11thanks KR this is what i was looking for :)
THIS IS IN REFERENCE WITH http://targetiit.com/iit_jee_forum/posts/help_3741.html
AND MANY SIMILAR POSTS
LAST TWO DIGITS OF 3100
3100 = (310)10 = ((32)5)10
(32)5 = 9x9x9x9x9 = 729x9x9// WE TAKE ONLY LAST 2 DIGITS OF 729
// AND OF OTHER 3 DIGIT NO.S
= 29x9x9 mod 100
= 261x9 mod100 = 61x9 mod 100 //ONLY 69 OF 269 TAKEN
= 49 mod 100
so 3100 = 4910 mod 100
49100 = ((492)5)
492 = 2401 = 1 mod 100
3100 = (015)1 = 01 mod 100
last two digits of 3100 are 01
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3 Answers
341If you are using congruences its easier to check mod 4 and mod 25
We have 3 \equiv -1 \bmod 4 \Rightarrow 3^{100} \equiv 1 \bmod 4
By Euler Totient Theorem 3^{20} \equiv 1 \bmod 25 \Rightarrow 3^{100} \equiv 1 \bmod 25
Since gcd(4,25) = 1, we have 3^{100} \equiv 1 \bmod 100
In case you are not familiar using congruences use binomial theorem 3^{100} = (10-1)^{50} = 100k - \binom{50}{1} 10 + 1 = 100k-500+1 This immediately tells us that the remainder on division by 100 is 1.