Data Sufficiency

sachin_yadav wrote:What is the value of the positive integer m?

(1) When m is divided by 6, the remainder is 3.
(2) When 15 is divided by m, the remainder is 6.

When it comes to remainders, we have a nice rule that says:
If N divided by D, leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc.
For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc.

Target question: What is the value of positive integer m?

Statement 1: When m is divided by 6, the remainder is 3
According to the above rule, we can write the following:
The possible values of m are: 3, 3+6, 3+(2)(6), 3+(3)(6)...
Evaluate to get: the possible values of m = 3, 9, 15, 21, etc.
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT


Statement 2: When 15 is divided by m, the remainder is 6
According to the above rule, we can conclude that....
Possible values of 15 are: 6, 6 + m, 6 + 2m, 6 + 3m, ...

Aside: Yes, it seems weird to say "possible values of 15," but it fits with the language of the above rule]

Now, let's test some possibilities:
15 = 6...nope
15 = 6 + m. Solve to get m = 9. So, this is one possible value of m.
15 = 6 + 2m. Solve to get m = 4.5
STOP. There are 2 reasons why m cannot equal 4.5. First, we're told that m is a positive INTEGER. Second, the remainder (6 in this case) CANNOT be greater than the divisor (4.5)

If we keep going, we get: 15 = 6 + 3m. Solve to get m = 3. Here, m cannot equal 3 because the remainder (6) CANNOT be greater than the divisor (3).
If we keep checking possible values (e.g., 15 = 6 + 3m, 15 = 6 + 4m, etc), we'll find that all possible values of m will be less than the remainder (6).

So, the ONLY possible scenario here is that m must equal 9
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer = B

Cheers,
Brent

sachin_yadav wrote:Why this cannot be E ? In both statements m can be 3 and 9. Am i missing anything ?

I think the mistake that you made here was simply to subtract 6 from 15, from which you got 9, and to think of the factors of 9: 3 and 9. The problem is that 3 goes evenly into 15, so there would be no remainder. (Incidentally, this would also be true for the other factor of 9: 1).

Remember with remainder problems:
- if a number is a multiple of the divisor, the remainder is 0.
- the remainder can never be larger than the divisor.