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If x is an integer, then x(x – 1)(x – k) must be evenly divisible by

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shameekv

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26 Oct 2015, 18:54

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VeritasPrepKarishma

I am providing the theoretical explanation below. Once you get it, you can solve such questions in a few seconds in future!

Notice a few things about integers:
-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16......

Every number is a multiple of 1
Every second number is a multiple of 2
Every third number is a multiple of 3
Every fourth number is a multiple of 4 and so on...

So if I pick any 3 consecutive integers, one and only one of them will be a multiple of 3: e.g. I pick 4, 5, 6 (6 is a multiple of 3) or I pick 11, 12, 13 (12 is a multiple of 3) etc..

x(x - 1)(x - k) will be evenly divisible by 3 if at least one of x, x - 1 and x - k is a multiple of 3. We know from above, (x - 2)(x - 1)x will have a multiple of 3 in it. Also, (x-1)x(x + 1) will have a multiple of 3 in it because they both are products of 3 consecutive integers. So k can be 2 or -1. Eliminate these options.
Now let me write down consecutive integers around x:

(x-5), (x - 4), (x - 3), (x - 2), (x - 1), x, (x + 1), (x + 2), (x + 3), (x + 4), (x + 5) etc

(x - 2)(x - 1)x will have a multiple of 3 in it. x could be the multiple of 3, (x - 1) could be the multiple of 3 or (x - 2) could be the multiple of 3, in which case (x - 5) will also be a multiple of 3.
So in any case, (x - 5)(x - 1)x will have a multiple of 3 in it. So k can be 5.

Similarly, (x-1)x(x + 1) will have a multiple of 3 in it. x could be the multiple of 3, (x - 1) could be the multiple of 3 or (x + 1) could be the multiple of 3, in which case (x + 4) will also be a multiple of 3.
So in any case, (x - 1)x(x + 4) will have a multiple of 3 in it. So k can be -4.

We cannot say whether (x-1)x(x + 2) will have a multiple of 3 in it and hence if k = -2, we cannot say whether the product is evenly divisible by 3.

Answer (B).

Quote:

Plz Could you please explain how x-5 will also be a multiple of 3. I couldnot understand that part.

If (x - 2) is a multiple of 3, (x - 5), a number 3 places away from (x - 5) will also be divisible by 3.

Say (x - 2) = 9 (a multiple of 3)
then (x - 5) = 6 (previous multiple of 3)

Hi Karishma, Initially I was trying the same method only, though I could not figure out the following scenario (x-1) x (x+1) (x+2), now say (x-1) is divisible by 3, in that case (x+2) i.e. k = -2 will also be divisible by 3 and hence in that case (x-1) x (x+2) would be evenly divisible by 3. Could you please tell me what am I missing here?

KarishmaB

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15 Mar 2016, 08:35

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shameekv

So the question is "... MUST BE divisible by ..."

Note that (x - 1)x(x + 2) may or may not be divisible by 3. This happens in case (x +1) is a multiple of 3.

e.g. say x = 5
(x - 1)x(x + 2) = 4*5*7 --> Not divisible by 3

Say x = 6
(x - 1)x(x + 2) = 5*6*8 --> Divisible by 3

So it is not necessary that (x - 1)x(x + 2) will be divisible by 3.

Why must the others be divisible by 3?
Say, for example take k = -4
(x - 1)x(x + 4)
This is equivalent to (x - 1)x(x + 1) in case we are talking about divisibility by 3. We know that (x - 1)x(x + 1) MUST have a multiple of 3.
So (x - 1)x(x + 4) MUST have a multiple of 3.

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enigma123

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

Show Spoiler

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.

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Bunuel

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 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.

which means that (x-1) and (x+2) will have the same remainder upon division by 3

Please explain the above line.

Bunuel

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mysteriouswoman

Bunuel

which means that (x-1) and (x+2) will have the same remainder upon division by 3

Please explain the above line.

Notice that (x-1) and (x+2) are 3 part each other, so the remainder upon division by 3 will be the same. For example, 5 and 8; 11 and 14; 12 and 15; ... each pair is 3 apart and the remainder for each pair upon division by 3 is the same.

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santorasantu

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)

Though this method gives the correct answer. There is a fundamental problem with the approach.

Here the x(x-1)(x-k) is the product of 3 numbers. I don't think we can take the sum of these 3 terms and say the product is divisible by 3 if the sum is divisible by 3.

With that same logic, 12 = 3x4, sum of factors: 3+4 = 7 (not divisible by 3). Then 12 should not be divisible by 3.

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CrackVerbal

Hi, can we not substitute values in the equation (x-K) from the answer options and see which one is divisible by 3? If we get x-k to be divisible by 3 we shall get to the answer. Is there any other way to solve the question quickly?

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Tarungaur

Since its a must be true type question and if you are using the answer options to solve it instead of seeing which one is divisible by 3 ,we should take each answer option and prove that in some particular case $ x(x – 1)(x – k)$ will not be divisible by 3. If you are using this approach, plugging in values for x can save a lot of time.

#Approach 1: Plugging in values for x.

Here we need to find a case where x(x – 1)(x – k) is not divisible by 3.

Lets take 3 consecutive numbers: 9,10,11

Lets start by assuming x =11 as x and x-1 is not a multiple of 3

x(x – 1)(x – k) = 11*10 *(11 - k) .

Since 11 and 10 are not divisible by 3, if (11- k) is also not divisible by 3 ,then we will get a scenario where x(x – 1)(x – k) is not divisible by 3.

Check each answer option and see for which one , 11-k is not divisible by 3.

A. -4 => (11--4) =15 is divisible by 3

B. -2 => (11--2) = 13 is not divisible by 3. That means 11*10*13 is not divisible by 3.
Since its a must be true question we can say that when k=-2 ,x(x – 1)(x – k) is not always divisible by 3. Option B is the correct answer.

C. -1 => (11--1) =12 is divisible by 3
D. 2 => (11-2) =9 is divisible by 3
E. 5 => (11-5) =6 is divisible by 3

#Approach 2:
We know that in every 3 consecutive numbers, there will be a multiple of 3.

Here it's given that x(x – 1)(x – k) must be evenly divisible by 3 and we need to find a value of k where x(x – 1)(x – k) is not always divisible by 3.
Let's say x,x-1,x-2 are 3 consecutive terms.

Case 1: if x is a multiple of 3, x(x – 1)(x – k) is always be divisible by 3 , irrespective of any values of k

Case 2: x is not a multiple of 3 and x-1 is multiple of 3.
Here also, x(x – 1)(x – k) is always be divisible by 3 , irrespective of any values of k

Case 3: x and x-1 is not a multiple of 3, That means x-2 should be a multiple of 3. As we know in every 3 consecutive numbers, there will be a multiple of 3.

If x-2 is a multiple of 3, then if we add or subtract 3 to x-2 each time, it should be also a multiple of 3.
So we can conclude that x-2, x-5, x-8, x-11... as well as x-2,x+1,x+4, x+7 will also be multiple of 3.
Hence we can conclude that value of K should be either 2,5,8,11.. or 2,-1,-4,-7.
Combining all the possible values of K in which x(x – 1)(x – k) must be divisible by 3 will be in an Arithmetic series ...-7,-4,-1,2,5,8.. with a common
difference 3.
If you analyse the answer options, Option B. -2 is not in list .Therefore, we can conclude that if k=-2 , we cannot say that x(x – 1)(x – k) must be evenly divisible by 3

#Approach 3: Spot the pattern in the answer options. Here in each answer options there is a common difference of 3 except Option B. i.e. -2. If you can apply the Arithmetic series logic explained in Approach 2 , we can easily figure that Option B would be the answer.

I hope this clears all your doubts.
Let me know in case if you need any further help on this.

Thanks,
Clifin J Francis,
GMAT SME

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KarishmaB

anilnandyala

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

KarishmaB please could you help me understand the following:

1) We know from above, (x - 2)(x - 1)x will have a multiple of 3 in it How did we get this? Is it because they are consecutive?
2) in (x-1)x(x+1) if we say (x-1) is a Multiple of 3 then (x+2) is also a multiple of 3. In that case, "k" could be -2. Why am I wrong?

Could you help me understand your process a bit better? thanks

KarishmaB

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31 Jan 2022, 23:06

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Hoozan

KarishmaB

anilnandyala

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

Hoozan

1. Pick any 3 consecutive integers. Exactly one of them will be a multiple of 3 because every third integer is a multiple of 3.

Say 1, 2, 3
or 5, 6, 7
or -1, 0, 1 (0 is a multiple of 3)
etc

Since (x-2), (x-1), x represent 3 consecutive integers, one of them will certainly be a multiple of 3.

Consider (x-2), (x-1), x
I say one of them must be a multiple of 3. It could be (x-2) or (x-1) or x.

What if I instead make it (x-2), (x-1), (x+3)
I can still say that one of them must be a multiple of 3. Why? Because (x-2), (x-1), x certainly has a multiple of 3. If it were (x-2), we still have (x-2). If it were (x-1), we still have (x-1). If it were x, then (x+3) is also a multiple of 3 so we would still have a multiple of 3.

2.
(x-1), x, (x+2)
needn't necessarily have a multiple of 3. What if (x+1) is a multiple of 3.
Say I select 4, 5, 7 (numbers of form (x-1), x, (x+2)). They don't have a multiple of 3.

Hence, it is not necessary that when k = -2, x(x – 1)(x – k) must be evenly divisible by 3. It may be but it is not necessary.

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KarishmaB So basically, we know that (x-1)(x)(x+1) are consecutive numbers and hence one of them will be a multiple of 3. Now if we have (x-1)x(x+4) we can still be sure that one of them will be a multiple of 3 since (x+1) and (x+4) mean the same.

But when we say (x-1)x(x+2) we aren't sure if one of them will be a multiple of 3 since (x-1) and (x+2) mean the same. Hence it is not necessary for k = -2 such that the expression is a multiple of 3

KarishmaB

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Updated on: 08 Aug 2023, 04:09

Originally posted by KarishmaB on 22 Mar 2022, 21:25.
Last edited by KarishmaB on 08 Aug 2023, 04:09, edited 1 time in total.

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Hoozan

Exactly.
This is discussed in a blog post on my website too: https://anaprep.com/arithmetic-factors- ... -integers/
and this and other number properties are discussed in detail in these videos:
https://www.youtube.com/watch?v=DxIH8rjhpKY
https://www.youtube.com/watch?v=Kd-4cH4cqHw

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santorasantu

Hello santorasantu and Bunuel - Don't intend to doubt the approach here. I did take random values and the values do hold true for this approach.

However, I'm not able to think through as to how can taking sum of three integers confirm that the product of these integers be a multiple of 3? Shouldn't the sum of digits of the product of three integers tell us that?

Hope my doubt is clear.

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santorasantu

I believe santorasantu incorrectly interpreted x(x – 1)(x – k) as a three-digit number, where x is the hundreds digit, (x-1) the tens digit, and (x-k) the units digit. This differs from the correct approach of viewing x(x – 1)(x – k) as the product of the three terms x, (x-1), and (x-k). Therefore, that method is incorrect.

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Bunuel

Bunuel, how come this incorrect method did work then? I still can't wrap my head around the logic.

Posted from my mobile device

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Bunuel

Bunuel, how come this incorrect method did work then? I still can't wrap my head around the logic.

Posted from my mobile device

It worked by a fluke. For the product of integers to be divisible by 3, it's not necessary for their sum to be divisible by 3. Ignore and move on.

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Similarly, (x-1)x(x + 1) will have a multiple of 3 in it. x could be the multiple of 3, (x - 1) could be the multiple of 3 or (x + 1) could be the multiple of 3, in which case (x + 4) will also be a multiple of 3.
So in any case, (x - 1)x(x + 4) will have a multiple of 3 in it. So k can be -4.

I am not able to understand, how you have concluded (x + 4) will also be the multiple of 3 ?

KarishmaB

anilnandyala

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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If x+1 = 3k, then x+4 => (x+1) + 3 = 3(k+1)

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hey how did you decide to take x=5?
say x = 10 , then for k=-1 it wont be divisible by 3

vjsharma25

Hi bunuel,
I don't understand the problem language, it says

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

how does it matter whats the value of K, i can choose x = 3 and the expression will always be divisible by 3.

Am i missing any minor yet important point?