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sidbidus
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PS: Odd Integer 05 Sep 2007, 08:43
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How do we solve this??



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PS: Odd Integer 05 Sep 2007, 09:19
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sidbidus
How do we solve this??
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quest stem say>> must be an odd integer

supp a=4 n b=2

1) 4/2=2- even
2)2/2=1 - odd
3) 4+2/2= 3 - odd
4)4+2/2=3 - odd
5)2+2/2=2 - even

supp a=8 n b=4
1) 8/2=4 - even
2)4/2=2 - even
3)8+4/2= 6 -even
4)8+2/2= 5 - odd
5)4+2/2 = 3 - odd

when u compare the results u'll get D as odd integer. similarly plug in any values D will always b an odd integer
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PS: Odd Integer 05 Sep 2007, 13:34
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can anyone explain this to me using even-odd properties instead of plugging in (just for the sake of principle)?
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KillerSquirrel
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PS: Odd Integer 05 Sep 2007, 14:53
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1. Since a-b is even , then a,b must be both odd or both even.

odd-odd = even or even-even = even

but not

odd-even = odd or odd-even = odd

2. knowing that, we can say for sure that a,b are both even - since even/even = even but odd/odd = odd.

so a,b are both even.

since a/b = even integer and we know that a=even and b=even then we can say that at least b*2/b = a/b but can be also 2*b*2/b = a/b and so on.

3. choice (D) is best since:

(b*2+2)/2 = 2(b+1)/2 = 2*(even+1)/2 = odd.

the other choices just don't cut it:

:)
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PS: Odd Integer 05 Sep 2007, 17:42
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KillerSquirrel
1. Since a-b is even , then a,b must be both odd or both even.

odd-odd = even or even-even = even

but not

odd-even = odd or odd-even = odd

2. knowing that, we can say for sure that a,b are both even - since even/even = even but odd/odd = odd.

so a,b are both even.

since a/b = even integer and we know that a=even and b=even then we can say that at least b*2/b = a/b but can be also 2*b*2/b = a/b and so on.

3. choice (D) is best since:

(b*2+2)/2 = 2(b+1)/2 = 2*(even+1)/2 = odd.

the other choices just don't cut it:

:)
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You cannot determine if a division is even or odd. Even/even can be 10/2 = odd or even/even = 16/2 = 8 = even.
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PS: Odd Integer Updated on: 05 Sep 2007, 22:54
Originally posted by KillerSquirrel on 05 Sep 2007, 22:35.
Last edited by KillerSquirrel on 05 Sep 2007, 22:54, edited 2 times in total.
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ywilfred
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1. Since a-b is even , then a,b must be both odd or both even.

odd-odd = even or even-even = even

but not

odd-even = odd or odd-even = odd

2. knowing that, we can say for sure that a,b are both even - since even/even = even but odd/odd = odd.

so a,b are both even.

since a/b = even integer and we know that a=even and b=even then we can say that at least b*2/b = a/b but can be also 2*b*2/b = a/b and so on.

3. choice (D) is best since:

(b*2+2)/2 = 2(b+1)/2 = 2*(even+1)/2 = odd.

the other choices just don't cut it:

:)

You cannot determine if a division is even or odd. Even/even can be 10/2 = odd or even/even = 16/2 = 8 = even.
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You are wrong !

lets say a = 28 and b = 14

a-b = 28 - 14 = 14 ---> even

a/b = 28/14 = 2 ---> even

so according to the stem those numbers are valid.

(28+2)/2 = 15 ---> odd

since b = 2*n (even integer) then a has to be 2*n*2 ! 14 isn't a multiplie of 4 !

:)
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ggarr
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PS: Odd Integer 05 Sep 2007, 22:39
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how about setting

a - b = 2i and

a/b = 2i and then solving that way?
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after15
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ggarr
how about setting

a - b = 2i and

a/b = 2i and then solving that way?
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I beleive that this is wrong. You assume that a - b = a/b, which does not hold. We can assume that a - b = 2k, a/b = 2m where k and m are integers. No way can we say that k = m.

As mentioned before, I beleive that plugging in values is the faster way to solve. Try the following values: (a,b) = (4,2), (8,4). (48,8), (80,10). Always the option (a+2)/2 is an odd number.

For sure, this is not the most appropriate way to solve anyway...
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PS: Odd Integer 05 Sep 2007, 23:12
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D. from the question stem, we know a and b both are even. since a/b is an even integer, a has to be a multiple of 4 but b could be a multiple of either of 2 or of 4.

a. a/2 could be even if a is a multiple of 4.
b. b/2 could be even if b is a multiple of 4.
c. (a+b)/2 could be even if a and b both are a multiple of 4
d. (a+2)/2 = (a/2 + 1). since a is a multiple of 4, it equals to odd integer.
e. (b+2)/2 = (b/2 + 1). b could be a multiple of 2 ( say 6), it could be even.

so D.



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