When the product $\left(5^{3}\right)\left(7^{52}\right)$ is expanded, the units digit (that is, the last digit) is
(A) 5
(B) 3
(C) 9
(D) 7
(E) 0
First, we see that both $5^{3}$ and $7^{52}$ are odd integers, so their product is odd. Also, $5^{3}$ is a multiple of 5, so $\left(5^{3}\right)\left(7^{52}\right)$ is a multiple of 5 , and thus is an odd multiple of 5 . Since all odd multiples of 5 end with the digit 5 , then the units digit of $\left(5^{3}\right)\left(7^{52}\right)$ is 5 .

ANSWER: (A)