Thanks for your attention to this question, here is the problem:
Compute the number of positive integers $x$ less than or equal to $1000$ that satisfy the following condition: $x! \text{ is divisible by } 1+2+3+...+x$
Thanks for your attention to this question, here is the problem:
Compute the number of positive integers $x$ less than or equal to $1000$ that satisfy the following condition: $x! \text{ is divisible by } 1+2+3+...+x$
We need to find $x$ such that $x!$ is divisible by $1+2+3+\cdots+x$. Note that $1+2+\cdots + x = \frac{x(x+1)}{2}$ Hence, we need $\frac{x!}{x(x+1)/2} = k \in \mathbb{Z}^+$ i.e. $\frac{2 \times (x-1)!}{x+1} = k \in \mathbb{Z}^+$.
Now we split it into five cases:
$1$. $x+1$ is an odd prime. If $x+1$ is an odd prime i.e. a prime greater than $2$, then we need $x+1$ to divide one of the numbers in the set $\{1,\ldots,x-1\} \cup \{2\}$ . But all the numbers in the set are smaller than $x+1$ and hence $x+1$ cannot divide any of the numbers. Hence, when $x+1$ is an odd prime, we cannot have $\frac{2 \times (x-1)!}{x+1}$ to be an integer whenever $x+1$ is an odd prime.
$2$. $x+1 = 2$. This can be easily checked that $x!$ is divisible by $1+2+3+\cdots+x$ i.e. $1!$ is divisible by $1$.
$3$. If $x+1$ is a composite number and is not of the form $p^2$, where $p$ is a prime, then it is not an irreducible. Hence $x+1 = ab$ where $1 < a < b
$4$. If $x+1$ is a composite number of the form $p^2$ where $p$ is an odd prime, then it immediately follows that $2p . Hence, a similar argument as $3$, by grouping $p$ and $2p$, gives us that $x+1$ divides $(x-1)!$. $5$. If $x+1$ is $2^2$, then again we can easily check that $4| \left(2 \times 2! \right)$. Hence, when $x+1$ is a composite number (or) $2$, we have that $(x+1) | \left( 2 \times (x-1)! \right)$. Hence, the numbers you are interested are $\{x \in \mathbb{Z}^+: x+1 \text{ is composite (or) }x+1 = 2\}$