2024 AMC 12A Problem 12

Attempt Problem 12 of the 2024 AMC 12A 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 2024 AMC 12A solutions, or check the answer key.

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12.

The first three terms of a geometric sequence are the integers a, 720,a,\ 720, and b,b, where a<720<b.a\lt720\lt b. What is the sum of the digits of the least possible value of b?b?

99

1212

1616

1818

2121

Answer: E
Concepts:geometric sequencefactoroptimization
Difficulty rating: 1630
Solution:

Since the terms are geometric, 7202=ab,720^2=ab, so ab=518400=283452.ab=518400=2^8\cdot3^4\cdot5^2. Because b=518400/a,b=518400/a, minimizing bb means finding the smallest divisor of 518400518400 greater than 720.720. There is no divisor strictly between 720720 and 768:768: if its exponent of 55 is 2,1,2,1, or 0,0, then after dividing by 25,5,25,5, or 1,1, respectively, a number of the form 2i3j2^i3^j would have to lie in (28.8,30.72), (144,153.6),(28.8,30.72),\ (144,153.6), or (720,768);(720,768); the allowed powers i8,j4i\le8,j\le4 give none. Since 768=283768=2^8\cdot3 is a divisor, it is the least possible b.b. Its paired divisor is a=518400/768=675,a=518400/768=675, and the digit sum of bb is 7+6+8=21.7+6+8=21. Thus, the correct answer is E.

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