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2018 AMC 12B Problem 14

Problem 14 of 25IntermediateNumber Theory

Joey and Chloe and their daughter Zoe all have the same birthday. Joey is 11 year older than Chloe, and Zoe is exactly 11 year old today. Today is the first of the 99 birthdays on which Chloe’s age will be an integral multiple of Zoe’s age. What will be the sum of the two digits of Joey’s age the next time his age is a multiple of Zoe’s age?

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Solution

Let Chloe be nn today, so she is n1n-1 years older than Zoe. In yy years Chloe’s age n+yn+y is a multiple of Zoe’s age 1+y1+y exactly when 1+y1+y divides n1.n-1. Having 99 such birthdays means n1n-1 has exactly 99 divisors. A number with exactly 99 divisors has the form p2q2p^2q^2 for distinct primes p,q,p,q, or p8.p^8. Because Joey’s age at the requested future birthday has two digits, n1<99;n-1\lt99; the only possibility is 2232=36.2^2\cdot3^2=36. So Chloe is 3737 and Joey is 38.38. Joey’s age 38+y38+y is a multiple of 1+y1+y exactly when 1+y1+y divides 37.37. The next time is y=36,y=36, making Joey 74,74, with digit sum 7+4=11.7+4=11. Thus, the correct answer is E.

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Concepts: factor counting · divisibility

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.