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Interesting GMATFix Problem-23

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Interesting GMATFix Problem-23

by arora007 » Tue Sep 21, 2010 1:00 pm
N is a finite set of distinct positive integers. How many even integers does N contain?
1)6.25% of the product of all numbers in N is an odd integer
2)The sum of all even integers in N is 10
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by clock60 » Wed Sep 22, 2010 3:29 am
too hard to me, with 1/5 likelihood my answer can be B

(1) 1/16( x1*x2*x3*....xn)=k-odd integer
x1*x2*x3*....xn=16*k
let it be k=1
x1*x2*x3*.....xn=16, as all integers are +ve and distinct, possible variation for even
2*8=16 here 2 even integers, or
1*16=16, here 1 even integer insufficient

(2) sum of all even=10
2+8, 4+6, the only value i found (i can`t employ 0+10 as all must be +ve) in any case two distinct evens
i hope that sufficient
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by arora007 » Wed Sep 22, 2010 3:35 am
OA is C, but i did not understand the explanation given... can somebody help?
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by clock60 » Wed Sep 22, 2010 3:45 am
also want to see other replies, i again don`t see my mistake
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by arora007 » Wed Sep 22, 2010 3:53 am
read this explanation, but then please explain it to me as well...

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by clock60 » Wed Sep 22, 2010 4:21 am
arora007 thank you for posting oa i see now my silly mistake, bad on me
(1) (1/16)*(x1*x2*...xn)=k*- where k is odd integer and x1, x2, xn are integers in the set
suppose k=1-odd integer, then
x1*x2*....xn=16*1
and now sets that are possible
n-{1,2,8} here we have two even integers, 2 and 8. but another set
n-{1,16} here we have only one even integer 16
1 st insuff

(2) how we can obtain 10. (2+8), (4+6) here two even integers, but we can the only 10 in the set ,i mean 10 can be the only even integer and the number of even is 1
insufficient
both:the sum of all even must be 10 (4+6) (2+8) (10) and this sum must be divisible by 16 the only pair that suits is 2+8=10 2*8=16

(x1*x2*x3..xn)/16=integers and the only even integers that are divisible by 16 and give 10 summing are 2 and 8

hope it makes sence
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by archiekins2007 » Wed Sep 22, 2010 2:10 pm
Statement 1) gives us (1/80)*Product of Set = Odd Number. The product of set could be 80,240,400 etc etc so clearly its insufficient.

Statement 2 is very intersting :) , it says that sum of even inetergers is 10
Hence the possibilities below
4,6
8,2
0,4,6
0,8,2
We can either have 2 or 3 even inetergers(0 is an even integer). the question stem doe snot say it has to be a positive even ineterger.

Combined,from statement 1 we know that the set cannot have a 0 since then the product will be 0.
Hence we have 2 Even integers in the set.

Answer is C
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by sadragas » Sat Sep 25, 2010 10:18 am
Let me have a shot at explaining the solution attached.

I'm directly skipping to the merge statements part.
I'm assuming you understood why A alone or B alone cannot be used to answer the question.

Statement A says 1/16th of the prod of the numbers is ODD.
Which means, the prod of the even numbers is an odd multiple of 16

Statement B says, sum of the even numbers is 10

sum is 10 and prod is an odd multiple of 16,
Which gives us the only combination of 8 and 2

Hope this was clear,
Cheers!
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