BTGmoderatorDC wrote:If x and y are integers between 10 and 99, inclusive, is (x - y)/9 an integer?
(1) x and y have the same two digits, but in reverse order.
(2) The tens' digit of x is 2 more than the units digit, and the tens digit of y is 2 less than the units digit.
OA A
Source: Official Guide
Let's take each statement one by one.
(1) x and y have the same two digits but in reverse order.
Say x = 10p + q, where p is tens digit and q is units digit; thus, y = 10q + p.
=> (x - y) = (10p + q) - (10q + p) = 9(p - q). It is divisible by 9. Sufficient.
(2) The tens' digit of x is 2 more than the units digit, and the tens digit of y is 2 less than the units digit.
Say the units digit of x is p, thus, its tens digit is p + 2. Thus, x = 10(p + 2) + p = 11p + 20;
Again, say the units digit of y is q, thus, its tens digit is q - 2. Thus, y = 10(q - 2) + q = 11q - 20
=> (x - y) = (11p + 20) - (11q - 20) = 11(p - q) + 40
Case 1: p = 1 and q = 3, then (x - y) = 11(p - q) + 40 = 11(1 - 3) + 40 = -22 + 40 = 18, divisible by 9. The answer is yes.
Case 2: p = 2 and q = 3, then (x - y) = 11(p - q) + 40 = 11(2 - 3) + 40 = -11 + 40 = 29, not divisible by 9. The answer is no.
Insufficient.
The correct answer:
A
Hope this helps!
-Jay
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