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

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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
Intern
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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.
Manager
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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]
Manager
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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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Thanks!! [#permalink]  15 Aug 2006, 20:05
D is right!! I did'nt catch no.1 condition.

Thanks! ^^
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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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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.

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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.

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...

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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...

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...

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
Senior Manager
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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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so stmt 2 holds and the result is NO

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