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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

The OA is B. I am trying to use the concept of consecutive numbers but got stuck. Can someone please help?

Question says x(x – 1)(x – k) must be evenly divisible by three which means x(x-1) (x-k) should be consecutive.

We have the product of 3 integers: (x-1)x(x-k).

Note that the product of 3 integers is divisible by 3 if at least one multiple is divisible by 3. Now, to guarantee that at least one integer out of x, (x – 1), and (x – k) is divisible by 3 these numbers must have different remainders upon division by 3, meaning that one of them should have remainder of 1, another reminder of 2 and the last one remainder of 0, so be divisible by 3.

Next, if k=-2 then we'll have (x-1)x(x+2)=(x-1)x(x-1+3) --> which means that (x-1) and (x+2) will have the same remainder upon division by 3. Thus for k=-2 we won't be sure whether (x-1)x(x-k) is divisible by 3.

Answer: B.

30 second approach: 4 out of 5 values of k from answer choices must guarantee divisibility of some expression by 3. Now, these 4 values of k in answer choices must have some pattern: if we get rid of -2 then -4, -1, 2, and 5 creating arithmetic progression with common difference of 3, so -2 is clearly doesn't belong to this pattern.

The rule that the product of three consecutive integers is a good start, but not the be all and end all.

Think about it this way:

number = x(x – 1)(x – k)

So far, we have integer x and one less that it (x - 1), so we could go down one more, or up one from x --

k = 2 ----> x(x – 1)(x – 2)

k = -1 ----> x(x – 1)(x + 1)

Now, we don't know which of the three factors are divisible by 3 -- x, or (x - 1), or the (x - k). If it's either of the first two, then we're golden, and k doesn't matter. But pretend that neither x nor (x - 1) is divisible by 3, then we are dependent on that last factor. Well, if (x - 2) is a multiple of three, we should be able to add or subtract three and still get a multiple of three.

(x - 2) - 3 = (x - 5)

(x - 5) - 3 = (x - 8)

(x - 2) + 3 = (x + 1), which we have already

(x + 1) + 3 = (x + 4)

(x + 3) + 3 = (x + 7)

So, for divisibility purposes, (x - 8), (x - 5), (x - 2), (x + 1), (x + 4), (x + 7) are all equivalent -- if any one of them is a multiple of three, all the others are. (You can check that the difference between any two is a multiple of 3.) BTW, if they are not divisible by three, then they all would have equal remainders if divided by three.

Back to the question: 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

All of those choices give us a term on our list except for (B) -2.

BTW, notice all the answer choices are spaced apart by three except for (B).

Does that make sense? Please do not hesitate to ask if you have any questions.

Mike
_________________

Mike McGarry Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]

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31 Jan 2012, 16:34

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This is my 1st post finally thought of jumping in instead of just being an observer I attacked this problem in a simple way. As it states it is divisible by 3 that means both x & (x-1) cannot be a multiple of 3 otherwise whatever the value of k it will be still divisible by 3 so plugging in number i chose 5 in this case you can establish answer is -2 does not fit...

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

The OA is B. I am trying to use the concept of consecutive numbers but got stuck. Can someone please help?

Question says x(x – 1)(x – k) must be evenly divisible by three which means x(x-1) (x-k) should be consecutive.

We have the product of 3 integers: (x-1)x(x-k).

Note that the product of 3 integers is divisible by 3 if at least one multiple is divisible by 3. Now, to guarantee that at least one integer out of x, (x – 1), and (x – k) is divisible by 3 these numbers must have different remainders upon division by 3, meaning that one of them should have remainder of 1, another reminder of 2 and the last one remainder of 0, so be divisible by 3.

Next, if k=-2 then we'll have (x-1)x(x+2)=(x-1)x(x-1+3) --> which means that (x-1) and (x+2) will have the same remainder upon division by 3. Thus for k=-2 we won't be sure whether (x-1)x(x-k) is divisible by 3.

Answer: B.

30 second approach: 4 out of 5 values of k from answer choices must guarantee divisibility of some expression by 3. Now, these 4 values of k in answer choices must have some pattern: if we get rid of -2 then -4, -1, 2, and 5 creating arithmetic progression with common difference of 3, so -2 is clearly doesn't belong to this pattern.

Hope it helps.

Bunuel,

Would this approach work for all integer divisors (NOT the 30 sec approach)? Say the divisor is 4 and the choices had four terms instead of three, e.g. (x-1)(x-2)(x+k)(x+1)?

Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]

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04 Oct 2012, 23:10

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enigma123 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

The OA is B. I am trying to use the concept of consecutive numbers but got stuck. Can someone please help?

Question says x(x – 1)(x – k) must be evenly divisible by three which means x(x-1) (x-k) should be consecutive.

Since this is a multiple choice GMAT question, you can pick a particular value for x such that neither x, nor x-1 is divisible by 3 and start checking the answers. In the given situation, choose for example x = 2 and check when 2 - k is not divisible by 3. (A) 2 - (-4) = 6 NO (B) 2 - (-2) = 4 BINGO!

Answer B.
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]

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05 Oct 2012, 03:56

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Avantika5 wrote:

I too solved it using a value for x =2, but I am not sure if is it better to solve using value for such questions or otherwise.

On the GMAT, is definitely the fastest way to solve it. Being a multiple choice question, you can be sure that there is a unique correct answer. And in the given situation, the only issue is to choose for x values such that neither x, nor x - 1 is divisible by 3. It won't harm to understand and know to use the properties of consecutive integers presented in the other posts . They can be useful any time.
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]

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06 Oct 2012, 02:10

Avantika5 wrote:

Thanks EvaJager, I was always thought it is not good way to solve and I should learn the better way. Thanks a lot...

Better is a relative word...Mathematicians always try to prove and justify everything in a formal, logical way. But GMAT is not testing mathematical abilities per se. If they wanted so, the questions would have been open and not multiple choice. Have a flexible mind, think out of the box. GMAT is not a contest for the most beautiful, elegant, mathematical solution... Get the correct answer as quickly as possible, and go to the next question without any feeling of guilt...:O)

Though, as I said, try to understand the other properties of the integer numbers, they can become handy and also, because they are so beautiful! Isn't Mathematics wonderful?
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Last edited by EvaJager on 06 Oct 2012, 02:46, edited 1 time in total.

Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]

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06 Oct 2012, 02:25

1. k=-4, x+4=y so x=y-4. Now ----- (y-4)(y-5)(y). Gives number divisible by 3 in all cases. 2. k=-2, , x+2=y so x=y-2. Now ----- (y-2)(y-3)(y). Not applicable when y is 5 or when x is 7 3. k=-1, , x+1=y so x=y-1. Now ----- (y-1)(y-2)(y). Consecutive 3 integers. Divisible by 3 4. k=2, , x-2=y so x=y+2. Now ----- (y+2)(y+1)(y). Consecutive 3 integers. Divisible by 3 5. k=5, , x-5=y so x=y+5. Now ----- (y+5)(y+4)(y). Consecutive 3 integers. Divisible by 3
_________________

I will rather do nothing than be busy doing nothing - Zen saying

The rule that the product of three consecutive integers is a good start, but not the be all and end all.

Think about it this way:

number = x(x – 1)(x – k)

So far, we have integer x and one less that it (x - 1), so we could go down one more, or up one from x --

k = 2 ----> x(x – 1)(x – 2)

k = -1 ----> x(x – 1)(x + 1)

Now, we don't know which of the three factors are divisible by 3 -- x, or (x - 1), or the (x - k). If it's either of the first two, then we're golden, and k doesn't matter. But pretend that neither x nor (x - 1) is divisible by 3, then we are dependent on that last factor. Well, if (x - 2) is a multiple of three, we should be able to add or subtract three and still get a multiple of three.

(x - 2) - 3 = (x - 5)

(x - 5) - 3 = (x - 8)

(x - 2) + 3 = (x + 1), which we have already

(x + 1) + 3 = (x + 4)

(x + 3) + 3 = (x + 7)

So, for divisibility purposes, (x - 8), (x - 5), (x - 2), (x + 1), (x + 4), (x + 7) are all equivalent -- if any one of them is a multiple of three, all the others are. (You can check that the difference between any two is a multiple of 3.) BTW, if they are not divisible by three, then they all would have equal remainders if divided by three.

Back to the question: 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

All of those choices give us a term on our list except for (B) -2.

BTW, notice all the answer choices are spaced apart by three except for (B).

Does that make sense? Please do not hesitate to ask if you have any questions.

Mike

Great posts everyone, including you Mike, who I am now quoting, but I would like to add that..

The first two integers for each answer choice cannot be divisible by 3 because all the answers choices would then be correct, thereby making the question invalid. Therefore, the last integer (x+k) must be divisible by 3, which helps in the theoretical approach...

Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]

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17 Oct 2013, 15:25

enigma123 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

The OA is B. I am trying to use the concept of consecutive numbers but got stuck. Can someone please help?

Question says x(x – 1)(x – k) must be evenly divisible by three which means x(x-1) (x-k) should be consecutive.

Hi enigma123. Thank you for the nice question. But you might wanna put a spoiler on this statement The OA is B so that others can solve the question in real conditions. Thanks

Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]

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27 Oct 2014, 06:45

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I tried the problem in a different way:

for any number to be divisible by 3, the sum of the integers in the number should be a factor of 3.

Taking the sum of x, (x-1) and (x-k) we have:

x+x-1+x-k = 3x-1-k

now looking at the choices

k = -4 => sum = 3x+3 --> divisible by 3 k = -2 => sum = 3x+1 --> not divisible by 3 k = -1 => sum = 3x --> divisible by 3 k = 2 => sum = 3x-3 --> divisible by 3 k = 5 ==> sum = 3x-6 --> divisible by 3

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 tried the problem in a different way:

for any number to be divisible by 3, the sum of the integers in the number should be a factor of 3.

Taking the sum of x, (x-1) and (x-k) we have:

x+x-1+x-k = 3x-1-k

now looking at the choices

k = -4 => sum = 3x+3 --> divisible by 3 k = -2 => sum = 3x+1 --> not divisible by 3 k = -1 => sum = 3x --> divisible by 3 k = 2 => sum = 3x-3 --> divisible by 3 k = 5 ==> sum = 3x-6 --> divisible by 3

so answer choice is (b)

Just refining the highlighted language and presenting another view to del with this problem

CONCEPT1:For any number to be divisible by 3, the sum of the Digits of the Number should be a Multiple of 3.

CONCEPT2: Product of any three consecutive Integers always include one muliple of 3 hence product of any three consecutive Integers is always a Multiple of 3

CONCEPT3: If a Number "w" is a Multiple of 3 then any number at a difference of 3 or multiple of 3 from "w" will also be a multiple of 3 i.e. If w is a multiple of 3 then (w+3), (w-3), (w+6), (w-6) etc. will all be Multiples of 3

Here, I see that x(x – 1) is a product of two consecutive Integers but if another Number next to them is obtained then x(x – 1)(x – k) will certainly be a multiple of 3 [as per Concept2 mentioned above]

for x(x – 1)(x – k) to be a product of 3 consecutive integers,

(x – k) should be either (x - 2) i.e. k=2 OR (x – k) should be either (x - 5) i.e. k=5 [Using Concept3] OR (x – k) should be either (x + 1) i.e. k=-1 OR (x – k) should be either (x + 4) i.e. k=-4 [Using Concept3]

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The OA is B. I am trying to use the concept of consecutive numbers but got stuck. Can someone please help?

Question says x(x – 1)(x – k) must be evenly divisible by three which means x(x-1) (x-k) should be consecutive.

A little bit faster approach.

We know from the start that product of \(3\) consecutive integers is divisible by \(3\), because it’s indeed a multiple of \(3\).

So we have \(x(x+1)(x+2)\) is definitely divisible by 3 and k=2 is our anchor point.

Now we can use principles of modular arithmetic to generate multiples of \(3\). We can either add or subtract \(3\) from our anchor number. And we got following string of numbers:

… -4, -1, 2, 5, 7 …

As you can see in this progression only answer B (-2) is absent.