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If the set S consists of consecutive integers, are there

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Director
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If the set S consists of consecutive integers, are there [#permalink] New post 05 Jan 2007, 20:10
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A
B
C
D
E

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If the set S consists of consecutive integers, are there more even than odd in S?

1. There are even number of integers in set S.
2. The largest integer is 3335.
Senior Manager
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 [#permalink] New post 05 Jan 2007, 21:23
A

(1)even number of integers => there will be equal number of even and odd

sufficient

(2) insufficient
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 [#permalink] New post 06 Jan 2007, 04:54
answer is D in my opinion....

st1: as AK noted, if there are even number of integers in S, and since they are consecutive, there will be equal number of even and odd numbers. hence sufficient (the answer is no).

st2: a bit more subtle, but the answer is no as well, and hence sufficient: the only way to get more even than odd in S it must start and end in even number. if it ends in odd number, either there are equal number of even an odd numbers or there is more odd than even.... in any case there are not more even than odd..... so sufficient.
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 [#permalink] New post 06 Jan 2007, 11:27
hobbit wrote:
answer is D in my opinion....

st1: as AK noted, if there are even number of integers in S, and since they are consecutive, there will be equal number of even and odd numbers. hence sufficient (the answer is no).

st2: a bit more subtle, but the answer is no as well, and hence sufficient: the only way to get more even than odd in S it must start and end in even number. if it ends in odd number, either there are equal number of even an odd numbers or there is more odd than even.... in any case there are not more even than odd..... so sufficient.


its my own question, so OA is D and OE is as explained above...
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 [#permalink] New post 06 Jan 2007, 19:48
hobbit wrote:
it is a good question....



Just curious the number o falls in even number group or not ?

If 0 falls in even group then A is sufficient .
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 [#permalink] New post 06 Jan 2007, 21:19
st2: a bit more subtle, but the answer is no as well, and hence sufficient: the only way to get more even than odd in S it must start and end in even number. if it ends in odd number, either there are equal number of even an odd numbers or there is more odd than even.... in any case there are not more even than odd..... so sufficient.[/quote]

thats a great thinking Hobbit. I appreciate it

I missed it :beat
  [#permalink] 06 Jan 2007, 21:19
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If the set S consists of consecutive integers, are there

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