Find the unique four-digit integer n with these properties: The last digit (the units digit) of n is 9. The digits of n add up to 27. Two digits of n are the same. n is a perfect square.

63^2 = 3,969 Meets ALL the conditions.

If x^2 ends in 9 then the x must end in 3 or 7

If x^2 is four digits then x must be between 32 and 99

So the only possibilities for x are

33, 37, 43, 47, 53, 57, 63, 67, 73, 77, 83, 87, 93, 97

Square each of those and find the one/s that fit the other two requirements :)