Odd Divisibility!

Algebra Level 2

$\large \left( { 65 }^{ 101 }+{ 23 }^{ 101 } \right) \left( { 32 }^{ 101 }+{ 56 }^{ 101 } \right) \$

The above product is divisible by?

88 42 24 Cannot be determined

2 solutions

Mohit Gupta
Jun 1, 2015

$a^n +b^n$ is divisible by $a+b$ when $n$ is odd

Moderator note:

Yes, but why is it always true?

Praful Jain
Jun 6, 2015

After multiplying all we get (2080)^101+(3640)^101+(736)^101+(1288)^101 now all these are not the multiple of 3 or 6 so in the options given 88 is the only no which is divisible by 2 and not by 6 so the answer is 88

Odd Divisibility!

2 solutions

Moderator note:

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