Among the following four numbers, which one is a perfect square?
(1) 921438
(2) 76186
(3) 750235
(4) 2660161
Number Theory, Perfect Squares.
(Method 1)
The square of an integer, when divided by 4, leaves a remainder of 0 or 1, i.e., a number that leaves a remainder of 2 or 3 when divided by 4 is definitely not a perfect square.
$$
\begin{array}{l}
921438 \equiv 38 \equiv 2(\bmod 4), \\
76186 \equiv 86 \equiv 2(\bmod 4), \\
750235 \equiv 35 \equiv 3(\bmod 4)
\end{array}
$$

Therefore, $921438$, $76186$, and $750235$ are not perfect squares;
while $2660161=1631^{2}$, is a perfect square;
hence, the answer is (4).
(Method 2)
If the tens digit of a perfect square is odd, then its units digit must be 6;
conversely, if the units digit of a perfect square is 6, then its tens digit must be odd.
921438 and 750235 have odd tens digits, but their units digits are not 6;
76186 has a units digit of 6, but its tens digit is not odd;
hence, $921438$, $76186$, and $750235$ are not perfect squares;
while $2660161=1631^{2}$, is a perfect square;
hence, the answer is (4).