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

Note by Anik Mandal
2 years, 3 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

- 2 years, 3 months ago

Thanks bro!

- 2 years, 3 months ago

Very easy problem bro :P

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

- 2 years, 3 months ago

Yes i know. But how to compute it?

- 2 years, 3 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)

- 2 years, 3 months ago

Yes correct! How you got that?

- 2 years, 3 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.

- 2 years, 3 months ago

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

- 2 years, 3 months ago

Comment deleted Nov 19, 2015

What is a more general way to do it using modulo operation? Yes i appreciate your process.Thanks for replying.

- 2 years, 3 months ago