Data Sufficiency

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dell2: Senior | Next Rank: 100 Posts; Posts: 36; Joined: Thu Jun 02, 2011 10:28 am

by dell2 » Wed Jan 18, 2012 12:42 pm

If y is an integer greater than 0, is (y3 - y) divisible by 4?

(1) y2 + y is divisible by 10.

(2) For a certain integer k, y = 2k + 1.

OA : B

Source : GrockiT

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Source: Beat The GMAT — Data Sufficiency | Wed Jan 18, 2012 12:42 pm

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Ian Stewart: GMAT Instructor; Posts: 2623; Joined: Mon Jun 02, 2008 3:17 amFollowed by: 355 members; Location: Montreal; Thanked: 1090 times; GMAT Score:780

by Ian Stewart » Wed Jan 18, 2012 3:18 pm

dell2 wrote:If y is an integer greater than 0, is (y3 - y) divisible by 4?

(1) y2 + y is divisible by 10.

(2) For a certain integer k, y = 2k + 1.

Since this is a divisibility question, we should certainly try to get factorizations. The expression y^3 - y can be factored first by factoring out one y, and then by using the difference of squares:

y^3 - y = y(y^2 - 1) = (y)(y-1)(y+1)

Now (y-1)(y)(y+1) is just the product of three consecutive integers. If, as Statement 2 tells us, y is odd, then y-1 and y+1 are both even, so their product must be divisible by 2^2. So Statement 2 is sufficient.

From Statement 1 we know that (y)(y+1) is divisible by 2*5. That will certainly be true if y = 10, but if y = 10, then (y-1)(y)(y+1) = 9*10*11 is *not* divisible by 4. So with Statement 1 alone, the answer to the question can be 'no'. It can also be 'yes' of course (let y = 5, say, or any other odd multiple of 5), so Statement 1 is not sufficient, and the answer is B.

For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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