Find the remainder when 12^107 is divided by 37? a)34 b)36 c)11 d)3 pls explain method u use to solve this...

The Euler of 37 is 36. Since 12 and 37 are co-prime, 12^36 mod 37 = 1
Similarly, 12^108 mod 37 = 1

So, 12 * $$12^{107}$$ mod 37 = 1
Let $$12^{107}$$ mod 37 be k
So, 12k mod 37 = 1

=> 12k = 37t + 1
The range of t is from 1 to (12-1) = 11

Substituting the values, we see that the equation is satisfied for t = 11
=> k = 34
Remainder = 34

10^3/37 Reminder wil be 1 >>>>>> En 12^107/13 left only  to 12^8/13 and now 12^2*12^2.../13 

12^2/13 reminder will be -4 hence try multiplying -4*-4*-4*-4 =256/13 Reminder will be -3 =34 :-)

from options , multiply by 12 and check whether it is divisible by 37.
Because 12^108/37 is 1.😉

The range of k is from 1 to (12-1) = 11 means
the range of 'k' or 't'