If n is an integer greater than 6, which of the following mu

Since 3 is a prime number then in order the product to be divisible by 3 either of the multiples must be divisible by 3. Now, to guarantee that at least one multiple is divisible by 3, these numbers must have different remainders upon division by 3, meaning that one of them should have the remainder of 1, another the reminder of 2 and the third one the remainder of 0, so be divisible by 3.

For option A: n and n+1 have different remainder upon division by 3. As for n-4, it will have the same remainder as (n-4)+3=n-1, so also different than the remainders of the previous two numbers.

Answer: A.

Similar question to practice: if-x-is-an-integer-then-x-x-1-x-k-must-be-evenly-divisible-126853.html

Hope it helps.
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anything in the form of (n-1) (n) (n+1) is divvisible by 3. in other word, a product of any 3 consecutie intevers is divisible by 3.

A. n (n+1) (n-4) = n (n+1) ((n-1)-3) is equivalant to (n-1) (n) (n+1)
B. n (n+2) (n-1) is equivalant to (n+1) missing.
C. n (n+3) (n-5) is equivalant to (n-1) missing and n repeating.
D. n (n+4) (n-2) is equivalant to odd/even consqcutive integers
E. n (n+5) (n-6) is equivalant to (n+1) missing and n repeating.


So A is good.

question testing multiples of 3.

plug in 7 and 8 as test values.

A. n (n+1) (n-4) = 7*8*3 and if n = 8 --> 8*9*4
B. n (n+2) (n-1) = 7*9*6 and if n = 8 --> 8*10*7
C. n (n+3) (n-5) = 7*10*5; eliminate as there are no multiples of 3
D. n (n+4) (n-2) = 7*11*5; eliminate as there are no multiples of 3
E. n (n+5) (n-6) = 7*12*1 and and if n = 8 --> 8*13*2

only A works.

Substituting 3 consecutive numbers might be the easiest way to solve this question as shown above.

Algebraic solution:

The product of 3 numbers to be divisible by 3 at least one of them must be divisible by 3. So, to ensure that the product of 3 integers shown is divisible by 3 all 3 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. We should have something like \(n(n+1)(n+2)\) (for example: if n divided by 3 yields remainder of 1, then n+1 yields remainder of 2 and n+2 yields remainder of 0, thus it's divisible by 3 OR if n divided by 3 yields remainder of 2, then n+2 yields remainder of 1 and n+1 yields remainder of 0, thus it's divisible by 3).

Only option A satisfies this, because \(n(n+1)(n-4)=n(n+1)(n-6+2)\) and \(n-6\) has the same remainder as \(n\) upon division by 3 thus we can replace it by \(n\).

Answer: A.

Hope it's clear.
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Same concept different way of thinking

How do I ensure that a product of three numbers is divisible by 3 ?
Step 1 : I divide the set of integers into three parts such that one of the parts contains exactly and only the multiples of 3
Step 2 : I ensure that I chose my numbers such that each belongs to a different part
Step 3 : If I achieve this, I must have exactly one number which belongs to the set of multiples of three. Hence their product is divisible by three.

How do I achieve this ?
Step 1 : The division I choose is numbers of the form \(3\alpha, 3\alpha+1, 3\alpha+2\). The union of these sets is all integers and they have no overlap between them, also one of them contains all and only multiples of 3.
Step 2 : Simplest way to choose three numbers such that exactly one belongs to each of the above sets is 3 consecutive numbers. This choice will always satisfy what I am looking for. Hence n*(n-1)*(n+1)

Important Note : If I shift any number by 3, its constituent set amongst \(\alpha, 3\alpha+1, 3\alpha+2\) does not change

So how do we solve this question ?
Start from the expression n*(n-1)*(n+1) and see if by adding or subtracting 3 or multiples of 3 from one or more of the three terms we can get any target expression.

A) Satisfies
B) n-1 repeats twice
C) n repeats twice
D) n+1 repeats twice
E) n repeats twice

Hence, solution is A
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Ok, so, MGMAT says that the product of k consec. int's is ALWAYS divisible by k!

Well, this ain't right. Q.82 of the OG 12 -->

If n is an integer greater than 6, which of the following must be divisible by 3?

a) n(n+1)(n-4)
b) n(n+2)(n-1)
c) n(n+3)(n-5)
d) n(n+4)(n-2)
e) n(n+5)(n-6)

I chose B because B represents 3 consecutive integers, my reasoning was that the product of B MUST be divisble by k!, which is 3! = 6... therefore by the Factor Foundation Rule, it MUST also be divisible by 3 because 3 is a factor of 6, right? WRONG.

Can anyone think why this is NOT the case? The official answer is A...

As for the rule you quote it's correct:

• If \(k\) is odd, the sum of \(k\) consecutive integers is always divisible by \(k\). Given \(\{9,10,11\}\), we have \(k=3\) consecutive integers. The sum of 9+10+11=30, therefore, is divisible by 3.

• If \(k\) is even, the sum of \(k\) consecutive integers is never divisible by \(k\). Given \(\{9,10,11,12\}\), we have \(k=4\) consecutive integers. The sum of 9+10+11+12=42, therefore, is not divisible by 4.

• The product of \(k\) consecutive integers is always divisible by \(k!\), so by \(k\) too. Given \(k=4\) consecutive integers: \(\{3,4,5,6\}\). The product of 3*4*5*6 is 360, which is divisible by 4!=24.
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Shortcut: In every set of 3 consecutive numbers, ONE of them must be divisible by 3 when we are multiplying each of the digits

1*2*3
2*3*4
3*4*5
4*5*6

(A) n(n + 1)(n – 4)

The easiest is if we have something like n(n+1)(n+2)
We know in this case we DEFINITELY have an expression that is divisible by 3.
n=1 => 1*2*3
n=2 => 2*3*4
n=3 => 3*4*5

All are divisible by 3.

Any expression must pass our 3 consecutive integer test.

n(n + 1)(n – 4)
n=1 => 1*2*-3
n=2 => 2*3*1
n=3 => 3*4*-1

Even if n = 4 we have:
4*5*0 = 0 which is divisible by 3.

If n = 16
16*17*12 is divisible by 3.

So (A) passes all the tests and one of numbers in the expression will be divisible by 3 so the whole expression when multiplied together will be divisible by 3.

Every three consecutive positive integers will have one value divisible by 3.

Which method is better in questions like these to apply on the test? to pick the numbers or to apply concept of divisibility by 3 ( Remainder 0,1,2)

Also, dear Bunuel,

"Now, to guarantee that at least one multiple is divisible by 3, these numbers must have different remainders upon division by 3"

Why is this? Why the numbers must have different reminders?

Hi Bunuel,

Can you explain why the remainder should be different upon division by 3? I atill dint understood uproach

Posted from my mobile device

An integer divided by 3 can have 3 possible remainders: 0, 1, or 2.

Now, consider the product of three numbers a*b*c. If we are told that a, b, and c have different reminders upon division by three, this would mean that one of the numbers yields the remainder of zero, thus it's a multiple of 3. Thus abc is a multiple of 3.

Hope it helps.
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In my opinion, using logic is almost always better than plugging in. Plugging in is full of possibilities of making mistakes - incorrect calculation, not considering all possibilities, getting lost in the options etc.

Logic is far cleaner.

I know that talking about positive integers, in any set of 3 consecutive positive integers, one integer will be divisible by 3 and the other 2 will not be.

So I am looking for 3 consecutive positive integers e.g. (n-1)n(n+1)

Note that (n-4) is equivalent to (n-1) since if (n-4) is divisible by 3, so is (n-1). If (n-4) is not divisible by 3, neither is (n-1) (because the difference between these two integers is 3)
Hence (A) is equivalent to 3 consecutive integers.
Answer (A)

On same lines, note that n is equivalent to (n-3), (n + 3), (n + 6) etc.
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Any positive number will take one of three forms: 3m, (3m+1) or (3m+2) i.e. either it will be divisible by 3, will leave remainder 1 or will leave remainder 2 when divided by 3. If the number takes the form 3m, the number after it is of the form 3m+1 and the one after it is of the form 3m+2.

If we have 3 consecutive numbers such as a, (a+1), (a+2), we know for sure that at least one of them is divisible by 3 since one of them will be of the form 3m. We don't know which one but one of them will be divisible by 3.

So given numbers such as (n-1)*n*(n+1), we know that the product is divisible by 3.
In the given options, we don't know whether n is divisible by 3 or not. We need to look for the option which has 3 consecutive numbers i.e. in which the terms leave a remainder of 0, 1 and 2 to be able to say that the product will be divisible by 3.

Note a product such as n(n+3)(n+6). When this is divided by 3, we cannot say whether it is divisible or not because all three factors will leave the same remainder, 1.
Say n = 4. Product 4*7*10. All these factors are of the form 3m+1. We don't have any 3m factor here.
So we need the factors to have 3 different remainders so that one of them is of the form 3m.
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Hi Bunuel,

I took the picking numbers method but I want to understand yours as well.
I understand that n (n+1), n (n+2), n (n+3)...etc will all have different remainders, and I understand the concept about the different remainders.
I don't understand how you reached this statement: "As for n-4, it will have the same remainder as (n-4)+3=n-1"
Can you please elaborate on that?

n-1=(n-4)+3. Now, since 3 IS divisible by 3, then the remainder when (n-4)+3 is divided by by 3 is the same as n-4 is divided by 3.

For example, say n=11:
The remainder when n-1=11-1=10 divided by 3 is 1.
The remainder when n-4=11-4=7 divided by 3 is also 1.

Does this make sense?
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Hi MensaNumber,

This question can be solved rather easily by TESTing VALUES, although the work itself will take a bit longer than average and it would help a great deal if you could spot the subtle Number Properties involved.

From the question stem, you can see that we're dealing with division by 3 (or the 'rule of 3', if you learned the concept that way). You don't actually have to multiply out any of the answer choices though - you just need to find the one answer that will ALWAYS have a '3' in one of its 'pieces.' The subtle Number Property I referred to at the beginning is the 'spacing out' of the terms.

(1)(2)(3) is a multiple of 3, since it's 3 times some other integers.

(5)(6)(7) is also a multiple of 3, since we can find a 3 'inside' the 6, so we have 3x2 times some other integers.

Looking at the answer choices to this question, we're clearly NOT dealing with consecutive integers, but the 'cycle' of integers is something that we can still take advantage of.

For example, we know that...
When n is an integer, (n+1)(n+2)(n+3) will include a multiple of 3, since it's 3 consecutive integers (one of those 3 terms MUST be a multiple of 3, even if you don't know exactly which one it is).

You can take this same concept and 'move around' any (or all) of the pieces:

(n+1)(n+2)(n+6) will also include a multiple of 3 (that third term is 3 'more' than 'n+3').

Instead of adding a multiple of 3 to a term, you could also subtract a multiple of 3 from a term.

eg. (n-2)(n+2)(n+3) will also include a multiple of 3 (that first timer is 3 'less' than 'n+1').

The correct answer to this question subtracts a multiple of 3 from one of the terms.

Final Answer:

All things being equal, I'd still stick to TESTing VALUES (and not approaching the prompt with math theory) - the math is easy and you can put it 'on the pad' with very little effort.

GMAT assassins aren't born, they're made,
Rich
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To solve semi logically, take n in each. In choice A, either adding 1 to n or subtracting 4 from n will give you a multiple of 3. If neither n will be a multiple of 3.So it has to be divisible by 3.

To illustrate take the extreme values. In A, (n-4) and (n+1) are the extreme values. There is a difference of 5. (3,8), (4,9), (5,10) are examples . (n-4), (n+1) and n resp. are the multiples of 3 in these examples, making the expression divisible by 3.

Seeing 2 values at a time reduces the clutter.
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Video solution from Quant Reasoning:

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