If n is an integer, is n/7 an integer?

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sachin_yadav Master | Next Rank: 500 Posts
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If n is an integer, is n/7 an integer?

Post Tue Dec 30, 2014 6:35 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    If n is an integer, is n/7 an integer?
    (1) 3n/7 is an integer.
    (2) 5n/7 is an integer.

    OA is D

    What is the correct way in solving this kind of question ? I am getting E. Below are the two methods

    Method 1:

    statement 1: 3n/7 = x (int)
    Next step, n/7 = x/3 ==> gives both integer and non integer

    same goes for statement 2. Hence E

    Method 2:
    Is n/7 an integer ? Is n/7 = m(int) ==> Is n = 7m ? (Is n a multiple of 7 ?)

    Statement 1: 3n/7 = x (int)
    Next step, 3n = 7x ==> n = 7x/3 (given n is an integer, so it is sufficient)

    Same goes for statement 2. Hence D

    Thanks & Regards
    Sachin

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    Marty Murray Legendary Member
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    Post Tue Dec 30, 2014 7:14 am
    sachin_yadav wrote:
    If n is an integer, is n/7 an integer?
    (1) 3n/7 is an integer.
    (2) 5n/7 is an integer.

    OA is D

    What is the correct way in solving this kind of question ? I am getting E. Below are the two methods

    Method 1:

    statement 1: 3n/7 = x (int)
    Next step, n/7 = x/3 ==> gives both integer and non integer

    same goes for statement 2. Hence E

    Method 2:
    Is n/7 an integer ? Is n/7 = m(int) ==> Is n = 7m ? (Is n a multiple of 7 ?)

    Statement 1: 3n/7 = x (int)
    Next step, 3n = 7x ==> n = 7x/3 (given n is an integer, so it is sufficient)
    Method 2 looks good to me.

    Still I would do it another way. Here it is, barely any math.

    Statement 1 says that 3n/7 is an integer. 3 is not a multiple of 7 and 7 is not a multiple of 3. So multiplying an integer by 3 changes nothing related to being a multiple of 7. So if n is an integer and 3n is a multiple of 7, n is a multiple of 7.

    Sufficient

    Statement 2 does the same thing with 5. 5 is not a multiple of 7 and 7 is not a multiple of 5. So the same logic applies.

    Sufficient

    Done.

    Choose D.

    Thanked by: sachin_yadav
    Post Tue Dec 30, 2014 9:53 am
    sachin_yadav wrote:
    If n is an integer, is n/7 an integer?
    (1) 3n/7 is an integer.
    (2) 5n/7 is an integer.
    Statement 1: 3n/7 is an integer.
    3n/7 = a, where a is an integer.
    Thus:
    3n = 7a
    n = (7/3)a.

    For n to be an integer -- as required by the prompt -- a must be a multiple of 3.
    If a=3, then n = (7/3)(3) = 7.
    If a=6, then n = (7/3)(6) = 14.
    If a=9, then n = (7/3)(9) = 21.
    The resulting values for n are all multiples of 7.
    Thus, n/7 must be an integer.
    SUFFICIENT.

    Statement 2: 5n/7 is an integer
    5n/7 = b, where b is an integer.
    Thus:
    5n = 7b
    n = (7/5)b.

    For n to be an integer -- as required by the prompt -- b must be a multiple of 5.
    If b=5, then n = (7/5)(5) = 7.
    If b=10, then n = (7/5)(10) = 14.
    If b=15, then n = (7/5)(15) = 21.
    The resulting values for n are all multiples of 7.
    Thus, n/7 must be an integer.
    SUFFICIENT.

    The correct answer is D.

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    Post Tue Dec 30, 2014 10:28 am
    sachin_yadav wrote:
    If n is an integer, is n/7 an integer?

    (1) 3n/7 is an integer.
    (2) 5n/7 is an integer.
    A lot of integer property questions can be solved using prime factorization.
    For questions involving divisibility, divisors, factors and multiples, we can say:
    If N is divisible by k, then k is "hiding" within the prime factorization of N

    Consider these examples:
    24 is divisible by 3 because 24 = (2)(2)(2)(3)
    Likewise, 70 is divisible by 5 because 70 = (2)(5)(7)
    And 112 is divisible by 8 because 112 = (2)(2)(2)(2)(7)
    And 630 is divisible by 15 because 630 = (2)(3)(3)(5)(7)
    --------------------------------
    Okay, onto the question:

    Target question: Is n/7 an integer?

    Statement 1: 3n/7 is an integer
    If 3n/7 is an integer, then we can also say that 3n is DIVISIBLE by 7.
    This means that there's a 7 HIDING in the prime factorization of 3n.
    Since there's no 7 HIDING in 3, there must be a 7 HIDING in the prime factorization of n.
    If there's a 7 HIDING in the prime factorization of n, then n must be divisible by 7
    If n is divisible by 7, then n/7 is DEFINITELY an integer.
    Since we can answer the target question with certainty, statement 1 is SUFFICIENT

    Statement 2: 5n/7 is an integer
    If 5n/7 is an integer, then we can also say that 5n is DIVISIBLE by 7.
    This means that there's a 7 HIDING in the prime factorization of 5n.
    Since there's no 7 HIDING in 5, there must be a 7 HIDING in the prime factorization of n.
    If there's a 7 HIDING in the prime factorization of n, then n must be divisible by 7
    If n is divisible by 7, then n/7 is DEFINITELY an integer.
    Since we can answer the target question with certainty, statement 2 is SUFFICIENT

    Answer = D

    Cheers,
    Brent

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    sachin_yadav Master | Next Rank: 500 Posts
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    Post Fri Jan 02, 2015 1:48 am
    Thank you all for your responses. Appreciate it.

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    Post Sat Jan 03, 2015 6:27 pm
    I'd recommend that you compare this question with OG #135, which tests a similar principle.

    In this problem, your second interpretation is much better: "is n/7 an integer -> is n divisible by 7." It's generally better to approach divisibility problems in conceptual terms rather than algebraic ones.

    If in this problem one of the statements had read "14n/7 is an integer," then the statement would be INSUFFICIENT. Since 14 itself is divisible by 7, n could be anything and the statement would still be true.

    Since 3 and 5 are not divisible by 7, then if 3n and 5n are each divisible by 7, n must be divisible by 7 - the 3 and 5 can't be helping to divide 7, as 14 was in the previous example.

    If in a case like this the coefficient in the numerator shares no factors with the denominator, then the variable must be divisible by the denominator.

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