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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.

Note by Anik Mandal
1 year 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 Shubhendra Singh · 1 year ago

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@Shubhendra Singh Thanks bro! Anik Mandal · 1 year ago

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Very easy problem bro :P

Hint : Find \({3}^{17} \mod 100\). Swapnil Das · 1 year ago

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@Swapnil Das Yes i know. But how to compute it? Anik Mandal · 1 year ago

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@Anik Mandal 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 · 1 year ago

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@Swapnil Das Yes correct! How you got that? Anik Mandal · 1 year ago

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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 ago

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@Kuldeep Sachan Suppose it is a big number.Then you can't calculate it..in this case you can Anik Mandal · 1 year ago

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Comment deleted Nov 19, 2015

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@Kay Xspre What is a more general way to do it using modulo operation? Yes i appreciate your process.Thanks for replying. Anik Mandal · 1 year ago

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