is a + b a multiple of 3?

[Ramesh]

If a and b are both single-digit positive integers, is a + b a multiple of 3?

(1) The two-digit number "ab" (where a is in the tens place and b is in the ones place) is a multiple of 3.

(2) a - 2b is a multiple of 3.

A Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient.
D EACH statement ALONE is sufficient.
E Statements (1) and (2) TOGETHER are NOT sufficient.

the answer to this is D. I got this right. But I am not sure about my explanation. I see a different approach in the solution given. Can anyone please say whether there is any flaw in my reasoning:

1. Since the 2 digit number 'ab' is divisible by 3 obviously its sum (a+b) has to be divisible by 3 (Since this is the general rule of divisibility for 3)
2. Since a-2b is divisible by 3, a and b have to be divisible by 3 (b is being only multipied by 2 which is not going to help he number being divisible by 3 in any way).
As a and b are divisible by 3 thier sum will definitely be divisible by 3.

Thus, I selected D. Please kindly point out the flaws (if any) in the above reasoning

[thesamet]

Your final answer is right, but the reasoning for statement 2 sufficiency is incorrect: a-2b divisibility by 3 does not imply that both a and b are divisible by three. For instance, a=5 and b=1.

If a-2b is divisible by three, then also (a-2b)+(3b) is divisible by three, since it is a sum of two numbers that are both divisible by three. Now, (a-2b)+3b=a+b, so we conclude that a+b is divisible by 3.

[esledge]

Good explanation, thesamet.

More generally, here are some rules about adding or subtracting multiples:

(Multiple of x) +/- (Multiple of x) = Multiple of x
(Multiple of x) +/- (NON-multiple of x) = NON-multiple of x
(NON-Multiple of x) +/- (NON-multiple of x) = could be either a (Multiple of x) or (NON-multiple of x). No guarantees.