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Number properties

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Number properties

by Uri » Fri May 01, 2009 11:21 pm

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If S is a finite set of consecutive even numbers, is the median of S an odd number?

(1) The mean of set S is an even number.

(2) The range of set S is divisible by 4.

Please explain your appraoch.

OA: [spoiler](D)[/spoiler]
Source: MGMAT
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by aj5105 » Sat May 02, 2009 3:44 am

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Statement (1)

S {0,2,4}

S {0,2,4,6,8}

Be it for any set, we get the answer NO.

SUFF

Statement (2)

S {0,2,4}

S {0,2,4,6,8}

S {0,2,4,6,8,10,12}

The answer is again NO.

SUFF.

(D)
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Re: Number properties

by Ian Stewart » Sat May 02, 2009 6:20 am

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Uri wrote:If S is a finite set of consecutive even numbers, is the median of S an odd number?

(1) The mean of set S is an even number.

(2) The range of set S is divisible by 4.

Please explain your appraoch.

OA: [spoiler](D)[/spoiler]
Source: MGMAT
In any 'equally spaced' set, three quantities are always equal:

mean = median = average of smallest and largest

Since a set of consecutive even integers is 'equally spaced' - consecutive even numbers are spaced by two - then in our set, the median, mean, and average of the smallest and largest are all equal.

So if, from 1), the mean is even, the median must be even. Sufficient.

For Statement 2), call the largest number in our set L, and the smallest S. We know that L - S = 4k, for some integer k. That is, L = 4k + S. So:

median = average of the smallest and largest
= (L + S)/2
= (4k + S + S)/2
= 2k + S

and since S is even, 2k + S is even. So again the median is even. Sufficient.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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Re: Number properties

by Uri » Sat May 02, 2009 6:57 am

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Ian Stewart wrote: In any 'equally spaced' set, three quantities are always equal:

mean = median = average of smallest and largest

Since a set of consecutive even integers is 'equally spaced' - consecutive even numbers are spaced by two - then in our set, the median, mean, and average of the smallest and largest are all equal.

So if, from 1), the mean is even, the median must be even. Sufficient.

For Statement 2), call the largest number in our set L, and the smallest S. We know that L - S = 4k, for some integer k. That is, L = 4k + S. So:

median = average of the smallest and largest
= (L + S)/2
= (4k + S + S)/2
= 2k + S

and since S is even, 2k + S is even. So again the median is even. Sufficient.
like always, this time also your explanation is excellent. thanks, Ian!
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by cramya » Sat May 02, 2009 9:33 am

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As always, thanks for a wonderful explanation, Ian.

Regards,
CR
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by singalong » Sat Jun 04, 2011 6:42 am

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I saw a similar question on the the web but statement 2 was changed to "(2) The range of set S is divisible by 6."
How would it affect the solution then?
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by magicalhat » Sat Jun 04, 2011 7:13 am

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Great explanation!
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by cans » Sat Jun 04, 2011 7:25 am

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s is set of consecutive even numbers. Thus all the numbers in the set are even only.
median of S is odd only f there are even no. of elements in the set
a) mean is even.
If consecutive numbers, mean= median (consider 2,4,6 mean=median=4)
(consider 2,4,6,8 mean=median=5)
as mean is even, median is also even
Sufficient
b)range is divisible by 4
let there be n numbers. first one is k (even), last one is k+(n-1)*2)
range = last-first = (n-1)*2
as it is divisible by 4, it means (n-1) is divisible by 2.
or n -1 is even and thus n is odd
and thus median is even
Sufficient
IMO D
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by bblast » Sat Jun 04, 2011 7:30 am

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singalong wrote:I saw a similar question on the the web but statement 2 was changed to "(2) The range of set S is divisible by 6."
How would it affect the solution then?
range of the set is divisible by 6 means set has exactly 4 consecutive even elements. hence median will always be equal to the mean

try the sets
{2,4,6,8}-> m=5,M=5
{14,16,18,20}->m=17,M=17
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by cans » Sat Jun 04, 2011 9:25 am

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bblast wrote:
singalong wrote:I saw a similar question on the the web but statement 2 was changed to "(2) The range of set S is divisible by 6."
How would it affect the solution then?
range of the set is divisible by 6 means set has exactly 4 consecutive even elements. hence median will always be equal to the mean

try the sets
{2,4,6,8}-> m=5,M=5
{14,16,18,20}->m=17,M=17
correction: range of sets is divisible by 6 and we are not sure if it is equal to 6.
If equal, then 4 integers.
But if its 12, then we have 7 consecutive even integers instead of 4.
But mean will be same as median. (because integers in set are consecutive)
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by bblast » Sat Jun 04, 2011 11:53 am

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hmmmmmm
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by vikram4689 » Sun Jun 05, 2011 5:24 am

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Nice explanation by Ian
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by sunilramu » Sun Jun 05, 2011 6:37 am

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1. Since Mean = Median on equally spaced sets. Sufficient.
2. Considering sets {2,4,6}, {2,4,6,8}, and {2,4,6,8,10} we can see that only sets with odd number of terms can have range divisible by 4. Which follows that Median of the set is even. Sufficient.

IMO D
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by nitesh50 » Thu Dec 13, 2018 3:47 am

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The set of consecutive even integers is an evenly spaced set (aka arithmetic progression). Now, for any evenly spaced set mean=median.

(1) The mean of set S is an even number --> mean=median=even. Sufficient.
(2) The range of set S is divisible by 6 --> if S={0, 2, 4, 6} then median=(2+4)/2=3=odd but if S={0, 2, 4, 6, 8, 10, 12} then median=6=even. Not sufficient.

Answer: A.
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