The miraculous integer

For the last digit of a perfect square to be $1$ , $A$ is either $1$ or $9$ , $111^2 = 12321 \ne 118001$ , therefore the number must be $\boxed{999}$ and $999^2 = 998001$ .

To rewrite the equation, we have:

Then using the quadratic equation, we get:

Since subtracting it will lead us to a negative number, we just use addition. Hence,

Therefore, .

The sum of the digits of $AAA$ is $3\times A$ , so it's clearly divisible by $3$ . The product $AAA \times AAA$ will therefor have to be divisible by $9$ .

$A + A + 1 + 8$ must be a multiple of $9$ , so $2 \times A$ has to be a multiple of $9$ . This is only posible if $A=9$ .

As the last digit of the square of the given 3-digit number ends in 1 so we can conclude that either A=1 or 9 . We can easily predict that 111 x 111 is a 5-digit number. So, we get A=9, giving us the 3-digit number = 999

999 999=998001 Because 999=1000-1 So 999 999=(1000-1)(1000-1) Turn the former (1000-1) to A. Then A(1000-1) So 1000A-A. 1000(1000-1)-(1000-1) 1000000-1000-1000+1=990000-1000+1=980000+1=998001

lol,the answer is 999

9^2=81 99^2=9801 999^2=998001 9999^2=99980001

The solution can be acquired by knowing this property of n digits of 9 being squared.