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

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.

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.

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?

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

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

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.

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

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

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.

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.

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.