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PS Problem

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by majsandip » Wed May 07, 2008 8:40 pm
If x and y are consecutive positive integer multiples of 3, what is the greatest integer J such that xy/J is always an integer?
a) 27
b) 18
c) 9
d) 6
e) 3.

My ans a.
Right choice B :shock:

Since the Ques stem says consecutive +ve integer it may be 3,6 or 6,9 or 12,15.......???
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by netigen » Wed May 07, 2008 9:17 pm
since, the two numbers are consecutive multiples of 3 they can be written as a and a+1

so x = 3a and y=3(a+1)

xy = 9a(a+1)

since, a and a+1 are consecutive one of them has to be even

so a(a+1) = 2W

xy = 9 * 2W = 18W
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by VP_RedSoxFan » Thu May 08, 2008 3:42 am
Alternatively, the question stem shares a few insights about the prime factors of x and y

Since they are both multiples of 3, they must each have a prime factor of 3. Since they are consecutive, one of them must be even and so one will have a prime factor of 2 as well.

Therefore, even without knowing any thing else about the numbers, you know that x * y will at least have 2, 3, 3 as 3 of its prime factors. No matter what the numbers are then, x * y will always be divisible by 2 * 3 * 3 = 18
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Re: PS Problem

by Stuart@KaplanGMAT » Thu May 08, 2008 10:36 am
majsandip wrote:If x and y are consecutive positive integer multiples of 3, what is the greatest integer J such that xy/J is always an integer?
a) 27
b) 18
c) 9
d) 6
e) 3.

My ans a.
Right choice B :shock:

Since the Ques stem says consecutive +ve integer it may be 3,6 or 6,9 or 12,15.......???
I'd have just done it by picking numbers.

Since xy/J MUST be an integer, and we want the greatest possible value of J, let's pick the smallest possible values of x and y.

So, let's use x=3 and y=6.

We know that 18/J MUST be an integer. So, the biggest possible J that will work is 18: choose (b).

If we're doubting that one pick is enough to draw that conclusion, try 6 and 9 as well.

54/J is an integer - 18 works again. Even the doubters should pick (b) now!
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by II » Thu May 08, 2008 12:36 pm
If we list the postive multiples of 3, we have:
3, 6, 9, 12, 15, 18, .... etc

Lets take the first two: 3 and 6.
What do we know about these numbers ? Well 3 is divisible by 3 (3 has a prime factor of 3), and 6 is an even number and hence is divisible by 2 as well as 3 (it has prime factors of 2 AND 3).
This fact applies to every pair of consecutve positive multiple of 3.
So this means that J must have a 2, a 3, and another 3 as its prime factors, if xy/J is always an integer.
2*3*3 = 18.

Hope this makes sense.

P.S. Most of the number properties questions (especially divisibilty and primes) are based upon prime factors, which are the base upon which all numbers are built. So it is helpful to break numbers down into their prime factors.

However, for this question, I agree with Stuart, plugging numbers is the best approach.
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Thanks!!!!

by majsandip » Fri May 09, 2008 4:33 am
Thanks to all of you :)
lived by chance!!
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