Kritisood
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?
(1) M = 13
(2) N = 14
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