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Consider a set of 729 consecutive integers whose smallest element is x

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Consider a set of 729 consecutive integers whose smallest element is x  [#permalink]

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New post 26 Nov 2019, 02:01
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Question Stats:

81% (01:54) correct 19% (02:00) wrong based on 31 sessions

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Re: Consider a set of 729 consecutive integers whose smallest element is x  [#permalink]

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New post 26 Nov 2019, 02:37
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I think this is a fairly straight forward question, although it took me close to 6 minutes to fully resolve .

One property of consecutive numbers is that they are evenly spaced

Because we don't know how many integers there are, we can assume that the arithmetic mean of 729 integers is the middle number

729/2 = 364.

This means that the arithmetic mean of 1072, is the 365th number.
Meaning that there are 364 numbers before 1072, and equally 364 numbers after it ..

Therefore to find the smallest, we simply subteact 364 from 1072.

1072-364 = 708.

If the question said we sbould find the greatest number, we simply add 364 to 1072 and get 1,436.

But since the question asked what the smallest integer is, we stick with our earlier answer of 708.

Answer = B

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Consider a set of 729 consecutive integers whose smallest element is x  [#permalink]

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New post 26 Nov 2019, 03:31
Bunuel wrote:
Consider a set of 729 consecutive integers whose smallest element is x. If the arithmetic mean of this set of numbers is 1,072, what is the value of x?

A. 672
B. 708
C. 780
D. 828
E. 1,236


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729x+( ( 728*729)/2 )= 1072*729
x= 708
IMO B
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Re: Consider a set of 729 consecutive integers whose smallest element is x  [#permalink]

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New post 26 Nov 2019, 04:46
The mean for odd consecutive integers is the middle term
The middle term for 729 consecutive integers is 365th term
If 365th term in consecutive integers is 1072 then the first integer in the set must be 1072-364=708

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Ans: B
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Re: Consider a set of 729 consecutive integers whose smallest element is x  [#permalink]

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New post 30 Nov 2019, 00:54
The sum of n integers in any Arithmetic Progression with first term a and last term l is given as (a+l)/2
The first term is x and the last terms is x+728 (since they are consecutive)

=> mean = (x+728 + x)/2 = 1072 => 2x + 728 = 2144 => 2x = 1416

=> x = 708

Ans B
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Re: Consider a set of 729 consecutive integers whose smallest element is x   [#permalink] 30 Nov 2019, 00:54
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