2016 AMC 10A Problem 25

Attempt Problem 25 of the 2016 AMC 10A 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 2016 AMC 10A solutions, or check the answer key.

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25.

How many ordered triples (x,y,z)(x,y,z) of positive integers satisfy lcm(x,y)=72,\text{lcm}(x,y) = 72,lcm(x,z)=600, \text{lcm}(x,z) = 600, and lcm(y,z)=900?\text{lcm}(y,z)=900?
1515
1616
2424
2727
6464
Answer: A
Concepts:least common multipleprime factorizationcasework
Small hint:
Treat each prime exponent independently
Big hint:
The powers of 33 and 22 each leave a small max-condition count
Solution:
Factor the given least common multiples: 72=23⋅32,600=23⋅3⋅52,900=22⋅32⋅52. \begin{gathered} 72=2^3\cdot3^2, \\ \quad 600=2^3\cdot3\cdot5^2, \\ \quad 900=2^2\cdot3^2\cdot5^2. \end{gathered} Since lcm⁡(x,y)\operatorname{lcm}(x,y) has no factor of 55, neither xx nor yy has a factor of 55, and zz must contain 525^2.
Write x=2a3bx=2^a3^b, y=2c3dy=2^c3^d, and z=2e3f52z=2^e3^f5^2. The power of 33 in lcm⁡(y,z)\operatorname{lcm}(y,z) forces d=2d=2, and the power of 22 in lcm⁡(x,z)\operatorname{lcm}(x,z) forces a=3a=3.
The remaining independent conditions are max⁡(b,f)=1\max(b,f)=1 and max⁡(c,e)=2\max(c,e)=2. These have 33 and 55 ordered choices, respectively, so there are 3⋅5=153\cdot5=15 triples.
Thus, the correct answer is A.
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