Thread: Finding last two digits of a large number

Hello, I have a problem solving this :

23^(24^25)

I am supposed to find the last two digits of this number. I know I have to find the remainder after the modulo 100 operation but all together I'm lost. Thank you for help.

Originally Posted by merlin1x1

Hello, I have a problem solving this :

23^(24^25)

I am supposed to find the last two digits of this number. I know I have to find the remainder after the modulo 100 operation but all together I'm lost. Thank you for help.

First find y(1)=23^24 mod 10. Now construct the sequence:

y(n)=y(n-1)^24 mod 10

so y(n)=23^(24^n) mod 10

CB