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The average of 5 distinct single digit integers is 5. If 2

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New post 31 Oct 2007, 10:05
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

The average of 5 distinct single digit integers is 5. If 2 integers are discarded, the new average is 4. What is the largest of the 5 integers?

1) Exactly (no more than) 3 of the integers are consecutive primes.
2) The smallest integer is 3.

The answer is A. It is clear that there is only one combination of numbers that meet the requirements of 1). However, it seems to me that there is only one combination of numbers that meet 2) also: 3,4,5,6,7, where (3+4+5)/3 = 4. So wouldn't the answer be d?

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New post 31 Oct 2007, 10:32
take 2,3,5,7,8
their average is 5.
now if i take out 8 and 5 from these
im left with 2,3,7 and their average is 4.
the smallest integer was 2, not 3
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New post 31 Oct 2007, 10:36
wait a second but doesnt this make A wrong too then? was A the OA?
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New post 31 Oct 2007, 10:38
or actually wait 8 is still the largest number in both cases.
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New post 31 Oct 2007, 12:09
The OA is A.

For 1) the only possible combination of numbers that works is 3,4,5,6,7. Since there are exactly (no more than) 3 consecutive primes, 2,3,5,7 cannot be in the same batch of numbers. So for both 1 and 2 independent of each other, the only group of numbers that work are 3,4,5,6,7.

So why is the answer not d?
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New post 31 Oct 2007, 12:25
ok my bad i see it now. D does make sense to me, can anyone else figure out why D is not right?
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