For any integer n geq 1,the remainder when th | vedclass

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(B) Let $f(n) = n^5 - 5n^3 + 4n + 139$. We can factor the expression $n^5 - 5n^3 + 4n$ as $n(n^4 - 5n^2 + 4) = n(n^2 - 1)(n^2 - 4) = n(n-1)(n+1)(n-2)(

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$19$

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(B) Let $f(n) = n^5 - 5n^3 + 4n + 139$. We can factor the expression $n^5 - 5n^3 + 4n$ as $n(n^4 - 5n^2 + 4) = n(n^2 - 1)(n^2 - 4) = n(n-1)(n+1)(n-2)(n+2)$. Rearranging the terms,we get $f(n) = (n-2)(n-1)n(n+1)(n+2) + 139$. The product $(n-2)(n-1)n(n+1)(n+2)$ is the product of $5$ consecutive integers,which is always divisible by $5! = 120$. Therefore,$f(n) = 120k + 139$ for some integer $k$. To find the remainder when $f(n)$ is divided by $120$,we calculate $139 \pmod{120}$. $139 = 120 	imes 1 + 19$. Thus,the remainder is $19$.

For any integer $n \geq 1$,the remainder when the expression $n^5-5n^3+4n+139$ is divided by $120$ is

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