As
$$\sqrt{1100a+11b}=\sqrt{11(100a+b)}$$
To make the expression an integer, we may write
$$100a+b=11k^2,k\in{\Bbb{N}}$$
Also as $a,b\le9$, we have
$$100a+b=11k^2\le909$$
Furthermore, as $1000<1089=33^2$, we have $a\ge3$.
So $300\le100a+b=11k^2\le909$
And with some easy observation, we know that $6\le k<10$.
Obviously, only when $k=8,11k^2=704\implies a=7,b=4$.
So the number is $7744.$
First observe that it suffices to check the integers from $0$ to $99$ as $(100x+y)^2 \equiv y^2 \pmod{100}$. Now, we consider $y \ge 50$. In that case,
$y = 50 +n$, where $0 \le n \le 49$. In that case,
$$y^2 = (50+n)^2 = 2500 + 100n + n^2 \equiv n^2 \pmod{100}.$$
Therefore, the numbers $n$ and $50+n$ share the same last two digits (check, for example, $n=0,1,2$ etc.). As a consequence, we do not need to check the last two digits of the numbers greater than $49$. Rather, we focus on the numbers $0, 1, \cdots , 49$.
Among them, we now consider $y \ge 25$. These numbers can be written as $y = 50-n$ for $0 \le n \le 25$. But
$$y^2 = (50-n)^2 = 2500 - 100n + n^2 \equiv n^2 \pmod{100}.$$
Therefore, the numbers $n$ and $50-n$ share the same last two digits (check, for example, $n=1,2$ etc.; i.e. $1^2$ and $49^2$ have the same last two digits). As a consequence, we do not need to check the last two digits of the numbers greater than $25$. Rather, we focus on the numbers $0, 1, \cdots , 25$.
For $0 \le n \le 25$, we compute their squares and observe the pattern of the last two digits. The pattern turns out to be $00, e1, e4, 25, o6, e9$.
Best Answer
Taking the last two digits of a number is equivalent to taking the number $\bmod 100$. You can write a large number as $100a+10b+c$ where $b$ and $c$ are the last two digits and $a$ is everything else. Then $(100a+10b+c)^2=10000a^2+2000ab+200ac+100b^2+20bc+c^2$. The first four terms all have a factor $100$ and cannot contribute to the last two digits of the square. The term $20bc$ can only contribute an even number to the tens place, so cannot change the result. To have the last digit of the square odd we must have $c$ odd. We then only have to look at the squares of the odd digits to see if we can find one that squares to two odd digits. If we check the five of them, none do and we are done.