MGMAT problem

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Redhorsep: Senior | Next Rank: 100 Posts; Posts: 49; Joined: Thu Jun 30, 2011 2:47 pm; Thanked: 1 times

by Redhorsep » Sat Oct 01, 2011 10:41 am

Hi,

I don't really understand the MGMAT explanation for this problem, it would be helpful if someone can solve this and provide a detailed explanation.

Thanks!

***
If x is an integer, then x (x-1) (x-k) must be evenly divisible by 3 when k is any of the following except:

A) -4
B) -2
C) -1
D) 2
E) 5

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Source: Beat The GMAT — Problem Solving | Sat Oct 01, 2011 10:41 am

GmatMathPro: GMAT Instructor; Posts: 349; Joined: Wed Sep 28, 2011 3:38 pmFollowed by: 54 members; Location: Austin, TX; Thanked: 236 times; GMAT Score:770

by GmatMathPro » Sat Oct 01, 2011 11:17 am

Every third integer is a multiple of 3: 1,2,3,4,5,6,7,8,9....

Any time you multiply integers together and one of them is a multiple of 3, the product will be divisible by 3.

x and x-1 are consecutive integers. Because they are factors of this expression, it will certainly be divisible by three if either one of them is a multiple of 3, like (6,7) or (9,10).

The only way it could fail to be divisible by three is if x and x-1 fall between the multiples of three, like if they were (7,8). In that case making the expression divisible by three would fall entirely on the shoulders of the third factor, (x-k). In this example, x=8 and x-1=7. Plugging in values of k from the answer choices with x=8 gives us 12, 10, 9, 6, 3 for A,B,C,D,E, respectively. B is the only choice for k that fails to make x-k a multiple of three, so it is the only choice of k that would fail to GUARANTEE that the product is ALWAYS divisible by 3.

If you don't like plugging in numbers, consider the following sequence:

x-5,x-4,x-3,x-2,x-1,x,x+1,x+2,x+3,x+4

Again, if x-1 and x are not multiples of 3, then x+1, x+4, x-2, and x-5 must be. These would correspond to A,C,D,E for x-k

Pete Ackley
GMAT Math Pro
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nimish: Newbie | Next Rank: 10 Posts; Posts: 8; Joined: Mon Nov 09, 2009 10:38 pm; Thanked: 2 times

by nimish » Sat Oct 01, 2011 6:21 pm

GmatMathPro wrote:Every third integer is a multiple of 3: 1,2,3,4,5,6,7,8,9....

Any time you multiply integers together and one of them is a multiple of 3, the product will be divisible by 3.

x and x-1 are consecutive integers. Because they are factors of this expression, it will certainly be divisible by three if either one of them is a multiple of 3, like (6,7) or (9,10).

The only way it could fail to be divisible by three is if x and x-1 fall between the multiples of three, like if they were (7,8). In that case making the expression divisible by three would fall entirely on the shoulders of the third factor, (x-k). In this example, x=8 and x-1=7. Plugging in values of k from the answer choices with x=8 gives us 12, 10, 9, 6, 3 for A,B,C,D,E, respectively. B is the only choice for k that fails to make x-k a multiple of three, so it is the only choice of k that would fail to GUARANTEE that the product is ALWAYS divisible by 3.

If you don't like plugging in numbers, consider the following sequence:

x-5,x-4,x-3,x-2,x-1,x,x+1,x+2,x+3,x+4

Again, if x-1 and x are not multiples of 3, then x+1, x+4, x-2, and x-5 must be. These would correspond to A,C,D,E for x-k

But, the question says, evenly divisible... So, does that mean, we have to look for even divisors(3*2, 3*4, 3*6..etc)

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GmatMathPro: GMAT Instructor; Posts: 349; Joined: Wed Sep 28, 2011 3:38 pmFollowed by: 54 members; Location: Austin, TX; Thanked: 236 times; GMAT Score:770

by GmatMathPro » Sat Oct 01, 2011 6:35 pm

In this context, "evenly" doesn't have anything to do with even integers. When you say x divides y evenly, it just means y divided by x has no remainder. For example, 5 divides 15 evenly, because 15/3=5, even though none of those are even numbers.

Pete Ackley
GMAT Math Pro
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rooster: Senior | Next Rank: 100 Posts; Posts: 44; Joined: Wed Aug 25, 2010 12:23 am; Thanked: 6 times

by rooster » Sun Oct 02, 2011 9:50 pm

This is a test of you knowledge of number properties: the product of a number of consecutive integers is divisible by its number of terms. it works to make a table listing x , x+1 and x+2 where it turns into this:

x-6, x-5, x-4
x-3, x-2, x-1
x, x+1, x+2
x+3, x+4, x+5

Thus, you need a number that fulfills the middle column, the only one that does not is B

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