Find the last $2$ digits of the value of $(1! + 2! + 3! + 4! + .... + 2016!)^{42}$.

Question

I am stuck at the very last step. Please help me on this. My solution approach is as below :-

I know that last two digits of all the factorials more than $9!$ will have two or more zeroes at the end so I had considered only the factorials' sum till $9!$ so;
$1!+2!+3!+4!+5!+6!+7!+8!+9!$
$\Rightarrow 01 + 02+ 06+24 +120+720+5040+...20+...80=.......13$

$(1!+2!+3!+4!+5!+6!+7!+8!+9!)^{42}$
$\Rightarrow (......13)^{42}$

Now when I check the pattern of the last two digits when $13$ is raised to some power then I can't find any pattern except that the last digit will be in the pattern that is $3,9,7,1$. Please see below :-
$13^1=13$
$13^2=169$
$13^3=2197$
$13^4=28561$
$13^5=371293$
$13^6=4826809$

Then how can we determine the $2nd$ last digit of the given number? Please help me on this !!!

Thanks in advance !!!

Answer 1

Your question is equivalent to finding the remainder when $13^{42}$ is divided by 100. $13^{42}=(10+3)^{42}$ which after binomial expansion takes the form $3^{42}+420.3^{41}+100k$ for some positive integer $k$. So, the remainder is $3^{41}.423$. Now, $3^{41}=3(80+1)^{10}$ so, it's remainder (after binomial expansion) when divided by 100 is 3. So, the remainder of $13^{42}$ is same as remainder of $3.423$ when divided by 100 which is equal to 69. Hence, the answer is 69.

Answer 2

You can use binomial formula like this. Just a shortcut that is also mentioned in the linked site

Multiply the tens digit of the number with last digit of exponent to get the tens digit. The units digits resulted from the product will the second last digit.

You can use the above method if you are not supposed to show your work in the exam otherwise you can use the binomial formula used in the above link.

$13^{42}$ In this last digit will be $9$ (how) and second last digit will be $3 \times 2 \implies 6$

Answer 3

Here is some general method using the CRT. Set $$N=(1! + 2! + ... + 2016!)^{42}$$ so we are seeking some number $x$ s.t. $$N \equiv x \mod 100 \, .$$ Now let's first calculate the remainders of $N$ w.r.t. $4$ and $25$, since $100=4\cdot 25$: $$N\equiv(1+2!+3!+...)^{42} \equiv (1+2+6+0)^{42} \equiv 1^{42} \equiv 1 \mod 4$$ and similarly $$N\equiv (1+2!+3!+4!+5!+6!+7!+8!+9!+...)^{42} \\ \equiv(1+2+6+24+120+720+5040+40320+362880+...)^{42} \\ \equiv (1+2+6-1-5-5+15+20+5+0)^{42}\equiv 13^{42} \\ \equiv 13^{2\cdot 20 +2} \equiv 13^2 \equiv 19 \mod 25=5^2$$ since $\varphi(5^2)=4\cdot 5=20$.

Now $N=100k+x$ for some $k$ and it follows $$x \equiv 1 \mod 4 \\ x \equiv 19 \mod 25$$ which is essentially the statement of the CRT. You can solve this system by setting $$x=4m+1 =25n+19$$ for yet unknown integers $m$ and $n$. Taking this equation $\bmod 4$ gives $n\equiv 2 \mod 4$ and so $$x\equiv 25\cdot 2 + 19 = 69 \mod 100 \, .$$