BTGmoderatorDC wrote:What is the sum of the digits of two-digit positive integer M?

(1) If the digits of M are reversed, the resulting integer is 27 greater than M.

(2) The difference between the squares of the digits of M is 39.

Statement 1:
Let M = 10T + U, where T = the tens digit and U = the units digit.
When the digits are reversed, we get:
10U + T.

Since the reversed integer is 27 more than M, we get:
(10U + T) - (10T + U) = 27
9U - 9T = 27
9(U-T) = 27
U-T = 3.
U = T+3

The resulting equation indicates that the units digit of M is 3 more than the tens digit.
Case 1: T=1 and U=4, implying that M=14
In this case, the sum of the digits = 1+4 = 5.
Case 2: T=2 and U=5, implying that M=25
In this case, the sum of the digits = 2+5 = 7.
Since the sum can be different values, INSUFFICIENT.

Statement 2:
Make a list of perfect squares:
1, 4, 9, 16, 25, 49, 64, 81
Only the perfect squares in color have a difference of 39.
The blue perfect square implies that one of the digits = 5.
The red perfect square implies that the other digit = 8.
Thus, the sum of the digits = 5+8 = 13.
SUFFICIENT.

The correct answer is B.