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If x is the sum of six consecutive integers, then x is divisible by wh
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21 Oct 2014, 07:35
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If x is the sum of six consecutive integers, then x is divisible by wh
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21 Oct 2014, 18:33
1 + 2 + 3 + 4 + 5 + 6 = 21 = Divisible by 3 2 + 3 + 4 + 5 + 6 + 7 = 20 + 7 = 27 = Divisible by 3 3 + 4 + 5 + 6 + 7 + 8 = 25 + 8 = 33 = Divisible by 3 The consecutive addition will just increase by 6 each time Answer = A One more method:Addition of six consecutive integers = odd + even + odd + even + odd + even = (odd + odd) + (even + even + even) + odd = even + even + odd = even + odd = odd The result will ALWAYS be ODD. So it cannot be divisible by 4 & 6. Answer = 3 = A
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Re: If x is the sum of six consecutive integers, then x is divisible by wh
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21 Oct 2014, 09:32
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 k2, k1, 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. hence option A



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Re: If x is the sum of six consecutive integers, then x is divisible by wh
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21 Oct 2014, 22:11
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.



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Re: If x is the sum of six consecutive integers, then x is divisible by wh
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01 Sep 2015, 06:16
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 3,2,1,0,1,2 or 2,1,0,1,2,3 Only option A



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Re: If x is the sum of six consecutive integers, then x is divisible by wh
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03 Sep 2015, 21:41
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
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Re: If x is the sum of six consecutive integers, then x is divisible by wh
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28 Nov 2016, 08:29
Sum of six consecutive numbers = 6 (6 + 1 ) \ 2 = 21 21 is divided by 3, not by 4 and 6 Answer : A



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



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Re: If x is the sum of six consecutive integers, then x is divisible by wh
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11 Dec 2017, 04:03
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. 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.
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Re: If x is the sum of six consecutive integers, then x is divisible by wh
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11 Dec 2017, 14:20
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
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Re: If x is the sum of six consecutive integers, then x is divisible by wh
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11 Dec 2017, 15:07
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. 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.
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Re: If x is the sum of six consecutive integers, then x is divisible by wh
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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. Thankyou again.



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If x is the sum of six consecutive integers, then x is divisible by wh
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13 Apr 2018, 16:26
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 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.
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Re: If x is the sum of six consecutive integers, then x is divisible by wh
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18 Aug 2018, 18:42
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 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
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