Question

Let $A$ be the largest positive integer that divides all the numbers of the form $3^k + 4^k + 5^k$, and $B$ be the largest positive integer that divides all the numbers of the form $4^k + 3(4^k) + 4^{k+2}$, where $k$ is any positive integer. Then $(A + B)$ equals

Accepted Answer

82. First, find $A$: $A$ divides $3^k + 4^k + 5^k$ for all positive integers $k$. - For $k = 1$: $3^1 + 4^1 + 5^1 = 12$. - For $k = 2$: $3^2 + 4^2 + 5^2 = 9 + 16 + 25 = 50$. $ 	ext{gcd}(12, 50) = 2$. Thus, $A = 2$. Next, find $B$: $B$ divides $4^k + 3(4^k) + 4^{k+2} = 4^k (1 + 3 + 16) = 20 	imes 4^k$. - For $k = 1$: $20 	imes 4 = 80$. - For $k = 2$: $20 	imes 16 = 320$. $	ext{gcd}(80, 320) = 80$. Thus, $B = 80$. Then $A + B = 2 + 80 = 82$.

Largest Integer Divisors Sum A + B

Explanation

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