For n in N,the largest positive integer that | vedclass

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(A) Let $f(n) = 81^n + 20n - 1$. For $n=1$,$f(1) = 81 + 20 - 1 = 100$. For $n=2$,$f(2) = 81^2 + 20(2) - 1 = 6561 + 40 - 1 = 6600$. The largest integer

Vedclass · Accepted Answer

$117$

Vedclass · Answer

(A) Let $f(n) = 81^n + 20n - 1$. For $n=1$,$f(1) = 81 + 20 - 1 = 100$. For $n=2$,$f(2) = 81^2 + 20(2) - 1 = 6561 + 40 - 1 = 6600$. The largest integer $k$ that divides $f(n)$ for all $n \in N$ is the greatest common divisor of $f(1)$ and $f(2)$,which is $\gcd(100, 6600) = 100$. Thus,$k = 100 = 2^2 	imes 5^2$. The sum of all positive divisors $S$ of $k = 2^2 	imes 5^2$ is given by: $S = (1 + 2 + 2^2) 	imes (1 + 5 + 5^2) = (7) 	imes (31) = 217$. Therefore,$S - k = 217 - 100 = 117$.

For $n \in N$,the largest positive integer that divides $81^n + 20n - 1$ is $k$. If $S$ is the sum of all positive divisors of $k$,then $S - k =$

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