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Re: If x is the sum of six consecutive integers, then x is divisible by wh [#permalink]

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21 Oct 2014, 09:32

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Bunuel wrote:

Tough and Tricky questions: properties of numbers.

If x is the sum of six consecutive integers, then x is divisible by which of the following:

I. 3 II. 4 III. 6

A. I only B. II only C. III only D. I and III E. I, II, and III

let the six numbers be k-2, k-1, k, k+1, k+2, k+3 then, there sum will be 6k+3 3(2k+1),which is clearly a multiple of 3. Also, 2k+1 will always be odd. thus the sum will never be divisible by 2.

On Test Day, the answers to Roman Numeral questions are almost always written in a way so that you can avoid some of the 'work.' Here, you should notice that at least 1 of the Roman Numerals appears in each answer, so we can use that to our advantage.

We're told that X is the SUM of 6 CONSECUTIVE INTEGERS. We're asked what X is divisible by....

Let's TEST VALUES....

IF we use the 6 consecutive integers: 1, 2, 3, 4, 5 and 6, then the sum = 21.

21 is divisible by 3 21 is NOT divisible by 4 21 is NOT divisible by 6

There's only one answer that 'fits' with these facts, so there's no additional work required...

Re: If x is the sum of six consecutive integers, then x is divisible by wh [#permalink]

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11 Dec 2017, 03:02

I understand this can be done by picking numbers and testing them out.

But can someone clarify this concept. It is said every 2 numbers, atleast one number is divisible by 2, for every 3 number atleast one number is divisble by 3. and so on so forth . Using this concept. Should the answer not be E?

It says consecutive six integers? so by the concept the factors should include 6 ,4 and 3.

I am sure i am making a very stupid mistake here. Kindly correct me where i am wrong.

I understand this can be done by picking numbers and testing them out.

But can someone clarify this concept. It is said every 2 numbers, atleast one number is divisible by 2, for every 3 number atleast one number is divisble by 3. and so on so forth . Using this concept. Should the answer not be E?

It says consecutive six integers? so by the concept the factors should include 6 ,4 and 3.

I am sure i am making a very stupid mistake here. Kindly correct me where i am wrong.

You should consider the sum of the numbers. What I mean that, yes from two consecutive integers one must be divisible by 2 (even) but does is mean that the sum of two consecutive integers is divisible by 2? No, for example, 1 + 2 = 3, which is not divisible by 2.

If x is the sum of six consecutive integers, then x is divisible by which of the following:

I. 3 II. 4 III. 6

A. I only B. II only C. III only D. I and III E. I, II, and III

(n - 2) + (n - 1) + n + (n + 1) + (n + 2) + (n + 3) = 6n + 3 = 3(2n + 1) = 3*odd. So, the sum of six consecutive integers, is an odd multiple of 3: ..., -9, -3, 3, 9, 15, ... It will always be divisible by 3 but not by 2 or any other multiple of 2.

I understand this can be done by picking numbers and testing them out.

But can someone clarify this concept. It is said every 2 numbers, atleast one number is divisible by 2, for every 3 number atleast one number is divisble by 3. and so on so forth . Using this concept. Should the answer not be E?

It says consecutive six integers? so by the concept the factors should include 6 ,4 and 3.

I am sure i am making a very stupid mistake here. Kindly correct me where i am wrong.

Hi mtk10,

You're "mixing" two different ideas that are NOT logically linked.

When dealing with 2 consecutive integers, ONE of the integers will be evenly divisible by 2. When dealing with 3 consecutive integers, ONE of the integers will be evenly divisible by 3 and AT LEAST ONE will be evenly divisible by 2. When dealing with 4 consecutive integers, ONE of the integers will be evenly divisible by 4, TWO of the integers will be evenly divisible by 2 and AT LEAST ONE of the integers will be evenly divisible by 3. Etc.

However, this question is asking about the SUM of six consecutive integers - and that is a different concept entirely. The concept involved here is "the SUM of ANY 3 consecutive integers will be evenly divisible by 3", so since we're dealing with six consecutive integers, we're dealing with two 'groups' of 3 consecutive integers - so that SUM will also be evenly divisible by 3.

I understand this can be done by picking numbers and testing them out.

But can someone clarify this concept. It is said every 2 numbers, atleast one number is divisible by 2, for every 3 number atleast one number is divisble by 3. and so on so forth . Using this concept. Should the answer not be E?

It says consecutive six integers? so by the concept the factors should include 6 ,4 and 3.

I am sure i am making a very stupid mistake here. Kindly correct me where i am wrong.

I tihnk you are mistakenly inferring that you can "factor" these numbers out of the SUM. You would be correct if the question said the "product of six consecutive integers". But, when adding the integeres, instead of multiplying them, you cannot simply factor out each individual integers.
_________________

Re: If x is the sum of six consecutive integers, then x is divisible by wh [#permalink]

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12 Dec 2017, 02:43

EMPOWERgmatRichC wrote:

mtk10 wrote:

I understand this can be done by picking numbers and testing them out.

But can someone clarify this concept. It is said every 2 numbers, atleast one number is divisible by 2, for every 3 number atleast one number is divisble by 3. and so on so forth . Using this concept. Should the answer not be E?

It says consecutive six integers? so by the concept the factors should include 6 ,4 and 3.

I am sure i am making a very stupid mistake here. Kindly correct me where i am wrong.

Hi mtk10,

You're "mixing" two different ideas that are NOT logically linked.

When dealing with 2 consecutive integers, ONE of the integers will be evenly divisible by 2. When dealing with 3 consecutive integers, ONE of the integers will be evenly divisible by 3 and AT LEAST ONE will be evenly divisible by 2. When dealing with 4 consecutive integers, ONE of the integers will be evenly divisible by 4, TWO of the integers will be evenly divisible by 2 and AT LEAST ONE of the integers will be evenly divisible by 3. Etc.

However, this question is asking about the SUM of six consecutive integers - and that is a different concept entirely. The concept involved here is "the SUM of ANY 3 consecutive integers will be evenly divisible by 3", so since we're dealing with six consecutive integers, we're dealing with two 'groups' of 3 consecutive integers - so that SUM will also be evenly divisible by 3.

GMAT assassins aren't born, they're made, Rich

Thank you and all other experts for such prompt replies

And yes i got to this conclusion the second i made the post lol. That its asking about SUM not the integers.