2016 AMC 10A Problem 25

Below is the professionally curated solution for Problem 25 of the 2016 AMC 10A, 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.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:least common multipleprime factorizationcasework

Difficulty rating: 2390

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

Solution:

Factor the given least common multiples: 72=2332,600=23352,900=223252.72=2^3\cdot3^2,\quad 600=2^3\cdot3\cdot5^2,\quad 900=2^2\cdot3^2\cdot5^2. 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 35=153\cdot5=15 triples.

Thus, the correct answer is A.

Problem 25 in Other Years