What is the remainder when the two-digit, positive integer.

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What is the remainder when the two-digit, positive integer x is divided by 3?

(1) The sum of the digits of x is 5
(2) The remainder when x is divided by 9 is 5

The OA is D.

I got confused. I could not determine the statements alone are sufficient. Can any expert help me here?

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by Jay@ManhattanReview » Wed Sep 20, 2017 11:52 pm
Vincen wrote:What is the remainder when the two-digit, positive integer x is divided by 3?

(1) The sum of the digits of x is 5
(2) The remainder when x is divided by 9 is 5

The OA is D.

I got confused. I could not determine the statements alone are sufficient. Can any expert help me here?
Statement 1: The sum of the digits of x is 5.

Divisibility rule of 3:

If the sum of the digits of a number is divisible by 3, then the number is divisible by 3.

However, if the sum of the digits of a number is NOT divisible by 3, then the remainder is the same when the number is divided by 3.

Thus, the remainder when the two-digit, positive integer x is divided by 3 = Remainder when 5 is divided by 3 = 2. Sufficient.

Statement 2: The remainder when x is divided by 9 is 5.

The rule for divisibility of 9 is same as that for 3.

The remainder when the two-digit, positive integer x is divided by 3 = Remainder when 5 is divided by 3 = 2. Sufficient.

Alternatively,

Say X = 9q + 5, where q is quotient

Thus, X/3 = (9/3)*q + 5/3

X/3 = 3q + 1 + 2/3

=> Remainder = 2. Sufficient.

The correct answer: D

Hope this helps!

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-Jay
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