Question on Number System.

Question

If $N = 1!+2!+3!+...+10!$. What is the last digit of $N^{N}$?

Answer 1

For finding the last digit of any number we generally use,

$( n )\mod 10$ = last digit of the number.

Now for this question, $n = 1! + 2! + 3! + \ldots + 10!$, and we need to find $n^n \mod 10$.

First closely look that after $4!$ the last digit is always $0$. ( eg, $5! = 120, 6! = 720, 7! = 5040 \ldots $)

So first we add up $1! + 2! + 3! + 4! $ which is pretty trivial and straightforward and gets us  $1! + 2! + 3! + 4!  = 33 $ and $ 33 \mod 10 = 3$.

So this makes the question simplified as:

$$ 33^{1! + 2! + 3! + \ldots + 10!} \mod 10  $$

Proceeding,

$$ 33^{1! + 2! + 3! + \ldots + 10!} \mod 10 $$

$$ =  33^{33} \mod 10 $$

$$ = 33^{32} \cdot 33 \mod 10 $$,

$$ = 33^{8 \times 4} \cdot 33 \mod 10 = 33^{4 \times 8} \cdot 33 \mod 10 $$

$$ = 1^8 \cdot 33 \mod 10 = 33 \mod 10  $$

$$ = 3 $$

Hence the last digit is $3$.

Brute Force to verify.

I put the question directly into Wolfram, and here's the result,

 $$4037913^4037913 = 1.144730641841134696286784629687671473303988068399 \ldots \times 10^{26675087} $$

and the last few decimal digits are $ \ldots 878102895\textbf{3} $

Sure do comment if you find out another way.

Comments

how have u written 33^(4*8) as 1^(8) plz explain

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$33^4 \mod 10 $.
Let's forget about the power, and look at $33 \mod 10  = 3$ ( Remainder when 10 divides 33 )
Now, raise this to the power of 2, $ 33 \mod 10 = 3 \mod 10 $ becomes $33^2 \mod 10 = 3^2 \mod 10$. Continue in this way, $ 33^4 \mod 10 = 3^4 \mod 10 = 81 \mod 10 $. Now what's the remainder when 10 divides 81? It's 1.

Thus, raising $33^4$ to $x$ is equivalent to $1^{x} \mod 10$.

Sorry for the late reply, I was travelling.

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thanks

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$33^{1! + 2! + 3! + \ldots + 10!} \mod 10$

 How did the above expression turn into the below expression?

$33^{33} \mod 10$