What is the tens digit of $3^{2016}$ ?
(A) 0
(B) 2
(C) 4
(D) 6
(E) 8
When two integers are multiplied together, the final two digits (the tens digit and the units digits) of the product are determined by the final two digits of each of the two numbers that are multiplied.

This is true since the place value of any digit contributes to its equal place value (and possibly also to a greater place value) in the product.

That is, the hundreds digit of each number being multiplied contributes to the hundreds digit (and possibly to digits of higher place value) in the product.

Thus, to determine the tens digit of any product, we need only consider the tens digits and the units digits of each of the two numbers that are being multiplied.

For example, to determine the final two digits of the product $1215 \times 603$ we consider the product $15 \times 03=45$. We may verify that the tens digit of the product $1215 \times 603=732645$ is indeed 4 and the units digit is indeed 5 .

Since $3^{5}=243$, then the final two digits of $3^{10}=3^{5} \times 3^{5}=243 \times 243$ are given by the product $43 \times 43=1849$ and thus are 49 .

Since the final two digits of $3^{10}$ are 49 and $3^{20}=3^{10} \times 3^{10}$, then the final two digits of $3^{20}$ are given by $49 \times 49=2401$, and thus are 01 .

Then $3^{40}=3^{20} \times 3^{20}$ ends in 01 also (since $01 \times 01=01$ ).

Further, $3^{20}$ multiplied by itself 100 times, or $\left(3^{20}\right)^{100}=3^{2000}$ also ends with 01 .

Since $3^{10}$ ends with 49 and $3^{5}$ ends with 43 , then $3^{15}=3^{10} \times 3^{5}$ ends with $49 \times 43=2107$ and thus has final two digits 07 .

This tells us that $3^{16}=3^{15} \times 3^{1}$ ends with $07 \times 03=21$.

Finally, $3^{2016}=3^{2000} \times 3^{16}$ and thus ends in $01 \times 21=21$, and so the tens digit of $3^{2016}$ is 2 .

ANSWER: (B)