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Integers - MGMAT

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by II » Mon Jul 07, 2008 3:16 pm
"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 "

Ok ... we can quickly identify that this problem is to do with consecutive integers. How ?
- x(x-1)(x-k) is most likely the product of 3 consecutive integers
- "must be evenly divisible by 3" --> recall the special products rule: if you have 3 consecutive integers, then the product of those 3 consecutive integers is divisible by 3. You will always have a factor of 3 in the product of 3 consecutive integers.

So we can say the following:
we know for sure that x and x-1 are consecutive. We only have a question about x-k ...

Basically we have 2 possible scenarios, with regards to x-k:
(1) x-k is the greatest of the 3 numbers: (x-1)(x)(x-k)
(2) x-k is the least of the 3 numbers: (x-k)(x-1)(x)

Lets take a look at scenario (1):
This means that x-k must be equal to x+1 (the next number up from x).
x-k = x+1
-k = 1
So k = -1 in this scenario.

Lets look at scenario (2):
This means that x-k must be equal to x-2 (the next number down from x-1).
x-k = x-2
-k = -2
So k = 2 in this scenario.

Lets take the values of k which we have found from considering the 2 scenarios:
we have k=-1, and k=2. The difference between these two values is 3.

In order to answer the question we just need to find the value in the answer choices which does not have a difference of 3, when taking one of the two possible values of k which we have found above (-1 and 2).

So lets go through the answer choices:
(A) -4 .... the difference between -1 and -4 is 3. So we can cross this out.
(B) -2 .... the difference between -1 and -2 is 1. So lets consider the other possible value of k. The difference between 2 and -2 is 4. So k cannot be -2 and B is the correct answer.

Just for completeness sake, we can go through the rest of the answer choices:
(C) -1 ... the difference between -1 and -1 is 0. So lets consider the other possible value of k. The difference between 2 and -1 is 3. So we can cross this out.
(D) 2 ... the difference between -1 and 2 is 3. So we can cross this out.
(E) 5 ... the difference between -1 and 5 is 6. Lets consider the other possible value of k. The diff between 5 and 2 is 3. So we can cross this out.

Hope this long explanation makes sense !
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Re: Integers - MGMAT

by raajan_p » Fri Nov 28, 2008 8:04 am
AleksandrM 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

a. -4
b. -2
c. -1
d. 2
e. 5
I thought of a different way of approaching it..can anyone confirm if my approach is ok...

Since the question says that X is an integer, Let me take X as 5(as an example)

Now, substitute the value of X in X(X-1)(X-K) for different values of K mentioned in answer choices.

Lets See...

When K = -4; X(X-1)(X-K) = (5*4*9)/3 -> divisible
When K = -2 ; X(X-1)(X-K) = (5*4*7)/3 -> Not divisible
When K = -1; X(X-1)(X-K) = (5*4*6)/3 -> divisible
When K = 2; X(X-1)(X-K) = (5*4*3)/3 - > divisible
When K = 5; X(X-1)(X-K) = (5*4*0)/3 -> divisible...

So B is the answer.
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by cramya » Fri Nov 28, 2008 9:37 pm
Raajan thats a good aproach and mine is similar!

Pick x and x-1 in such a way that neither is divisible by 3. We can clearly see x+2 i.e x-(-2) is the only one that will fit the bill

Eg: x=2 x-1=1 x-(-2) = x+2 which is 4

x=5 x-1=4 x-(-2) = x+2 = 7


Done

B)[/b]
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by bhumika.k.shah » Sun Mar 21, 2010 8:45 am
II wrote: Hope this long explanation makes sense !
yes it does. thanks :)

i dint quite understand MGMAT's explanation.
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by Stuart@KaplanGMAT » Sun Mar 21, 2010 1:28 pm
AleksandrM 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

a. -4
b. -2
c. -1
d. 2
e. 5
We can solve this question in 10 seconds if we understand the concepts and work with the choices, playing everyone's favourite Sesame Street game, "one of these things is not like the others, one of these things just doesn't belong" (https://www.youtube.com/watch?v=tZIvgQ9i ... re=related).

We're dividing our product by 3. Therefore, the correct answers will all be similarly related to 3.

c is 3 more than a; d is 3 more than c; e is 3 more than d. Accordingly, a, c, d and e are all in the same "cycle" of 3 and will all generate the same answer to the question.

b is the odd answer out: choose b.
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by kstv » Sun Mar 21, 2010 11:24 pm
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
a. -4 b. -2 c. -1 d. 2 e. 5
This concept is tested repeatedly.
x(x-1)(x-k ) will be divisible by 3 if they are consecutive no cos' one of the values has to be a multiple of 3.
Take any no 5 6 7 7 8 9 2 3 4
observe the numbers in the extreme which are underlined if you replace them with value +- 3 the divisibility by 3 holds
2 6 7 or 5 6 9
because if x x+1 x+2 is divisible by 3 so change x or x+2 by a value 3 and the product of three no. will still be divisible by 3
cos x(x+1) (x+2) is of 3a form as it is divisible by 3
(x+3)(x+1)(x+2) or x(x+1)(x+2+3) will be divisible by 3
This is the tedious part, but if we are convinced about this concept the actual calc. is easy
x (x-1) (x-k) notice x (x-1) are consequitive no. so if (x-k) is (x-2) it will be divisible by 2 , option d. (x-2-3) is (x-5) option e.
the series can also be (x-k) x(x-1) so if (x-k) is (x+1) is is valid (x-(-1)) option c or (x-(-1-3)) option aSo b remains , and it is the right answer.
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by kstv » Mon Mar 22, 2010 5:06 pm
"Stuart Kovinsky
We're dividing our product by 3. Therefore, the correct answers will all be similarly related to 3.

c is 3 more than a; d is 3 more than c; e is 3 more than d. Accordingly, a, c, d and e are all in the same "cycle" of 3 and will all generate the same answer to the question.

b is the odd answer out: choose b.
But nothing to beat this approach. Have to form the habit of looking very closely at the options (like the monster in the video) before we race to solve the Q.
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by tyronetan82 » Mon Jun 28, 2010 10:59 pm
raajan_p wrote:
AleksandrM 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

a. -4
b. -2
c. -1
d. 2
e. 5
I thought of a different way of approaching it..can anyone confirm if my approach is ok...

Since the question says that X is an integer, Let me take X as 5(as an example)

Now, substitute the value of X in X(X-1)(X-K) for different values of K mentioned in answer choices.

Lets See...

When K = -4; X(X-1)(X-K) = (5*4*9)/3 -> divisible
When K = -2 ; X(X-1)(X-K) = (5*4*7)/3 -> Not divisible
When K = -1; X(X-1)(X-K) = (5*4*6)/3 -> divisible
When K = 2; X(X-1)(X-K) = (5*4*3)/3 - > divisible
When K = 5; X(X-1)(X-K) = (5*4*0)/3 -> divisible...

So B is the answer.

Yeah, plugging in a number is what I did as well, but I started out with x = 10, it made all the answer choices divisible by 3, can someone tell me how to choose for the best number to plug in?

Thanks,
Tyrone
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