How Would u solve the following problem

Is (26^5+ 27^5+ 28^5+29^5) divisible by 110.

Rule no simplifying into simple number and checking it 😀

lets see who can solve it the ramanujan way

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July 5, 2007 by Dilip Maripuri

How Would u solve the following problem

Is (26^5+ 27^5+ 28^5+29^5) divisible by 110.

Rule no simplifying into simple number and checking it 😀

lets see who can solve it the ramanujan way

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on July 5, 2007 at 9:17 pm |Sanjay110 = 2 * 55.

And these two are coprime.

Now,

26^5 + 27^5 + 28^5 + 29^5 is divisible by 2.

(a^n + b^n is always divisible by a+b if n is odd)

26^5 + 29^5 is divisible by 55

27^5 + 28^5 is divisible by 55.

26^5 + 27^5 + 28^5 + 29^5 is divisible by 55.

There fore 26^5 + 27^5 + 28^5 + 29^5 is divisible by 110 😀