Question

A perfect square number has four digits, none of which is zero. The digits from left to right have values that are: even, even, odd, and even. Find the number.

Answer 1

The square root of the smallest $4$-digit number, $1000$, is somewhere between $31$ and $32$.

So the smallest $4$-digit square has to be the square of $32$ is $1024$ (not 31 because $31$ square is $961$)

The largest $4$-digit number is $9999$, which is one less than the square of $100$.

It is therefore the square of $99$, or$9801$.

This implies that the answer is the square of a number between $32$ and $99$, inclusive of both.

Given that

The unit digit is even, so the number is even.

Thus, it must the square of an even number.

Let us assume $PQRS$ is a perfect square,

Where, $P=even,Q=even,R=odd,S=even$

So it can be the square of numbers in the range $32,34,36,\dots ,96,98.$

The leftmost digit has to be $2,4,6or8.$

Now, will check of square starting with even numbers that is: the square of numbers starting from 2000, 4000, 6000,8000.

The square root of the number $2000$ is between $44$ and $54$

The next range is $4000$ to $4999$, and the square root of $4000$ is between $63$ and $70$.

The next range is $6000$ to $6999$, and the square root of $6000$ is between $77$ and $83$.

The next range is $8000$ to $8999$, and the square root of $8000$ is between $90$ and $99$.

On evaluating all the squares between $90$ and $94$

Number	Square
$90$	$8100$
$91$	$8281$
$92$	$8464$
$93$	$8649$
$94$	$8836$

So, $8836$ is the perfect square as per the given conditions

Therefore, the square that satisfies all the conditions is $8836$.