If m = n*(n + 7)*(n + 8), where n is an integer, which of the followin

m=n∗(n+7)∗(n+8)
n = odd = 3: 3(10)(11) = divisible by 6
n = even = 10: 10(17)(18) = divisible by 6 and 4

Ans (B)

Now m=n(n+7)(n+8)....
As far as the property of multiple etc is concerned
I. n+7 and n+8 are consecutive, so surely a multiple of 2
II. n and n+3 and n+6 will have same property when divisiblity by 3 is concerned. So n can be written as n+6. Now this is nothing but product of THREE consecutive integers. So divisible by 3

Since we are looking for MUST, these 3 integers have to be multiple of 2 and 3, so m must be multiple of 2*3=6...

I. Divisible by 4..
If n+7 is a multiple of just 2....say n is 7, so 7(7+7)(7+8)=7*14*15....NO
II. Divisible by 6
YES we have already seen above
III. Divisible by 9
Same as n is 7...7*14*15...NO

II is MUST be true.

B

Posted from my mobile device


I. m is divisible 4......for n=3....it's not divisible
II. m is divisible by 6.......for n from 1 to 10....so on all values of m are divisible by 6
III. m is divisible by 9....for few values m is not divisible by 9


So only B is correct

OA:B

We are given that m=n*(n+7)*(n+8), we are to determine which of the following must be true. We are also given that n is an integer.

The best way to test is to set n=-1, n=1, n=2, n=-2, and check which of the conditions remains true.
If n=-1, m=(-1)(6)(7)
from the above, we know that m is divisible by 6 and not divisible by 4 and 9. Since statements I and III are not always true for n=-1, then I and III are not necessarily true for all possible values of m. Since the requirement for must be true that the statement must be true for all possible values, then discard I and III.

Because the answer choices do not include none, we can conclude that m is divisible by 6 for all integers. So II. must be true for all integer values of n.

The answer must, therefore, be option B.

The Answer is B

To be divisible by 4 we need 2 even numbers irrespective if n is odd or even. Like n and n+2 or n+1 and n+3. This case doesnt work if n is odd.

To be divisible by 6 , we need one number divisible by 3 and one by 2.
In this case if n is odd and not divisible by 3, every n+2,N+5, n+8 is divisble by 3. If n is odd and divisible by 3, every n, n+3 is divisble by 3.
Every n+1,n+3,n+5,n+7 is divisible by 2

If n is even, either n+7(as in case if 2) will be divisible by 3 or n+8(as in case of 4)

M cannot be divisible by 3 as we will never always have two odd numbers

Therefore answer is B

Posted from GMAT ToolKit

If m=n∗(n+7)∗(n+8), where n is an integer, which of the following must be true?

I. m is divisible 4.
II. m is divisible by 6.
III. m is divisible by 9.

--> if n= (-1), then m=(-1)*(6)*(7)=-42

(-42) is not divisible by 4 and 9.
--> That's it. We're done. We don't need to check any other values of n.
In answer choices, it says that one of I,II or III must be true.

The answer is B.

If n = 1, we have m = 1 x 8 x 9.

If n = 2, we have m = 2 x 9 x 10.

If n = 3, we have m = 3 x 10 x 11.

We see that when n = 3, m is not divisible by 4 or 9. However, in all instances, m is divisible by 6. Indeed, m is even (since one of the consecutive factors n + 7 or n + 8 must be even) and m is also divisible by 3 (because either n, n + 1 or n + 2 is divisible by 3 and we can write m = n * (n + 1 + 6) * (n + 2 + 6)). Thus, m is divisible by 6 regardless of what value we pick for n.

Answer: B

Its is not correct that the product is divisible by 6 for any 2 consecutive integers. the product is divisible by 6 only when the least integer is >7.
E.g: 4*5=20 -> not divisible by6
7*8 =56 -> not divisible.

It says n is an integer. Well if n = -7, then m is divisible by every number. I think some qualification should go into that statement.

That is exactly what the question asks. If you do not know the exact value of n, what must be true in the given cases.
You don’t know whether n=-7

If n=-7, then yes divisible by say 41,43 etc
But if not -7, then answer is no.
Combined we do not have a MUST be true scenario.

You're right. I missed that :-/ Thank you for correcting me.

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.