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A set of consecutive positive integers beginning with 1 is

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A set of consecutive positive integers beginning with 1 is [#permalink] New post 11 Nov 2008, 01:02
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A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

83% (01:01) correct 17% (03:34) wrong based on 12 sessions
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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Re: PS: Average [#permalink] New post 11 Nov 2008, 02:38
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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Re: PS: Average [#permalink] New post 11 Nov 2008, 02:43
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

Answer: A
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Re: PS: Average [#permalink] New post 11 Nov 2008, 03:03
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

Answer: A



Could you elaborate the highlighted step further?
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Re: PS: Average [#permalink] New post 11 Nov 2008, 03:42
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

Answer: A



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)
Re: PS: Average   [#permalink] 11 Nov 2008, 03:42
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