Remainders or what?

Number Theory Level 2

What is the remainder obtained when 4 3 101 + 2 3 101 43^{101} +23^{101} is divided by 66?


The answer is 0.

Yash Singhal by Yash Singhal , 22, India

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3 solutions

Sujoy Roy
Jul 6, 2014

a n + b n a^n+b^n is divisible by a + b a+b when n n is an odd number. Here, n = 101 n=101 .

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i do the same soo easy

Harshi Singh - 6 years ago

Write the sum as (33 + 10)^101 + (33 - 10)^101. When we expand using the binomial formula we find that all the remaining terms have 2 and 33 as factors, and hence the sum is divisible by 66.

Note that the fact that the exponent is odd is important, since then the expansion terms 10^101 and (-10)^101 cancel each other out.

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Federico Firoozi
Oct 26, 2014

Go here for a video solution -> http://youtu.be/tToYIv7rpDQ

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