Data Sufficiency

silvia928 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



Ans....

....

(A)

That should be D. I'm not sure where this question is from, but it's unusual to see irrelevant information in a real GMAT question; here, for example, it doesn't matter that the number has two digits.

On to the question:

1) Hopefully everyone here will know how to test whether a (positive) whole number is divisible by 3 (or 9): add the digits, and if the sum is divisible by 3 (or, respectively, 9), so is the original number. Less well known, but sometimes useful: you can use that test to find the remainder when you divide by 3 (or 9). Let's look at a different number:

3,472

If I add the digits, I get 16. The remainder is 1 when I divide 16 by 3, and that guarantees that the remainder will be 1 when I divide 3,472 by 3. The remainder will be 7 when I divide 3,472 by 9, because that's the remainder when I divide 16 by 9. One word of caution- these tests only work when you are dividing by 3 or 9, so only apply them in these cases.

In any event, if you know the above, you know immediately that Statment 1) is enough.

2) This tells us that x is five larger than a multiple of 9. But every multiple of 9 is certainly a multiple of 3, so x is five more than a multiple of 3. But the remainder can't be 5; it must be 0, 1 or 2 when we divide by 3. You can just subtract 3 from 5, in fact, and here's why:

x = 3q + 5 = 3q + 3 + 2 = 3(q+1) + 2

So x is 2 larger than a multiple of 3; the remainder is 2 when you divide x by 3. So the statement is sufficient.

chidcguy wrote:A) The possible numbers are 23,32,41,14,50,05

Ah, good point. In my post, I said that the fact that the number had two digits wasn't important- and it's not, the answer is D regardless of the number of digits- but if you know it has two digits, it is possible to list all the possibilities, as you've done.

chidcguy wrote:A) The possible numbers are 23,32,41,14,50,05

Here, note that 05 is not considered a two-digit number; it's a single digit number (5).

chidcguy wrote: B) 5,14,23,32,41,50,59,68 all leave remainder 5 when divided by 9. All of them seem to leave a remainder 2 as well.

You're missing a few possibilities here. 5 is only a single digit, so you could ignore it in this question. And you should have 77, 86, and 95 on your list. But yes, they all leave a remainder of 2 when you divide by 3.

chidcguy wrote: To me the answer looks like D. Am I missing some thing?

The answer is D, yes!