Math Book Cambodia: Find the last two digits of : `N=(1!+2!+3!+.......+101!)^101`

Saturday, August 28, 2021

Note: This is equivalent to finding $N(mod11)$ .

ie: The remainder when dividing $N$ by $100$ .

Observation: $10!\equiv0(mod100)$ Because, $10!=10...5...2$

Therefore, $N\equiv(1!+2!+3!+......+9!)^101(mod100)$

$N\equiv(1+2+6+24+20+20+40+20+80)^101(mod100)$

$N\equiv13^101(mod100)$

As we can see:

$13^2=169$

$13^4=..61$

$13^5=....93$

$13^20=(13^5)^4=....01$

Then, $N\equiv13(13^20)^5(mod100)$

Hence, The last two digits of $N$ is $13$ .

Math Book Cambodia