Question 538468
Find a four-digit perfect square, the sum of whose digits is 22 and
 which has its first two digits equal to each other, and it's last two digits equal to each other.
:
Let x = the first two digits
Let y = the last two digits
:
2x + 2y = 22
simplify, divide by 2
x + y = 11
:
Try and see method, I came up with 7744:
{{{sqrt(7744)}}} = 88