What is the remainder when $N = (1! + 2! + 3! + 4! + ........... + 1000! )^{40}$ is divided by $10?$

Question

What is the remainder when $N = (1! + 2! + 3! + 4! + ........... + 1000! )^{40}$ is divided by $10$ ?

My try:

On watching the pattern as it grows, after $4!$ all are divisible by $10$.

So, infact I am just left with $N = (1! + 2! + 3! + 4! + 0)^{40}$ and I need to check the remainder when this $N$ is divisible by $10$.

Hence, the $N$ sums up to $33^{40}$ when divided by $10$ .

Now, after this I can simply apply Euler's Theorem such that

$33^{4} = 1 (mod 10)$

After all, the remainder comes out to be $1$.

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I don't have an answer for this. Is my understanding right or did I miss something?

Answer 1

Your answer is correct. A few pointers, however:

1. Note that you can reduce $33$ to just $3$
2. Euler's theorem says that $3^{4}\equiv 1\pmod{10}$

Answer 2

Go to the basics.

Let $(1! + 2! + 3! + 4! ........+ 1000!)=x$.

Now it is clear that unit digit of $x$ will be $3$. (Why??)

Also, If a number is divided by $10$, the remainder is the unit digit.

let us see what will be the unit digit of $x^{40}$.

Notice that unit digits of powers of $3$ get repeated in pattern, as $3^0=1, 3^1=3, 3^2=9, 3^3=27$. Follow this pattern and you will find that unit digit of something like, $3^{40}$ will be $1$. I shall let you conclude now.