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Here the statement 1 is sufficient as when we interchange the digits the the form is => 10n +m + 10m+n = 11(m+n) m,n being the unit and tens digits..
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Bunuel
If a and b are both two digit integers, is a + b a multiple of 11?

(1) The tens digit of a is equal to the units digit of b, and the tens digit of b is equal to the units digit of a
(2) Both a and b are odd

Required: Is a + b a multiple of 11?

Statement 1: The tens digit of a is equal to the units digit of b, and the tens digit of b is equal to the units digit of a
This means the numbers will be of the form xy and yx.

On adding, we get (10x + y) + (10y + x) = 11x + 11y
This clearly is divisible by 11
SUFFICIENT

Statement 2: Both a and b are odd
The different values of a and b can be:
a = 23 and b = 35
Here a + b = 58. Not a multiple of 11

a = 11 and b = 33
a + b = 44. Multiple of 11
INSUFFICIENT

Option A
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Bunuel
If a and b are both two digit integers, is a + b a multiple of 11?

(1) The tens digit of a is equal to the units digit of b, and the tens digit of b is equal to the units digit of a
(2) Both a and b are odd

Official solution

Solution: A.

Start with the easier of the two statements, which in this case is statement (2). If a and b are odd, then a+b could be a multiple of 11 (say a = 91 and b = 19) or a+b could be something else (say 11 + 13); INSUFFICIENT. Now consider the first statement. Plugging in numbers should reveal that every sum of a and b is a multiple of 11, so the statement appears sufficient. If you want further justification, play around with number properties: say that the tens digit of a is x and the units digit of a is y. That means a = 10x + y. Since b is merely a "flipped around", b = 10y + x. The sum of a and b is thus (10x + y) + (10y + x), or 11x + 11y, or 11(x+y). Since x and y are both integers, 11(x+y) must be a multiple of 11, so a + b must be a multiple of 11 and statement (1) is SUFFICIENT. A
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