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 ðŸ™‚ 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

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 ðŸ™‚

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