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If x, y, and z are 3 positive consecutive integers such that

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Director
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If x, y, and z are 3 positive consecutive integers such that [#permalink] New post 23 Sep 2006, 04:50
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A
B
C
D
E

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If x, y, and z are 3 positive consecutive integers such that x < y < z, what is the remainder when the product of x, y, and z is divided by 8?

(1) (xz)^2 is even

(2) 5y3 is odd
Director
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 [#permalink] New post 23 Sep 2006, 04:58
D?

Case 1: (xz)^2 is even
=> xz is even.
From given data, x<y<z are consecutive, we can conclude that x and z should both be even numbers.
=> Remainder should be zero.

Case 2: 5*(y^3) is odd
=> y is odd.
=> From data given in the question, that x and z are even.
=> Remainder should be zero.
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 [#permalink] New post 23 Sep 2006, 09:33
If x, y, and z are 3 positive consecutive integers such that x < y < z, what is the remainder when the product of x, y, and z is divided by 8?

(1) (xz)^2 is even

(2) 5y3 is odd

n,n-1,n+1 = y,x,z respectively

the prodyct is multiple of 3 and 2 (at least one is even)

from one

xz is even thus the product is divisible by 8.....suff remainder is 0

from 2

y is odd thus x,z are even one a multiple of two and the other of 4...suff remainder is zero

My answer is D
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Re: DS : Crazy Eights [#permalink] New post 23 Sep 2006, 18:08
rxs0005 wrote:
If x, y, and z are 3 positive consecutive integers such that x < y < z, what is the remainder when the product of x, y, and z is divided by 8?

(1) (xz)^2 is even

(2) 5y3 is odd



Yes, it should be D

n(n+1)(n+2) for any integer n
Is divisible by 8 if n is even

A We know that x and z are even , so xyz is divisible by 8. Suff
B We know that y is odd so x and y should be even since xyz are consecutive. Suff
Re: DS : Crazy Eights   [#permalink] 23 Sep 2006, 18:08
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