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Hello J, Say, last two digits of 55^29 (50+5)^29 [29C1 * 50^1 * 5^28] + 5^29 => [29 * 50 * 5] + 25 => 50 + 25 = 75 Have I understood it? Regards 🙂 Harsh Reply
Do you know the binomial theorem? If not, it won’t make any sense. I have assumed thorough knowledge of the same… regards J Reply
CAT-holics 
Hello J,
Say, last two digits of 55^29
(50+5)^29
[29C1 * 50^1 * 5^28] + 5^29
=> [29 * 50 * 5] + 25
=> 50 + 25 = 75
Have I understood it?
Regards 🙂
Harsh
sir i did not understand the binomial expansion approach could you explain it in better way
Do you know the binomial theorem? If not, it won’t make any sense. I have assumed thorough knowledge of the same…
regards
J
for last 3 digit
We need to add the last 3 terms of binomial expression, right?
Yes, which is why it is unlikely to be asked except in the simplest of cases.
regards
J