Adhiraj's aabb number

$n^2 = \overline{aabb} = 11 \cdot (100a + b) \quad (1 \leq a \leq 9, 0 \leq b \leq 9).$

So,

$11 \mid n^2 \implies 11 \mid n \implies 11^2 \mid n^2 \implies 11 \mid 100a+b = \overline{a0b}.$

In order for $\overline{a0b}$ to be a multiple of $11$ , its alternating sum needs to be a multiple of $11$ , so $a+b=11$ . Therefore,

$\overline{aabb} \in \{ 2299, 3388, 4477, 5566, 6655, 7744, 8833, 9922 \}.$

The only quadratic residues mod $4$ are $0,1$ , so

$\overline{aabb} \in \{ 3388, 4477, 7744, 8833 \}.$

The only quadratic residues mod $5$ are $0,1,4$ , so

$n^2 = 7744 \implies n = \boxed{88}.$

Since the number n^2 is in the form aabb, n^2 is divisible by 11. Therefore n is also divisible by 11. So, we have to check all the squares of two digits numbers divisible by 11. These numbers are 11, 22, 33, 44, 55, 66, 77, 88, 99

88^2=7744

Therefore n=88

Since the first two digits equal to each other and the last two digits equal to each other, we can denote $n^2= 1000c+100c+10d+d=1100c+11d$ , where $0 \le c, d \le 9$ and $c \ne 0$ . Now our task is to find $c$ and $d$ . Notice that $1100c+11d=11(100c+d)$ and since $11(100c+d)$ is a perfect square, we can conclude that $100c+d=11.p^2$ (where $p \in \mathbb{N}$ ). Now we just need to construct a table for $p$ ranges from $4$ to $9$ (because $100 \le 100c+d \le 999$ ), we obtain the value $p=8$ and hence, $c=7$ and $d=4$ . Therefore, the number we want to find is $\sqrt{7744}=88$ .

As it is given $n^2 \geq 1000$ , and $n$ is two digit number, so minimum value of n which satisfies is $n = 32$ .

So $32 \leq n \geq 99$ , as $n$ need to be $4$ digits not $5$ .

We need to observe another property of the said four digit number $n^2$ , is that first two ditigs are equal and last two digits are equal, $i.e$ $n^2 = aabb$ , this property leads us to result that $n^2$ is divisible by $11$ , if $n^2$ is divisible by $11$ then $n$ must also be divisible by $11$ .

As a number is divisible by $11$ if of difference of sum of even positions and sum of odd positions is either $0$ or $-11$ or $11$ . Here $(a+b) - (a+b) = 0$ .

So for $n$ the numbers which we need to consider are $\{33, 44, 55, 66, 77, 88, 99\}$ . Out of which $88$ satisfies our criteria, $88 = 7744$ .

$n^{2} = \overline{aabb} = 1100 * a + 11 * b = 11 * (100 * a + b)$

This means that $11$ is a divisor of $n^{2}$ and $11$ is a divisor of $n$ . Furthermore, since $n^{2}$ is a 4 digit number, $32 \leq n \leq 99$ , so this means that $n$ must be one of the numbers $33, 44, 55, 66, 77, 88$ or $99$ . It turns out that $88$ is the only solution to this problem. This aproach is very simple, although it is not reliable, and I only wrote it because I found it interesting.

88^2 = 7744

Note that for $n$ to be $4$ -digited, $99 \ge n \ge 32$ . Also, $n^2=10^3a+10^2a+10^2b+b \equiv 0 \mod 11$ , thus $n^2=k \cdot 11 \Rightarrow n^2 = m \cdot 11^2 \Rightarrow n=r \cdot 11$ . Checking the $7$ cases individually, $n=88$ .

88 x 88 = 7744

use the calculator to solve this

n^2 = 1000 a + 100 a + 10*b + b

Then n^2 = 11(100*a + b)

From hear it can de seen that n must be a multiple of 11 and since it has two digits and its square must have four digits we have seven posibilities (33, 44, 55, 66, 77, 88, 99), from hear it can be seen that the answer is 88

88^2=7,744

Let the square of $n$ be written as a four digit number with the given property be written as $xxyy$ . Then, we must have $n^2 = xxyy = 1000 x + 100 x + 10 y + y = 11 (100 x +y).$ Since $11 (100 x +y)$ is a perfect square, and 11 is a prime number, $(100 x +y)$ must contain a factor of 11, i.e., 11 must divide $(100 x +y)$ or $x+y$ since 11 divides $99 x$ . The only possibility is that $x+y = 11$ with $2 \le x \le 9$ and $2 \le y \le 9$ . Further, since last digit of a perfect square must be one of $0, 1, 4, 5, 6, 9$ , $y$ must be one of them. This reduces the possible combinations of ordered pair $(x,y)$ to just four: $\{ (2,9), (5,6), (6,5), (7,4) \}$ . The corresponding numbers that multiply 11, namely $100 x + y$ , are $209 = 11 \times 19$ $506 = 11 \times 46$ $605 = 11 \times 55$ $704 = 11 \times 64$ Only the last combination results in a perfect square. Thus $n^2$ is $7744 = 11^2 \times 8^2 = 88^2$