Integers, is there a better explanation for this?

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ru2008: Senior | Next Rank: 100 Posts; Posts: 58; Joined: Tue Aug 26, 2008 9:42 am

by ru2008 » Mon Sep 06, 2010 5:11 am

If x is an integer, then x(x - 1)(x - k) must be evenly divisible by three when k is any of the following values EXCEPT
-4
-2
-1
2
5

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kmittal82: Master | Next Rank: 500 Posts; Posts: 324; Joined: Mon Jul 05, 2010 6:44 amFollowed by: 3 members; Location: London; Thanked: 70 times

by kmittal82 » Mon Sep 06, 2010 5:43 am

I think it should be (B) i.e. -2

If x = 8
x - 1 = 7
x-(-2) = 8+ 2 = 10

8 x 7 x 10 is not divisible by 3

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Maciek: Master | Next Rank: 500 Posts; Posts: 164; Joined: Sun Jul 18, 2010 5:26 amFollowed by: 4 members; Thanked: 49 times; GMAT Score:710

by Maciek » Mon Sep 06, 2010 5:55 am

Hi!

Integer is divisible by 3 if the sum of all the individual digits is evenly divisible by 3.
If the number is given as multiplication of factors, then any of them must be divisible my 3.

integer x(x-1)(x-k) is divisible by three if k has remainder 2 after dividing by 3.

a. -4
-4/3 = -2 remainder 2
b. -2
-2/3 = - 1 remainder 1
c. -1
-1/3 = -1 remainder 2
d. 2
2/3 = 0 remainder 2
e. 5
5/3 = 1 remainder 2

Hence, the answer is b

Hope it helps!
Best,
Maciek

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CompBanker: Newbie | Next Rank: 10 Posts; Posts: 7; Joined: Sat Jul 11, 2009 5:48 am; Thanked: 1 times; GMAT Score:760

by CompBanker » Mon Sep 06, 2010 12:54 pm

This question is testing your knowledge of consecutive integers. If three consecutive integers are multiplied together, the result MUST be divisible by 3. (1*2*3, 2*3*4, 3*4*5, etc.). This is true because 3 divides into every third number (3,6,9,12, etc.).

For this problem:

x represents a number and (x - 1) represents 1 less than a number. In order for you to be SURE that the product is divisible by 3, the third number must represent either a consecutive integer or a consecutive integer +/- a multiple of 3.

Plugging in some numbers:

x = 11
(x-1) = 10
(x-?) = [The result is divisible by 3 as long as this number is +/- a multiple of 3 away from 9.]

So, the possible values of [?] in this case can be: -4: (x-(-4)=15); -1: (x-(-1)=12); 2: (x-2=9); 5: (x-5=6)

Notice that this pattern can go on forever. -4; -1; 2; 5; 8; 11; 14; 17; 20 are all viable numbers.

CompBanker

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Mon Sep 06, 2010 3:13 pm

ru2008 wrote:If x is an integer, then x(x - 1)(x - k) must be evenly divisible by three when k is any of the following values EXCEPT
-4
-2
-1
2
5

Our goal is to make each of the 3 factors -- x, (x-1), and (x-k) -- NOT divisible by 3.

This means that x cannot be a multiple of 3 and x-1 cannot be a multiple of 3.

Let x = 5. This works because x = 5 is not divisible by 3 and x-1=4 is not divisible by 3.

Now we just plug in the answer choices, which represent the value of k. We're looking for the answer choice that doesn't yield a multiple of 3.

A) If k = -4, x-k = 5 - (-4) = 9. 9 is a multiple of 3. Eliminate A.
B) If k = -2, x-k = 5 - (-2) = 7. 7 is not a multiple of 3. Success!

The correct answer is B.

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gig92: Senior | Next Rank: 100 Posts; Posts: 78; Joined: Fri Jul 23, 2010 2:22 amFollowed by: 1 member; Thanked: 2 times

by gig92 » Tue Sep 07, 2010 4:42 am

ru2008 wrote:If x is an integer, then x(x - 1)(x - k) must be evenly divisible by three when k is any of the following values EXCEPT
-4
-2
-1
2
5

The product of any three consecutive integers must be divisible by 3. In the question above, as it says, choose ANY integer as x (say 11) and then plug the values around:
e.g. if x = 11 -> 11(10)(11-k), we can plug each value one by one. Even if some values of k don't give the next consecutive integer, we have enough information base to check some mutiple of 3. The product is divisible by for any value of k EXCEPT -2 and it stands true for any integer value of x.

gig92

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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members; Location: India; Thanked: 267 times; GMAT Score:760

by sanju09 » Tue Sep 07, 2010 5:06 am

ru2008 wrote:If x is an integer, then x(x - 1)(x - k) must be evenly divisible by three when k is any of the following values EXCEPT
-4
-2
-1
2
5

I am purposely taking x = 5 as reference, because the possible product

5 Ã— 4 Ã— (5 - k)

would have 5 Ã— 4 as already indivisible by 3.

Hence, we only need to check with which choice for k, (5 - k) is not divisible by 3.

A. Gives 9, proceed

B. Gives 7, [spoiler]STOP

B[/spoiler]

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Tue Sep 07, 2010 5:07 pm

ru2008 wrote:If x is an integer, then x(x - 1)(x - k) must be evenly divisible by three when k is any of the following values EXCEPT
-4
-2
-1
2
5

Here's a 10 second solution, based on understanding of divisibility concepts (and an application of an old Sesame Street game, "one of these things is not like the others").

When we divide by 3, we "cycle" in 3s.

-4, -1, 2 and 5 are all in the same place in the 3 cycle (i.e. each one is a multiple of 3 apart from the others).

Accordingly, -2 MUST be the correct answer to the question.

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