*Posted by Albert in : potd , trackback
*
Let $x$ be the highest positive integer such that $x!$ divides $$\sum\limits_{n=k}^{2011} n!$$ for $0 \leq k \leq 2011$. Find the number of integers $k$ such that $x \neq k$.

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**Solution:**
We can expand the sum to be: $\sum\limits_{n=k}^{2011} n! = k! (1 + (k+1)(1 + (k+2)(\cdots)))$, so clearly, $k!$ divides $\sum\limits_{n=k}^{2011} n!$.

To find if there is any $x!$ larger than $k!$ that divides $\sum\limits_{n=k}^{2011} n!$, we notice that if such a value exists, then $(k+1)!$ must also divide $\sum\limits_{n=k}^{2011} n!$, since $(k+1)!$ divides $x!$ for integer $x > k$. This means $k+1$ must divide $(1 + (k+1)(1 + (k+2)(\cdots)))$.

However, $k+1$ divides $(k+1)(1 + (k+2)(\cdots))$, which means $k+1$ must also divide $1$. The only value for which this is true (which is also in the range $0 \leq k \leq 2011$) is when $k = 0$.

Therefore, there is $\boxed{1}$ value of $k$ such that $x \neq k$.

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