Challenge Problem 09/30/02

[aaa]

What if n = 1? Wouldnt n-1 = 0? A would no longer be the correct answer. Where am I off?

Question
Given that n is an integer, is n "” 1 divisible by 3?

(1) n^2+n is not divisible by 3

(2) , where k is a positive multiple of 3

(A) Statement (1) alone is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) alone is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) Each statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.




Answer
Statement (1) gives us information about , which can be rewritten as the product of two consecutive integers as follows:





Since the question asks us about n "” 1, we can see that we are dealing with three consecutive integers: n "” 1, n, and n + 1 .

By definition, the product of consecutive integers is divisible by the number of terms. Thus the product of three consecutive integers must be divisible by 3.

Since we are told in Statement (1) that the product is not divisible by 3, we know that neither n nor n + 1 is divisible by 3. Therefore n "” 1 must be divisible by 3.

Statement (1) is therefore sufficient.

Statement (2) can be rewritten as follows:





Given that k is a positive multiple of 3, we know that n must be greater than or equal to 2. This tells us that the members of the consecutive set n "” 1, n, n + 1 are nonzero integers.

By itself, however, this information does not give us any information about whether n "” 1 is divisible by 3. Thus Statement (2) alone is not sufficient.

The correct answer is A.

[rfernandez]

I think you're assuming that 0 is not divisible by 3. But it is! 0/3 = 0, with no remainder.

Rey

[jigar24]

Hi,

The answer given in MGMAT is as follows:


Statement (1) gives us information about , which can be rewritten as the product of two consecutive integers as follows:

Since the question asks us about n "” 1, we can see that we are dealing with three consecutive integers: n "” 1, n, and n + 1 .

By definition, the product of consecutive nonzero integers is divisible by the number of terms. Thus the product of three consecutive nonzero integers must be divisible by 3.

Since we are told in Statement (1) that the product is not divisible by 3, we know that neither n nor n + 1 is divisible by 3. Therefore it seems that n "” 1 must be divisible by 3.

However, this only holds if the integers in the consecutive set are nonzero integers. Since Statement (1) does not tell us this, it is not sufficient.


Statement (2) can be rewritten as follows:
Given that k is a positive multiple of 3, we know that n must be greater than or equal to 2. This tells us that the members of the consecutive set n "” 1, n, n + 1 are nonzero integers.

By itself, this information does not give us any information about whether n "” 1 is divisible by 3. Thus Statement (2) alone is not sufficient.

When both statements are taken together, we know that the members of the consecutive set n "” 1, n, n + 1 are nonzero integers and that neither n nor n + 1 is divisible by 3. Therefore, n "” 1 must be divisible by 3.

The correct answer is C: both statements together are sufficient but neither statement alone is sufficient to answer the question.

However, according to MGMAT strategy guide, 0 is a multiple of all numbers. Hence 0/3 is 0 and has no remainder. So even if consecutive integers are 0,1,2 [(n-1), n, (n+1)] .... n-1 is still a multiple of 3. Why does the solution state otherwise?? The answer should be A and NOT C Please some MGMAT prof. reply.

[jigar24]

There is an error in the solution..isn't it?

[mschwrtz]

You're right, as written S1 is sufficient, so the answer should be A. Well, assuming that S2 isn't sufficient, which I can't really tell from the post, since I can't see S2. This does make we wonder whether the rest of the question is correctly transcribed.

[jigar24]

I should get a prize for finding such errors and wasting so much of my time over these wrong questions.. I have found 2 now..

[mschwrtz]

Eek. I'll see if we can't edit those questions.