GMAT Challenge Question: Divisibility

One of the most common concepts tested on the GMAT quantitative section is divisibility, and the data sufficiency problem below is a prime example of how the exam can ask you to use your factoring/divisibility skills.  Take a shot at this question, post your answers (and show your work!) in the comments field, and check back later for a full solution.

For positive integers x and y, x^2 = 350y.  Is y divisible by 28?

(1)    x is divisible by 4

(2)    x^2 is divisible by 28

UPDATE: Solution!

Solution: A.  In order to be divisible by 28, a number must have the prime factors 2*2*7.  From the given information we can conclude that x^2 = 2*5*5*7*y  (the factors indicate the prime factorization of 350), and that, because we need the factors to come in pairs in order for the square root of x^2 to be an integer, y is then divisible by at least 2 and 7.  From that, we need the statements to supply one more factor of 2 for y.  Statement 1 does exactly that: because x is divisible by 4, then x^2  will be divisible by 4*4, or 2*2*2*2.  As 350 only has one 2, y must supply the rest, so we can conclude that y is divisible by 2*2*2*7.  As only 2*2*7 was required for y to  be divisible by 28, statement 1 is sufficient.  Statement 2 only tells us that y is divisible by 2*7 – and we already knew that. Accordingly, statement 2 is not sufficient, and the correct answer is A.

9 Responses

  1. GMATBOY says:

    I will go with 1 . because .. if X^2 is having 16 only then we are assured that 7X5^2X2 needs to be multiplied by atleast 8 X 7 to make primes factors exponent equal on both sides.
    For B) ARGUMENT we can have just 7X2 multiplied to Right side. So not enough

  2. 144144 says:

    IMO A.

    x= root7*5*root2*rooty
    because x and y are integers:
    y needs to have – at least: 7, 2 so X will be integer.

    stat 1 tells us that x is dividable by 4 so it must have 2*2 right now we have
    root7*root7*root2*root2*5 —> y have to have at least a couple more 2*2 so that root x would stay integer and would be dividable by 4.
    so y=7*2*2*2 at least. therefore y will always be dividable by 28 (7*2*2).

    stat 2 – x^2 is dividable by 28 is insufficient.
    x^2= 7*5*5*2*y —> Y can be 2 (cannot be dividable by 2) or 28.

    so IMO A.

    hope im not wrong. thanks for this great question.

  3. 144144 says:

    x^2= 7*5*5*2*y —> Y can be 4 (cannot be dividable by 2) or 196.

  4. SonyGmat says:

    Based on the answer posted [quote]Statement 2 only tells us that y is divisible by 2*7 [/quote]

    From the approach I followed I found out that Statement 2 tells us that y is divisible by 2.

    Could you please explain how you got that y is divisible by 2*7 from statement 2?


    • Brian Galvin says:

      Hey Sony,

      Sure thing! So statement 2 says:

      x^2 is divisible by 28

      Which we can rephrase as:

      x*x = 2*2*7*(?) (where the question mark could be “anything else” – we know that x^2 is divisible by, but not limited to, 28).

      Because there are two x’s, we need to give each x an equal set of factors. The twos evenly divide so that each x gets its own 2. But the 7 does not. However, look at the question stem, which says “for integers x and y”. Sqrt(7) is not an integer, so the only way for x to be an integer is for each x to get its own 7…any other way and we don’t have an integer. So we now know that each x gets a 2 and a 7 (and maybe something else), so at minimum x = 2*7.

  5. SonyGmat says:

    perfect! i get it so far.. and then how do we continue our reasoning to conclude that y has only 2 and 7 as factors?

    Thanks again!

  6. SonyGmat says:

    sorry I meant at least 2 and 7 as factors..

  7. SonyGmat says:

    Questions with no answers yet.

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