161. Find the greatest integer $x$ such that the number

$$
4^{27}+4^{1000}+4^{x}
$$

is a perfect square.
161. Answer: $x=1972$. Since $4^{27}+4^{1000}+4^{x}=$ $=2^{54}\left(1+2 \cdot 2^{1945}+2^{2 x-54}\right)$, the expression in parentheses will be a perfect square when $2 x-54=2 \cdot 1945$, i.e., when $x=1972$. If $x>1972$, then $2^{2(x-27)}<1+2 \cdot 2^{1945}+2^{2(x-27)}<\left(2^{x-27}+1\right)^{2}$, i.e., the given number is between the squares of two consecutive natural numbers.