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.

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.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestLast 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 · 1 year, 10 months ago

Log in to reply

– Anik Mandal · 1 year, 10 months ago

Thanks bro!Log in to reply

Very easy problem bro :P

Hint: Find \({3}^{17} \mod 100\). – Swapnil Das · 1 year, 10 months agoLog in to reply

– Anik Mandal · 1 year, 10 months ago

Yes i know. But how to compute it?Log in to reply

– Swapnil Das · 1 year, 10 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)Log in to reply

– Anik Mandal · 1 year, 10 months ago

Yes correct! How you got that?Log in to reply

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 · 1 year, 10 months ago

Log in to reply

– Anik Mandal · 1 year, 10 months ago

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

Log in to reply

– Anik Mandal · 1 year, 10 months ago

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