Hrm. This seems to be incorrect. The remainder cannot be bigger than the divisor minus 1.

revised

kaka46 wrote:Hi All,
Came across this question today and upon solving, I got the answer as 6. However, the OA is 31. Can you please help me understand it ?

Q) What is the remainder when (37)^2012 is divided by 25 ?
Ans. a) 31
b) 21
c) 11
d) 1
OA : a

I would approach this question holistically from the given expression's divisor and dividend perspectives.

(37)^2012 divided by 25 is equivalent to (37/5)^2 * 37^2010 or (7+2/5)^2 * 37^2010 or
(49+28/5+4/25) * 37^2010 or ((49 + (28*5+4)/25) * 37^2010. We further check the right part of an expression for divisibility by 25. So, (49+144/25)* 37^2010

Now by looking up the left part of expression we can see that the remainder could be obtained using (28*5+4)/25=144/25 as well - not exclusively though, we should check divisibility of 37^2010 by 25 too. We notice that 144/25 is 5 + 19/25 and the remainder in this instance is 19 (not for the question, only here). 19 is a prime number, and if rewrite the other part such as 37^2010 divided by 25 = (49+144/25)* 37^2008 we spot that the remainder in this instance is again 19. So we have all 19s multiplied by each other and the other whole numbers, these are divided by 25. Each time 19 is obtained as a remainder the power of whole expression goes down by 2 - this means we have 2012/2 total power for 19. So the final solution will be obtained from calculating (19^1006)/25 or (19/5)^2 * 19^1004. In this new instance, from the left part we obtain a remainder 4 and the calculation follows through 4^503. The calculation task (and seemingly difficulty of finding the last two numbers by raising 19 to the powers) ends here, as we simplified everything to one digit remainder, 4. In the power of 503 we have 50 tens and three altogether giving as 4^10 (the last two digits are 76) * 4^3 = 76*64 = 4864. We have the the last two digits obtained as 64 and 64/25 leaves a remainder of 14 which is our answer.