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 Post subject: Arithmetic I Question 80 pg 96Posted: Mon Feb 07, 2011 8:58 pm

Joined: Mon Mar 02, 2009 8:18 pm
Posts: 27
Question 80 says the product of integers x,y,z is even. is z even?

1) x/y = z

2) z=xy

the back of the book says for explanation 1) x and y must both be even. I dont understand why they must both be even. X and/or Y can be even to still conform to the question stem statement that product xyz = even.
Please tell me what I am missing.

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 Post subject: Re: Arithmetic I Question 80 pg 96Posted: Tue Feb 08, 2011 7:21 pm

Joined: Thu Feb 12, 2009 6:32 pm
Posts: 497
if xyz is even, then we know that all three numbers can not be odd,

since odd * odd * odd = odd

statement 1 tells us that x = yz

statement 2 tells us that z = xy

since, per the above, all three numbers can't be odd there is no way to have the product of two of the numbers equal an odd number

for example, if x were odd, y or z would have to be even. but the product of even * odd = even

Veritas Help

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 Post subject: Re: Arithmetic I Question 80 pg 96Posted: Tue Feb 08, 2011 11:01 pm

Joined: Mon Mar 02, 2009 8:18 pm
Posts: 27
I understand. My point is the answer in the back of the book seems to have a false statement in the explanation (although the answer is correct). Please confirm I am correct.

statement 1) x/y = z or x=YZ

From the answer in the back of the book: "In the first statement, we know that x and y are even, but we do not know whether z is odd or even"

I believe that we do NOT know that X and Y are even for sure. We only know that X is even for sure.
Y and/or Z can be even.

For example (the three sceanrios are):
X (even) = Y (odd) * Z (even)
or
X (even) = Y (even) * Z (odd)
or
X (even) = Y (even) * Z (even)

this is why I believe the statement in the book "In the first statement, we know that x and y are even, but we do not know whether z is odd or even" is incorrect. as you can see above, Y has a possibiilty of being odd.

Please confirm I am not missing anything.

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 Post subject: Re: Arithmetic I Question 80 pg 96Posted: Wed Feb 09, 2011 5:53 pm

Joined: Thu Feb 12, 2009 6:32 pm
Posts: 497

We also have to consider x/y = z

z must be an integer. it must be even if either x or y is odd

scenario A: x even, y even, z even ... this is possible

scenario B: x even, y odd ... this is possible 100/5 = 20

scenario C: x odd, y even .. this scenario is not possible

scenario D: x even, y even, z odd ... this is possible 100/20 = 5

so, you are correct x and y do not both need to be even

Veritas

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