For all positive integers n,if 3(5^{2n+1}) + | vedclass

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(C) Let $f(n) = 3(5^{2n+1}) + 2^{3n+1}$.For $n=1$,$f(1) = 3(5^3) + 2^4 = 3(125) + 16 = 375 + 16 = 391$.Since $391 = 17 	imes 23$,the expression is di

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$7$

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(C) Let $f(n) = 3(5^{2n+1}) + 2^{3n+1}$.For $n=1$,$f(1) = 3(5^3) + 2^4 = 3(125) + 16 = 375 + 16 = 391$.Since $391 = 17 	imes 23$,the expression is divisible by $k=17$.For $n=2$,$f(2) = 3(5^5) + 2^7 = 3(3125) + 128 = 9375 + 128 = 9503$.$9503 \div 17 = 559$,so it is divisible by $17$.Thus,$k=17$.The prime numbers less than or equal to $17$ are $2, 3, 5, 7, 11, 13, 17$.Counting these,we get $7$ prime numbers.

For all positive integers $n$,if $3(5^{2n+1}) + 2^{3n+1}$ is divisible by $k$,then the number of prime numbers less than or equal to $k$ is:

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