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If X and Y are prime numbers, is y(x-3) odd?? 1) X>10 2) [#permalink ]
15 Aug 2006, 17:24

If X and Y are prime numbers, is y(x-3) odd??
1) X>10
2) Y<3

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let's think about the 2) condition only.
y<3 -> y=2 (prime number)
even x (x-3) always equals even.
So, B is the answer, I think.

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B...
B pretty much tells us that y has to be 2 (y<3; =2)
You can also factor this equation out to: xy - 3y
Two rules actually apply here:
Any number multiplied by any even will be even
xy - 3y
This means that both terms will be even.
Second rule is:
even +/- even = Even
odd +/- odd = Even
Just a couple of cool tricks to remember. Hope they help.[/b]

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It's D.
1) is sufficient. X is prime and X>10 means that x odd. 3 is odd. Odd-Odd=even. If we have one number in multiplication even the product is even. => y(x-3) is even.
2) is sufficient. y<3 =>y=2.

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Good catch Natalya!

I disregarded A without even working it out.

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D is right!! I did'nt catch no.1 condition.
Thanks! ^^

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Answer is D.
since both statements can independently lead to answer.
if x> 10 and prime .. x will always be odd and in that case
x-3 will always be even and hence. we can come to answer
y < 3 and a prime number y = 2
y ( x-3) .. will always be even :D

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D

In both cases at least one of the term is even so the result will be even.

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Can anyone explain please? [#permalink ]
16 Aug 2006, 03:37

Please explain how this could be D.
a) X > 10. Agree that A is sufficient
b) y< 3 i.e y=2 which is prime number. But we don't know the value of X.
X could be any number say 3
Y(x-3)=0. so expression gives 0. so B is insufficient.
So Answer is A

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Re: Can anyone explain please? [#permalink ]
16 Aug 2006, 05:37

baski6 wrote:

Please explain how this could be D. a) X > 10. Agree that A is sufficient b) y< 3 i.e y=2 which is prime number. But we don't know the value of X. X could be any number say 3 Y(x-3)=0. so expression gives 0. so B is insufficient. So Answer is A

Whatever is the value of other term, anything multiplied my an even number is EVEN.

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If you plug in 11 for X, and 2 for Y you get:
I) x> 10 2(11-3) = 2(8) 16 which is not even.. How is A sufficient here?
II) y<3 2(11-3) = 2(8) 16 which again is not even..
What am I missing in understanding this problem...
Thanks for all your help....

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gobucks wrote:

If you plug in 11 for X, and 2 for Y you get: I) x> 10 2(11-3) = 2(8) 16 which is not even .. How is A sufficient here? II) y<3 2(11-3) = 2(8) 16 which again is not even .. What am I missing in understanding this problem... Thanks for all your help....

16 is NOT even???

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E,
S1)
If you use X = 11 and Y = 2, y(x-3) = Even
If you use x = 11 and Y = 1, y(x-3) = Odd
Insufficient
S2)
If you use X = 11 and Y = 2, y(x-3) = Even
If you use X = 11 and Y = 1, y(x-3) = Odd
Insufficient
Both 1 and 2 are prime and they are less then 3.
With both S1&S2:
If you use X = 11 and Y = 2, y(x-3) = Even
If you use X = 11 and Y = 1, y(x-3) = Odd
Insufficient

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ps_dahiya wrote:

gobucks wrote:

If you plug in 11 for X, and 2 for Y you get: I) x> 10 2(11-3) = 2(8) 16 which is not even .. How is A sufficient here? II) y<3 2(11-3) = 2(8) 16 which again is not even .. What am I missing in understanding this problem... Thanks for all your help....

16 is NOT even???

SORRY, Meant to say that it is NOT ODD?? So, I am still confusused because the question asks if it is ODD...??

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The answer in the book is ..... D

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Whoa...I was way off, i see what the problem is now. Thanks.

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agsfaltex wrote:

Whoa...I was way off, i see what the problem is now. Thanks.

How do you get an ODD number to in order to get the answer D?? I substituted 11 for X, and 2 for Y

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gobucks wrote:

agsfaltex wrote:

Whoa...I was way off, i see what the problem is now. Thanks.

How do you get an ODD number to in order to get the answer D?? I substituted 11 for X, and 2 for Y

You have to answer if the term is odd or not.

Answer is it is NOT ODD. That means you can answer the question using both statements. Hence it is D.

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baski6 - zero is EVEN
so stmt 2 holds and the result is NO
So the answer is D