Let n = 1! + 4! + 7! + ldots + 400!. Then the | vedclass

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(B) We are given $n = 1! + 4! + 7! + \ldots + 400!$. First,calculate the values of the factorials: $1! = 1$ $4! = 24$ $7! = 5040$ $10! = 3628800$ For

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$6$

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(B) We are given $n = 1! + 4! + 7! + \ldots + 400!$. First,calculate the values of the factorials: $1! = 1$ $4! = 24$ $7! = 5040$ $10! = 3628800$ For any $k \ge 10$,$k!$ ends with at least two zeros,meaning the last two digits of $k!$ are $00$. Thus,the sum $n$ modulo $100$ is determined by the sum of the first few terms: $n \equiv 1! + 4! + 7! + 10! + \ldots + 400! \pmod{100}$ $n \equiv 1 + 24 + 40 + 0 + \ldots + 0 \pmod{100}$ $n \equiv 65 \pmod{100}$ The last two digits of $n$ are $65$. Therefore,the ten's digit of $n$ is $6$.

Let $n = 1! + 4! + 7! + \ldots + 400!$. Then the ten's digit of $n$ is

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