The sum 1! + 4! + 7! + 10! + 12! + 13! + 15! | vedclass

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(C) We know that for any $n \ge 10$,$n!$ ends with at least two zeros,meaning $n!$ is divisible by $100$. Specifically,$10! = 3628800$. Calculating th

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$5$

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(C) We know that for any $n \ge 10$,$n!$ ends with at least two zeros,meaning $n!$ is divisible by $100$. Specifically,$10! = 3628800$. Calculating the sum: $1! = 1$ $4! = 24$ $7! = 5040$ Sum $= 1 + 24 + 5040 + 10! + 12! + 13! + 15! + 16! + 17!$ Sum $= 5065 + (10! + 12! + 13! + 15! + 16! + 17!)$ Since $10!, 12!, 13!, 15!, 16!, 17!$ are all divisible by $3! = 6$,we check $5065$. $5065$ is not divisible by $6$ (as it is odd). However,checking divisibility by $3! = 6$: $1! + 4! + 7! = 1 + 24 + 5040 = 5065$. $5065 \pmod{6} \equiv 5065 \pmod{2 	imes 3} \equiv 1 \pmod{2}$ and $5065 \pmod{3} \equiv 1$. Wait,let us re-evaluate: $1! = 1, 4! = 24, 7! = 5040$. $1 + 24 + 5040 = 5065$. $5065$ is divisible by $5$ because it ends in $5$. Thus,the sum is divisible by $5$.

The sum $1! + 4! + 7! + 10! + 12! + 13! + 15! + 16! + 17!$ is divisible by

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