Number no.113
Number Theory
Level
pending
What is the remainder when 2^113+3^113 + 4^113 + 5^113 + 6^113 + . . . . . . + 112^113 is divided by 113?(x^y means 'x' raised to the power 'y'.)
The answer is 112.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
What is the remainder when 2^113+3^113 + 4^113 + 5^113 + 6^113 + . . . . . . + 112^113 is divided by 113?(x^y means 'x' raised to the power 'y'.)-----> This is also = (1^113+2^113+3^113+.........112^113 - 1)/113 = {(1^113 +112^113 )+(2^113 +111^113)+ ........+(57^113+56^113) - 1} /113. Now a^n + b^n is divisible by (a+b) if 'n' is odd. So the above expression will leave remainder (-1) that is, remainder will be 112.