Suppose that $\left(2^{3}\right)^{x}=4096$ and that $y=x^{3}$. What is the ones (units) digit of the integer equal to $3^{y}$ ?
We note that $\left(2^{3}\right)^{x}=2^{3 x}$ and that $4096=4 \cdot 41024=4 \cdot 4 \cdot 256=4 \cdot 4 \cdot 16 \cdot 16=2^{2} \cdot 2^{2} \cdot 2^{4} \cdot 2^{4}=2^{12}$.

Thus, $2^{3 x}=2^{12}$ which gives $3 x=12$ and so $x=4$.

Since $y=x^{3}$, then $y=4^{3}=64$.

We need to consider the integer $3^{64}$.

The first few powers of 3 are

$$
3^{1}=3, \quad 3^{2}=9, \quad 3^{3}=27, \quad 3^{4}=81, \quad 3^{5}=243, \quad 3^{6}=729, \quad \ldots
$$

The units digits of powers of 3 repeat in a cycle of length 4 , namely $3,9,7,1,3,9,7,1, \ldots$.

(Since each power is obtained by multiplying the previous power by 3 , then the units digit of each power is obtained by multiplying the units digit of the previous power by 3 (possibly keeping only the units digit). This means that once a units digit recurs, then the units digits will form a cycle.)

Since 64 is a multiple of 4 , then the units digit of $3^{64}$ will be the last in the cycle, or 1 .

ANSWER: 1