If x is the sum of six consecutive integers, then x is divisible by wh

Sum of six consecutive integers = 6y + 15 =3(2y+5) =x

Clearly, there are two important observations from above information :
1) x is a multiple of 3
2) (2y + 5) is an odd term

Therefore, the most relevant option to the question is option A.

-3,-2,-1,0,1,2 or -2,-1,0,1,2,3

Only option A

Hi All,

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...

Final Answer:

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

Sum of six consecutive numbers = 6 (6 + 1 ) \ 2 = 21
21 is divided by 3, not by 4 and 6
Answer : A

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.

Answer: A.

Hope it's clear.

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

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.

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.

Thankyou again.

I did not use this solution to find the answer, but I should have. Very fast if the principle is memorized and fresh on your mind.

As a GMAT Club Math Book Principle:
- If n is even, the sum of consecutive integers is never divisible by n. Given {9,10,11,12}, we have n=4 consecutive integers. Sum is 9+10+11+12=42, which is not divisible by 4.

For this problem, if the above principle was memorized, we could immediately rule out III. Then realize that 6 consecutive integers will lead to an odd number, so rule out II.

Only possible answer left would be I.

If we let the six consecutive integers be 1, 2, 3, 4, 5, and 6, then x = 21. We see that x is divisible only by 3, which eliminates Roman numerals II and III. Since there is no other option that excludes both II and III, the answer must be I only.

Answer: A

Even if we take the least possible consecutive integers (positive) we get 1+2+3+4+5+6 as 21.
21 is only divisible by 3. So A.

I am confused. The given information does not say that whether this is an even consecutive, multiple of 3 consecutive, or so on... Then 0 2 4 6 8 10 should also divided by 6. Right? Am I doing any thing wrong here?

Well, consecutive integers are consecutive. Why do you need to assume more than that?
If they were even consecutive then the question will mention that.!

Please correct me If I am wrong. In data sufficiency, If a given piece of information is not given, then possibility is there. Doesn’t it apply here as well?

Posted from my mobile device

What you are saying is very broad in nature.
I will take an example and try to explain.

Problem Solving reads : M is the sum of 6 consecutive odd integers.

That means M cannot be sum of consecutive even integers or consecutive primes. Though it can be any set of 6 consecutive odd integers (positive, negative) but cannot be other than that.

Data Sufficiency question : Is M>6?

Then you can fairly assume M can be anything in the world. Positive/Negative/Real/Even/Odd.
But again if Let's say Statement 1 reads : M is a prime number.

Then you know M is prime for a fact and nothing else. {2,3,5,7,11.......}

is this faster??

imagine sum of numbers from 1 to 6. do it like this :
(1+6)/2 [1 is the first term and 6 is the last term]
=3.5*6 [there are 6 terms in series]
=21
21 is divisible by 3.

again, imagine sum of numbers from 2 to 7. do it like this :
(2+7)/2 [2 is first term and 7 is the last term]
=4.5*6 [there are 7 terms in series]
=27
27 is divisible by 3.

while solving i did only 1st case [1to6] and got the answer in 35 seconds.