d chloe and their daughter zoe all have the same birthday. joey is 1 year older than chloe, and zoe is exactly 1 year old today. today is the first of the 9 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?

Question

27-10-2022
Mathematics

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d chloe and their daughter zoe all have the same birthday. joey is 1 year older than chloe, and zoe is exactly 1 year old today. today is the first of the 9 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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Qwtiger Qwtiger · Answer 1 · 2022-10-31T14:16:32-04:00

11 years is the sum of Joey's two digits the next time his age is a multiple of Zoe's age.

Assume Chloe is c years old today, and Joey is c+1 years old today. Chloe and Zoe will be n+c and n+1 years old after n years.

Given,

(n + c) ÷ (n + 1) = 1 + ((c - 1) ÷ (n + 1))

For 9 non-negative integers, n is an integer. As a result, c - 1 has 9 positive divisors. c-1's prime factorization is either [tex]p^8[/tex] or p²q². Because c - 1 < 100, the only possible solution is c - 1 = 2² × 3² = 36, from which c = 37. We may deduce that Joey is 38 years old today.

Assume Joey's age is a multiplier of Zoe's age after k years, resulting in Joey and Zoe being k + 38 and k + 1 years old, correspondingly.

Given,

(k + 38) ÷ (k + 1) = 1 + ((38 - 1) ÷ (k + 1))

For some positive integers, k is an integer. As 37 is divisible by k + 1, the only possible solution is k = 36. Infer that Joey will be k + 38 = 74 years old at that time, yielding the value 7 + 4 = 11.

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