Carcass wrote:

If x is a positive integer, what is the units digit of \((24)^{5+2x}\) + \((36)^6\) + \((17)^3\)

(A) 2

(B) 3

(C) 4

(D) 6

(E) 8

a n

Any hint for this one? I computed 36^6 which terminates with a 6, 17^3 terminates with a 3 and 24^(5+2x) can be rewritten as 24^5*24^2x where 24^5 terminates with a 4 and 24^2x terminates with a 6 for whatever value of x. Thus, the 24 term terminates with a 4. Summing three numbers terminating with 6, 3 and 4, it results a number terminating with 3, thus answer B seems right to me. However, the answer seems to be A. why?