Latest self Aptitude Question SOLUTION: The remainder when 17^37+29^37 is divided by 23 is

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(#M40013807) SELF question Self Keep an EYE

The remainder when 17^37+29^37 is divided by 23 is

(11 years ago)

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Vinay Sipani   (11 years ago)

Remainder is 0

(17)^x + (29)^x is always divisible by 23 where x is an odd number

Eg:- 17^3 + 29^3 divisible by 23

=> (17)^37 + (29)^37 divisible by 23
=> remainder is 0.

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Garima  (11 years ago)

If n is odd, a^n + b^n = (a + b)(a^(n - 1) - a^(n - 2)*b + a^(n - 3)*b^2 -... + b^(n - 1)).
Therefore, a^n + b^n is divisible by (a + b) when n is odd

so when 17^37+29^37 is divided by 23, remainder is zero as (17^37+29^37) is divisible by (17+29) or 46 which is divisible by 23.

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M.R.kartheek  (11 years ago)

when we apply a remainder theorem then we will get (-6)^37 + (6)^37 and the power is 37 it is odd so (-6)^37 is negative value and
(6)^37 is positive value, both magnitudes are equal then the remainder is zero...

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seema   (11 years ago)

0
37(17+29)/23=74
remainder 0

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