What is the remainder when $1\times 1! + 2\times 2! + 3\times 3! +\ldots+ 11\times 11! + 3$ is divided by $12!$?

Question

What will be the remainder when $$1\times 1! + 2\times 2! + 3\times 3! +\ldots+ 11\times 11! + 3$$ is divided by $12!$ ?

I found out that the remainder will be same in case of $12!$ as well as $12$, but don't know how?

Kindly help me with this question.

Note: ! means Factorial

Answer 1

Notice that$$1*1! + 2*2! + 3*3! +\ldots+ 11*11! + 3=3+\sum_{n=1}^{11}n!*n\\=3+\sum_{n=1}^{11}n!*(n+1-1)\\=3+\sum_{n=1}^{11}(n+1)!-n!\\=3+12!-1=12!+2$$so the remainder is obviously $2$.

Answer 2

Write $1!$ as $(2-1)1!$. Leave out the $-1$ and the rest is a $2!$. Add this to the existing $2*2!$ which makes $3*2!=3!$ and so on with every next term, which means that your sum is $$12!-1+3$$ which means that the remainder is $2$.