The remainder when ...
.^5
x= .^.
5^5 (24 times 5 )
is divided by 24 is _______ .
2152=1mod24
taking any power n on both sides
52n=1mod24
multplying by 5
52n+1=5mod24
2n is even
2n+1 is odd
thus for even power we hav remainder as 1
and for odd power we hav remainder as 5
here we are having odd power so remainder shud be 5 [1]
13Thnx bhai, u're right...... But am ZERO in congruences...[2]
How cud it be done any other way ?
21using congruences is the shortest way to do it....i dun no of any other method......may be someone else here may give u another method......
btw for learning congruences check this thread---http://targetiit.com/iit-jee-forum/posts/congruences-a-z-11709.html.......a popular one dese dayz[3]
62even without congruences it is easy to argue..
think of this like:
5even number = 25number = (24+1)number which leaves a remainder 1 on division by 24
5Odd number= 5x5even number = 25numberx5 = 5 x (24+1)number which leaves a remainder 5 on division by 24
here the power is odd.. hence the answer is 5
62btw the congruences proofs are equivalent to binomial theorem most of the times[3]
1this same question was asked by karan a long time ago its easy
the rem is 5 when power is odd and 1 when its even its a tata mcgraw hill question