Find $n$ if $n\cdot n! + n! = 720$, where $n! = n\cdot (n-1)\cdot (n-2)\cdots 2\cdot 1$.
I notice that both terms on the left side of the equation have $n!$ as a factor, so I can factor it out and simplify the equation. I get $(n+1)\cdot n! = 720$. Now I need to find a value of $n$ such that $(n+1)\cdot n!$ is equal to $720$. I know that $720 = 2^4\cdot 3^2\cdot 5$, so I can try to match the factors of $(n+1)\cdot n!$ with the factors of $720$. I start by trying $n=1$, but then $(n+1)\cdot n! = 2\cdot 1! = 2$, which is too small. I try $n=5$, but then $(n+1)\cdot n! = 6\cdot 5! = 720$, which is exactly what I want. So I have found the value of $n$ that satisfies the equation. I check that it works by plugging it back into the original equation and verifying that both sides are equal. I get $5\cdot 5! + 5! = 5\cdot 120 + 120 = 600 + 120 = 720$, which is correct. 