Sum divides product?

The problem can be rewritten as:

$n!\not \equiv 0\quad \left( \text{mod }{\frac { n\left( n+1 \right) }{ 2 } } \right)$

---

For $n$ is odd:

Substitute $n=2k+1$ , it becomes obvious that

$(2k+1)! \equiv 0 \quad \left( \text{mod }{ \left(2k+1\right)\left(k+1\right) } \right)$ is always the case

---

For $n$ is even, it becomes trickier.

Substitute $n = 2k$ :

$(2k)! \equiv 0 \quad \left( \text{mod }{k\left(2k+1\right) } \right)$

Now, the above is only true if $2k+1$ is divisible by any number smaller than $k$ , and will not be true if and only if $2k+1$ is prime.

So, the only $n$ that will satisfy $n!\not \equiv 0\quad \left( \text{mod }{\frac { n\left( n+1 \right) }{ 2 } } \right)$ are when $n=\text{Prime} - 1$

Since there are $42$ primes between $2$ and $181$ , the answer is therefore $\boxed{42}$

---

UPDATE: The answer should be $41$ since $n=1$ isn't one of the solutions since it is an odd, despite it being a $\text{Prime}-1$

This problem is equivalent to finding the number $n$ in range $1 \le n \le 180$ such that the products of $1 ... n$ which are NOT divisible by sum of $1 ... n$ .

First simply the problem,

$\prod_{j=1}^n j = j!$

and

$\sum_{j=1}^n j = \frac{j (j-1)}{2}$

So that the problem is to find the number of $1 \le n \le 180$ such that $k$ is NOT integer

$n! = \frac{n (n + 1)}{2} k$

$(n-1)! = \frac{n + 1}{2} k$

$k = \frac{2(n - 1)!}{n + 1}$

Notice the denominator $n + 1$ . If it is composite number, it can be divided by $(n-1)!$ so that $k$ is integer. If it is $2$ (prime number), it can be canceled by the $2$ in numerator.

So the problem is to find the number of prime number in $[ 1, 180 + 1 ]$ , which is $42$ . And minus 1 because when $n = 1$ the denominator is canceled.

Therefore the answer is $41$

the product is j!; the sum is j(j+1)/2. j divides j!, and if j+1 is composite then j+1 divides j! since it is the product of q and r less than j. If j+1 is prime, however, the sum does not divide j! so if j+1 is prime the non-congruence will be satisfied. There are 42 primes 2<=p<=181, so the non congruence is satisfied 42 times