Help needed!

Hey friends, I am stuck with a problem.Please help me in solving it out.

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

Thanks in advance for your time.

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
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`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

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

Shubhendra Singh - 5 years, 6 months ago

Thanks bro!

Anik Mandal - 5 years, 6 months ago

Very easy problem bro :P

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

Swapnil Das - 5 years, 6 months ago

Yes i know. But how to compute it?

Anik Mandal - 5 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)

Swapnil Das - 5 years, 6 months ago

@Swapnil Das – Yes correct! How you got that?

Anik Mandal - 5 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.

Kuldeep Sachan - 5 years, 6 months ago

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

Anik Mandal - 5 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$