIME Problem 14

Attempt Problem 14 of the 1996 AIME 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 1996 AIME solutions, or check the answer key.

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14.

A 150×324×375150\times324\times375 rectangular solid is made by gluing together 1×1×11\times1\times1 cubes. An internal diagonal of this solid passes through the interiors of how many of the 1×1×11\times1\times1 cubes?
Answer: 768
Concepts:rectangular prisminclusion-exclusiongreatest common divisor
Small hint:
Count the coordinate-plane crossings of the space diagonal
Big hint:
Correct for crossings of two or three grid planes at once using greatest common divisors
Solution:
For an a×b×ca\times b\times c array, the diagonal crosses a−1,a-1, b−1,b-1, and c−1c-1 internal grid planes of the three orientations. Crossings of two orientations coincide gcd⁡(a,b)−1,\gcd(a,b)-1, gcd⁡(a,c)−1,\gcd(a,c)-1, and gcd⁡(b,c)−1\gcd(b,c)-1 times, and triple crossings occur gcd⁡(a,b,c)−1\gcd(a,b,c)-1 times. Adding one for the initial cube and applying inclusion-exclusion gives a+b+c−gcd⁡(a,b)−gcd⁡(a,c)−gcd⁡(b,c)+gcd⁡(a,b,c).\begin{gathered}a+b+c-\gcd(a,b)\\-\gcd(a,c)-\gcd(b,c)\\+\gcd(a,b,c).\end{gathered} Here the pairwise gcds are 6,6, 75,75, and 3,3, and the triple gcd is 3.3. Thus the number of cube interiors met is 150+324+375−6−75−3+3=768.\begin{gathered}150+324+375\\-6-75-3+3=768.\end{gathered}
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