IIT JEE , AIEEE Forum - Mathematics , Algebra

umang

What is the remainder when 5 to the power 99 is divided by 13 ????

ankit_harry1992

The answer will be 5 only.

rate my answer or ask again

gcch29

5^99=5.5^98
5.25^49=5.(26-1)^49
=5.(26N-1)             N belongs tonatural no.
=5.(13P-1)             P belongs to natural no.
=13R-5                 R belongs to natural no
remainder =8.

shine

gcch29 is right/.........
we can also move as.......
25= 12(mod 13)
squaring
5^4= 144(mod 13)= 1(mod 13)
raising the equation to the power of 24
5^96=1(mod 13)
multiplying by 5^3
5^99= 125(mod 13)= 8(mod 13)
thus the remainder is 8.....
i hope u got this..............if not knock back
i'll try 2 help out

do rate the efforts

umang

Hey shine !

Pls explain how u said " 25= 12(mod 13) " ?????

nitin62225

ans=8

soln:

=5(26-1)⁴⁹=M(26)-5 i.e M(13)-5 ---------M stands for multiple of ()

=M'(13)+13-5=M'(13)+8 --------M' is another multiple,one less than M

so the remainder is 8.

nick

here is the shortest method

5 leaves remainder 5 when divided by 13

5^2=25 leaves 12 remainder

5^3= 125 leaves 8 as remainder when divided by 13

after this it becomes PERIODIC
which means
5^3n where n is any integer leaves remainder 8

5^99 comes in this category,it leaves remainder 8....

aaaaaaaaa objective approach-NEED OF THE HOUR....

umang

Hey nick !

very well solved !!!!!

I agree with u that shortcuts r very useful , but knowledge of proper method makes ur base strong . And I hav a gr8 trouble solving such kind of problems !!!!!!

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