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integers

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by pemdas » Tue Jan 03, 2012 1:50 pm
A set of five different positive integers has a mean equal to the sum of its two smallest integers. What is the largest possible integer in the set?

reposted (interested in solution)
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by neelgandham » Tue Jan 03, 2012 2:12 pm
Are you sure you have posted the correct question ? Or is it just the Beer

2,4,6,8,10, Sum = 30, Average = 6 (2+4)
1,100,101,102,201, Sum = 505 ,Average = 101 (100+1)
1,100000,100001,100002,200001, Sum = 500005, Average = 100001(1+100000) and it goes on !
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by chieftang » Tue Jan 03, 2012 2:23 pm
There is not a unique solution.

Consider this set: {1,2,3,4,5}

Or this set: {3,4,7,10,11}

Etc, etc.
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by pemdas » Tue Jan 03, 2012 2:40 pm
Any more views on this question

I have noticed the trend (almost mechanical) to set median as arithmetic mean value in the given sets. Is this something which proceeds from your solutions or the only way you see this question can be resolved to find the largest integer in the set.
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by chieftang » Tue Jan 03, 2012 2:58 pm
pemdas wrote:Any more views on this question

I have noticed the trend (almost mechanical) to set median as arithmetic mean value in the given sets. Is this something which proceeds from your solutions or the only way you see this question can be resolved to find the largest integer in the set.
I set median as mean in order to make a couple scenarios easy to derive.

Example, start with {4,5,?,?,?}

Sum of first two is now 9. Make that the next number (why not?), which will also now need to be the mean:

{4,5,9,?,?}

Now, to make 9 the mean the next two numbers need to have the same distance from 9 in the positive direction.

I.e. 9-5=4, so the next number can be 9+4=13

{4,5,9,13,?}

Complete the set the same way and get:

{4,5,9,13,14} -> 4+5=9, (4+5+9+13+14)/5 = 9
Last edited by chieftang on Tue Jan 03, 2012 3:40 pm, edited 1 time in total.
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by pemdas » Tue Jan 03, 2012 3:21 pm
very interesting and alike the way I approached this question which has answer choices (not posted) and hence is limited to the given no-s

Nevertheless, I found the largest integer as per rule: the total deviations below the mean should be equal to the total deviations above the mean

I set mean as the value greater than the fourth no (the only possible, because the fifth cannot be greater :) ) and obtained

a, a+C, a+C+1, a+C+2, E

a is the first term, C is the constant (likewise number with zeros in Nil's solution, 100000) and I accrued everything by 1 (the different positive integers), E is the largest integer

As the result I have obtained the function >>>
mean=a+a+C
Deviations below the mean:
a+a+C-a=a+C
a+a+C-(a+C)=a
a+a+C-(a+C+1)=a-1
a+a+C-(a+C+2)=a-2

summing up (finding total deviations)=4a+C-3

E=4a+C-3 +(a+a+C)=6a+2C-3

To simplify, if we are not using C as the constant, then we get the largest integer E=6a-1

and you are right, this set without the given answer choices is meaningless for GMAT as it continues to accrue and accrue the largest integers
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by chieftang » Tue Jan 03, 2012 3:35 pm
pemdas wrote:
and you are right, this set without the given answer choices is meaningless for GMAT as it continues to accrue and accrue the largest integers
What did the actual question with answer choices look like?
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by pemdas » Tue Jan 03, 2012 4:03 pm
a)10,b)14,c)16,d)18,e)20 and the correct answer is e

such questions could be resolved in several ways; the mean could be less than the fourth no or it could be equal to the third no (your case)
chieftang wrote:
pemdas wrote:
and you are right, this set without the given answer choices is meaningless for GMAT as it continues to accrue and accrue the largest integers
What did the actual question with answer choices look like?
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by chieftang » Tue Jan 03, 2012 4:28 pm
pemdas wrote:a)10,b)14,c)16,d)18,e)20 and the correct answer is e

such questions could be resolved in several ways; the mean could be less than the fourth no or it could be equal to the third no (your case)
OK. I'd probably take one derived example:

{4,5,9,13,14}

Then test the largest value given in the answer choices (20) by adding 6 to the largest in the derived set, and subtracting 6 from the second largest in the derived set to get:

{4,5,9,7,20}

Looks fine... So the answer must be E.
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