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# A set of consecutive positive integers beginning with 1 is

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SVP
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A set of consecutive positive integers beginning with 1 is [#permalink]

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11 Nov 2008, 02:02
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Question Stats:

42% (02:26) correct 58% (02:44) wrong based on 134 sessions

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A set of consecutive positive integers beginning with 1 is written on the blackboard. A student came along and erased one number. The average of the remaining numbers is 35 7/17. What was the number erased?

A. 7
B. 8
C. 9
D. 15
E. 17
[Reveal] Spoiler: OA
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11 Nov 2008, 03:38
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This one sure looks tricky. I am not able to reach the solution, but this is what i think. I am sure you must have done all this by now, but no harm in putting it down.
35 7/17 , means 602/17 , earlier i thought there must be 18 numbers, which is 17 + 1 that is taken out. But if there were, then their sum is 171 which is far too low. The average of 35 shows that the numbers were little higher, ( their sum should be 602 ) . The answer choices are very low numbers. I don't know how this is possible at all.

Sorry mate, i really doubt if the qs is correct ! Can you pls provide the source of this qs and check whether the numbers you mentioned are all correct ?
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11 Nov 2008, 03:43
My method, I may not use this during the test, so i will still wait for another better solution.

set of consecutive positive integers beginning with 1, if n is the last number => SUM(1) = n(n+1)/2

If 1 number was taken out, so the SUM(2) then will be (n - 1) * average = (n - 1) * (35 + 7/17)

As SUM(2) is an integer so (n-1) is divisible by 17, the original average is in this range (34+7/17, 36 + 7/17)
=> n is in (2*(34+7/17), 2*(36 + 7/17)) or n is in (68.8, 72.8)

(n-1) is divisible by 17 and n is in (68.8, 72.8) => n = 69

The number was erased = SUM(1) - SUM(2) = 69*70/2 - (69-1) * (35 + 7/17) = 7

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11 Nov 2008, 04:03
lylya4 wrote:
My method, I may not use this during the test, so i will still wait for another better solution.

set of consecutive positive integers beginning with 1, if n is the last number => SUM(1) = n(n+1)/2

If 1 number was taken out, so the SUM(2) then will be (n - 1) * average = (n - 1) * (35 + 7/17)

As SUM(2) is an integer so (n-1) is divisible by 17, the original average is in this range (34+7/17, 36 + 7/17)
=> n is in (2*(34+7/17), 2*(36 + 7/17)) or n is in (68.8, 72.8)

(n-1) is divisible by 17 and n is in (68.8, 72.8) => n = 69

The number was erased = SUM(1) - SUM(2) = 69*70/2 - (69-1) * (35 + 7/17) = 7

Could you elaborate the highlighted step further?
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11 Nov 2008, 04:42
scthakur wrote:
lylya4 wrote:
My method, I may not use this during the test, so i will still wait for another better solution.

set of consecutive positive integers beginning with 1, if n is the last number => SUM(1) = n(n+1)/2

If 1 number was taken out, so the SUM(2) then will be (n - 1) * average = (n - 1) * (35 + 7/17)

As SUM(2) is an integer so (n-1) is divisible by 17, the original average is in this range (34+7/17, 36 + 7/17)
=> n is in (2*(34+7/17), 2*(36 + 7/17)) or n is in (68.8, 72.8)

(n-1) is divisible by 17 and n is in (68.8, 72.8) => n = 69

The number was erased = SUM(1) - SUM(2) = 69*70/2 - (69-1) * (35 + 7/17) = 7

Could you elaborate the highlighted step further?

SUM (2) = (n - 1) * (35 + 7/17) = 35(n-1) + 7/17(n-1) as SUM(2) is an integer => 7/17(n-1) is integer or n-1 is divisible by 17

The range for the original average is just for to limit the result of n, if the largest number n was taken out, the average will be decreased at max n / (n-1) = 1 + 1/(n-1) ~ 1, so to be safe I place the range for this average +-1 to 35 7/17. This step can be ignored, instead we know that the average is ~35, so n should be around ~70 (average for consecutive integers = (n + 1)/2)
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Re: A set of consecutive positive integers beginning with 1 is [#permalink]

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08 Nov 2014, 11:48
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A better and easier method is as follows:
First of all, when ever a number is deleted from consecutive numbers, avg cant change more than 1/2.
So in this case new avg is 35 7/17, so old average was 35. Now, 35 is avg of first 69 numbers, so previously there were 69 num, and now there are 68(after deletion).
Now, 7/17 = 28/68 . So effectively, the number which was deleted gave 28/68 to each 68 num left, so the number should have been 28 less than 35(old avg)
Ans= 35-28=7.
Advantage with this method is that you can do it verbally.
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Re: A set of consecutive positive integers beginning with 1 is [#permalink]

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02 Jan 2015, 11:24
Bunuel

this part is little confusing to me. .

The range for the original average is just for to limit the result of n, if the largest number n was taken out, the average will be decreased at max n / (n-1) = 1 + 1/(n-1) ~ 1, so to be safe I place the range for this average +-1 to 35 7/17. This step can be ignored, instead we know that the average is ~35, so n should be around ~70 (average for consecutive integers = (n + 1)/2)
thanks
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Re: A set of consecutive positive integers beginning with 1 is [#permalink]

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02 Jan 2015, 13:31
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Hi All,

This question is layered with subtle "clues" as to how you can find the correct answer:

1) We're asked which number was removed from the list and the answers ARE numbers. We can use them against the prompt.
2) The average of the new group of numbers is 35 7/17. The denominator of THAT fraction tells us that the total number of terms in the new list MUST be a multiple of 17:

17, 34, 51, 68, 85, etc.

3) Since we're removing 1 number from a list of consecutive positive integers, we could quickly limit down the possible number of terms in the ORIGINAL list:

If new = 17, old = 18, but the numbers from 1 to 18 would NOT have an average in the mid-30s (it would be much smaller). ELIMINATE this option.
If new = 34, old = 35, but we run into the same problem. The average won't match here either. ELIMINATE this option.
If new = 51, old = 52, same problem here. ELIMINATE this option.

If new = 68, old = 69....HERE we have a group of numbers that might just be what we're looking for.

In the "old group", the average is the "middle term" = 35, which is REALLY close to the new average once we remove a number.

Now, we just have to figure out which of the 5 answers was removed (and that changed the new average to 35 7/17). There are a couple of ways to do this math, but I'm going to use the answers to save myself some steps:

Sum of 1 to 69, inclusive = 69(35) = 2415

Remove Answer B....2415 - 8 = 2407

New average = 2407/68 = 35 27/68

Now, we can compare 27/68 to 7/17

27/68 and 28/68

This is REALLY close, but is not the answer. Eliminate B. The answer is either A or C.

Since 28/68 is a BIGGER fraction than 27/68, we need the sum (of the 68 integers) to be BIGGER...which means we need to remove a SMALLER number.

[Reveal] Spoiler:
A

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

Co-Founder & GMAT Assassin

# Special Offer: Save $75 + GMAT Club Tests 60-point improvement guarantee www.empowergmat.com/ ***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*********************** Intern Joined: 23 Aug 2014 Posts: 10 Followers: 0 Kudos [?]: 1 [0], given: 6 Re: A set of consecutive positive integers beginning with 1 is [#permalink] ### Show Tags 02 Apr 2015, 23:33 EMPOWERgmatRichC wrote: Hi All, This question is layered with subtle "clues" as to how you can find the correct answer: 1) We're asked which number was removed from the list and the answers ARE numbers. We can use them against the prompt. 2) The average of the new group of numbers is 35 7/17. The denominator of THAT fraction tells us that the total number of terms in the new list MUST be a multiple of 17: 17, 34, 51, 68, 85, etc. 3) Since we're removing 1 number from a list of consecutive positive integers, we could quickly limit down the possible number of terms in the ORIGINAL list: If new = 17, old = 18, but the numbers from 1 to 18 would NOT have an average in the mid-30s (it would be much smaller). ELIMINATE this option. If new = 34, old = 35, but we run into the same problem. The average won't match here either. ELIMINATE this option. If new = 51, old = 52, same problem here. ELIMINATE this option. If new = 68, old = 69....HERE we have a group of numbers that might just be what we're looking for. In the "old group", the average is the "middle term" = 35, which is REALLY close to the new average once we remove a number. Now, we just have to figure out which of the 5 answers was removed (and that changed the new average to 35 7/17). There are a couple of ways to do this math, but I'm going to use the answers to save myself some steps: Sum of 1 to 69, inclusive = 69(35) = 2415 Remove Answer B....2415 - 8 = 2407 New average = 2407/68 = 35 27/68 Now, we can compare 27/68 to 7/17 27/68 and 28/68 This is REALLY close, but is not the answer. Eliminate B. The answer is either A or C. Since 28/68 is a BIGGER fraction than 27/68, we need the sum (of the 68 integers) to be BIGGER...which means we need to remove a SMALLER number. Final Answer: [Reveal] Spoiler: A GMAT assassins aren't born, they're made, Rich Thanks for providing this easy to understand solution. Just one step which can reduce effort at the end 69*35 = 2415 and our earlier average is 35 7/17 = 602/17, we know total numbers here in this case (after one got erased is 68) so 4*602/4*17 = 2408/68. now since total should have been 2415, the deleted number is 2415-2408=7 GMAT Club Legend Joined: 09 Sep 2013 Posts: 12228 Followers: 542 Kudos [?]: 151 [0], given: 0 Re: A set of consecutive positive integers beginning with 1 is [#permalink] ### Show Tags 25 Jul 2016, 21:54 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. _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6971 Location: Pune, India Followers: 2030 Kudos [?]: 12761 [0], given: 221 Re: A set of consecutive positive integers beginning with 1 is [#permalink] ### Show Tags 25 Jul 2016, 22:36 Expert's post 2 This post was BOOKMARKED tarek99 wrote: A set of consecutive positive integers beginning with 1 is written on the blackboard. A student came along and erased one number. The average of the remaining numbers is 35 7/17. What was the number erased? A. 7 B. 8 C. 9 D. 15 E. 17 The average of consecutive positive integers will be either an integer (if we have odd number of integers) or integer.5 (if we have even number of integers). The maximum change that can happen to the average when one integer is deleted is 0.5. New average = 35 (7/17) Old average = Either 35 or 35.5 So there were either 69 consecutive integers starting from 1 (in which case avg was 35) or 70 consecutive integers starting from 1 (in which case avg was 35.5) . If the average was 35 (and number of integers was 69) and it increased to 35 (7/17), it means a small number was removed which was taking the average down by 7/17 for the rest of the 68 numbers. So it must be 68 *(7/17) = 28 less than the average. That is, the number removed must have been 35 - 28 = 7. It is one of the options so answer (A). For more details on this method, check: http://www.veritasprep.com/blog/2015/09 ... -the-gmat/ _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: A set of consecutive positive integers beginning with 1 is   [#permalink] 25 Jul 2016, 22:36
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