Problem:
Joey and Chloe and their daughter Zoe all have the same birthday. Joey is year older than Chloe, and Zoe is exactly year old today. Today is the first of the 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?
Answer Choices:
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Solution:
Let Chloe be years old today, so she is years older than Zoe. For integers , Chloe's age will be a multiple of Zoe's age years from now if and only if
is an integer, that is, is a divisor of . Thus has exactly positive integer divisors, so the prime factorization of has one of the two forms or . There are no two-digit integers of the form , and the only one of the form is . Therefore Chloe is years old today, and Joey is . His age will be a multiple of Zoe's age in years if and only if is a divisor of . The nonnegative integer solutions for are and , so the only other time Joey's age will be a multiple of Zoe's age will be when he is years old. The requested sum is .
The problems on this page are the property of the MAA's American Mathematics Competitions