If M and N are positive integers greater than 1, what is the remainder

Hi, I had mistakenly not put the powers correctly GMATinsight Kinshook
Have made the edit

If M and N are positive integers greater than 1, what is the remainder when the expression 22^3M × 39^2N + 14^2(M+N) is divided by 5?

Are the powers there 3M, 2N and 2(M+N)? If yes, then ti should be written as 22^(3M) × 39^(2N) + 14^(2(M+N)) or using formatting as \(22^{3M} × 39^{2N} + 14^{2(M+N)}\)

Done.. apologies for the confusion!!

using cyclicity rule we can solve this question
for given expression \(22^{3M} × 39^{2N} + 14^{2(M+N)}\)
we can note that the unit digit value of 39^{2N} would always be 1 ; because its 9^ even power and unit digit value of 14^{2(M+N) would always be 6 as its also raised to even power
so the remainder value mainly depends what is the value of M for \(22^{3M}\) ; as M>1 ; we can get 4,1,6,8 as digits for M= 2,3,4,5
#1
M=13
unit digit value of \(22^{3*13}\) ; 8
so for given expression we get 22^(3M) × 39^(2N)+ 14^(2(M+N)
22^39 * 1 + 6
8*1+6 ; 14 ; remainder will be 4 when divided by 5
sufficient
#2
N=14
as stated in above ; knowing value of N is not required as it would always be even power raised or say 6
so insufficient as M is not know
IMO A is correct

Kritisood

Without edit too the answer to the question was Option A and the question was good.

I guess I will make post of that question which I solved.

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