# Help needed!

How to find the last two digits of $3^{17}$?

Note by Anik Mandal
4 years, 6 months ago

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Last two digits of a number is the remainder obtained on dividing that number by 100

$3^{17}=3.9^{8}=3.(10-1)^{8}$

In The expansion of $(10-1)^{8}$ all terms are divisible by 100 except two terms which are

$^{8}C_{7}.10.(-1)^{7}+(-1)^{8}=-79$ Remember the 3 which we had separated before, multiply it by - 79 to get - 237

On dividing - 237 by 100 we get the remainder as 63

- 4 years, 6 months ago

Thanks bro!

- 4 years, 6 months ago

Very easy problem bro :P

Hint : Find ${3}^{17} \mod 100$.

- 4 years, 6 months ago

Yes i know. But how to compute it?

- 4 years, 6 months ago

Hmm.. answer comes $63$. Is that correct? ( I wasted $2$ pages of my copy to discover that ${ 3 }^{ 15 }\equiv 7 \mod 100$ :P)

- 4 years, 6 months ago

Yes correct! How you got that?

- 4 years, 6 months ago

If you calculate 3^17,you will get the answer,129140163,and to extract its 2 digits from right,mod it by 100 that is, number%100.Then you would get the answer 63.

- 4 years, 6 months ago

Suppose it is a big number.Then you can't calculate it..in this case you can

- 4 years, 6 months ago