Answer :
The following time Joey's age is greater than Zoe's, his two numbers add up to 11 years.
Consider Chloe to be c years old and Joey to be c+1 years old. After n years, Chloe and Zoe will be n + c and n + 1 respectively.
Based on the given conditions,
(n + c) ÷ (n + 1) = 1 + [(c - 1) ÷ (n + 1)]
n is an integer for 9 non-negative integers.
C – 1 hence has 9 positive divisors.
Either [tex]p^{8}[/tex] or [tex]p^{2} q^{2}[/tex] is the prime factorization of c-1 < 100
Since c - 1 < 100, is
c - 1 = [tex]2^{2} *3^{2}[/tex]
c - 1 = 36,
c = 36 + 1
We can calculate the coefficient,
c = 37
We can infer that Joey is 38 years old as of right now.
Assume that after k years, Joey's age is multiplied by Zoe's age, making Joey and Zoe k + 38 and k + 1 years old, respectively.
So,
We can write,
(k + 38) ÷ (k + 1) = 1 + [(38 - 1) ÷ (k + 1)]
K is an integer for some positive integers.
Since 37 can be divided by k + 1,
k = 36 is the only viable option.
Using the assumption that Joey will be k + 38 = 74 years old, the result is 7 + 4 = 11.
Therefore,
11 years is the sum of Joey's two digits the next time his age is a multiple of Zoe's age.
To learn more about information visit Age problems :
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