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

Question

x, y and z are consecutive positive integers such that x < y < z; which of the following must be true?

1. xyz is divisible by 6
2. (z-x)(y-x+1) = 4
3. xy is odd

A) I only
B) II only
C) III only
D) I and II only
E) I, II, and III

beyondgmatscore · Answer

Since x, y and z are consecutive integers such that x < y < z, we can say x = y-1 and Z = y+1

Statement 1 would be true as at least one of the three numbers is divisible by 2 and one by 3 so xyz would be divisible by 6.

Statement 2 can be simplified if we write everything in terms of y as ((y+1)-(y-1))*(y-(y-1)+1) = 2*2 = 4 So, always true

Statement 3 talks about xy Since x and y are consecutive integers one of them is odd and other is even so product would always be even and hence not true.

So, I and II are always true and hence answer is D.

subhashghosh · Answer

y = x + 1, z = x + 2

so xyz = x * (x+1) * (x+2) hence it will have a 2 and 3 as factor, so it's divisible by 6.

(I) is true

(z-x)(y-x+1) = (2) (1 + 1) = 4

(II) is true

xy can't be odd, because if x is odd y is even, so odd * even = even

if x is even y is odd, so even * odd = even

(if x is even there is no need to see what y is though)

Answer - D

IanStewart · Answer

It can sometimes be useful to know that the product of n consecutive integers is always divisible by n!. For example, if you have 3 consecutive integers, their product will always be divisible by 3! = 6. This is true because multiples of 3 are three apart, so among any three consecutive integers, you will always find you have exactly one multiple of 3. You'll also always have at least one multiple of 2, so the product of three consecutive integers must be divisible by 3*2. You can use the same logic to prove that the product of 4 consecutive integers is divisible by 4! = 24 (among any four consecutive numbers, you always have exactly one multiple of 4, at least one multiple of 3, and another number which is a multiple of 2), and so on.

If you know this, you can see instantly that (1) is true. Further, xy is the product of 2 consecutive integers so must be divisible by 2! = 2, and there is no way that (3) is true. As others did above, you can establish that (2) is true by substitution.

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x, y and z are consecutive positive integers such that x < y < z

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