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What is the value of the positive integer m ?

This topic has 2 expert replies and 1 member reply
sachin_yadav Master | Next Rank: 500 Posts
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What is the value of the positive integer m ?

Post Tue Dec 30, 2014 5:00 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    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.

    OA is B

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

    Thanks & Regards
    Sachin

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    Post Tue Dec 30, 2014 5:55 am
    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.

    OA is B

    Why this cannot be E ? In both statements m can be 3 and 9. Am i missing anything ?
    Ok, you got it right regarding Statement 1. With the information Statement 1 we can only tell that m = 6k + 3, with k being unknown, and therefore we cannot determine the value of m from this information.

    So Statement 1 is insufficient.

    However, Statement 2 gives us a key piece of information, because there is only one number such that dividing 15 by that number gives a remainder of 6, and that number is 9. If we divide 15 by 3, as you suggested, the remainder is 0.

    So Statement 2 gives information sufficient to answer the question.

    Choose B.

    Thanked by: sachin_yadav
    Post Tue Dec 30, 2014 10:46 am
    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

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    Post Sat Jan 03, 2015 6:01 pm
    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.

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