If P is the greatest divisor of 49^{n}+16n-1 | vedclass

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(C) Let $f(n) = 49^n + 16n - 1$. For $n=1$,$f(1) = 49^1 + 16(1) - 1 = 49 + 16 - 1 = 64$. For $n=2$,$f(2) = 49^2 + 16(2) - 1 = 2401 + 32 - 1 = 2432$. W

Vedclass · Accepted Answer

$7$

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(C) Let $f(n) = 49^n + 16n - 1$. For $n=1$,$f(1) = 49^1 + 16(1) - 1 = 49 + 16 - 1 = 64$. For $n=2$,$f(2) = 49^2 + 16(2) - 1 = 2401 + 32 - 1 = 2432$. We check the divisibility: $2432 / 64 = 38$. Since $f(n) = (1+48)^n + 16n - 1 = 1 + n(48) + \frac{n(n-1)}{2}(48^2) + \dots + 48^n + 16n - 1$. $f(n) = 48n + 16n + \frac{n(n-1)}{2}(2304) + \dots = 64n + 1152n(n-1) + \dots$. Since $1152$ is divisible by $64$ $(1152 = 64 	imes 18)$,the expression is always divisible by $64$. Thus,the greatest divisor $P = 64$. The factors of $64 = 2^6$ are $1, 2, 4, 8, 16, 32, 64$. The total number of factors is $6+1 = 7$.

If $P$ is the greatest divisor of $49^{n}+16n-1$ for all $n \in N$,then the number of factors of $P$ is

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