Find the last two digits of 2^2016

What are the last two digits in 22016 ?

We essentially have the question find (22016)mod100 , now we know that if (22016)mod25=n , then (22016)mod100 must be either n,n+25,n+50 , or n+75 .

Think about this step and understand why n,n+25,n+50 , or n+75 must be the answer mod 100 before you go on.

Now since (22016)mod4 is obviously 0 , and we know that adding by 25 increases a number in modulo 4 by 1, we can deduce that only 1 out of n,n+25,n+50 , or n+75 will be divisible by 4 and the one that is divisible by 4 will be our answer since we determined that our final answer has to be divisible by 4. So for now we can simplify our question down to (22016)mod25 now if we take 2(10) which is 1024 and mod it by 25 we get 24 or −1 . so we can rewrite our equation as (((210)201)∗26))mod25 in other words (((−1)201)∗26))mod25 which eventually boils down to - 64mod25 which simplifies to (−64+25+25+25)mod25 or 11mod25 . Now since 11 is not divisible by 4 we have to keep adding 25’s until it is. Ex. 11,36 and since 36 is divisible by 4 we have an answer.

Thus (22016)mod100=36

v,jhvljh,b.b; hk; ljujb,vuvccklokll; k,m,km,

step-by-step explanation: