Find the remainder when 9^1990 is divided by 11 number theorem #mod(11) #mathematics

[daedora]

you can use fermat's little theorem to prove it quite easily.

[RupamKumari-jdro]

Bhai sahi explain Kiya mgar Fermat 's little theorem se jaldi kar sakte ho

[RupamKumari-jdro]

a^p-1=1(mod p) p is prime
a&p are co-prime

[parvmangal-qufd]

BRO THIS QUESTION CAN BE EASILY FINISHED IN 2 STEPS BY USING BINOMIAL THEOROM-
9 CAN BE WRITTEN AS 10-1
SO THE EQUATION WILL BE (10-1)^1990/11
AFTER SOLVING YOU WILL GET -(1)^1990 ==> 1
THEREFORE 1 THE REMAINDER.