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Re: If x,y, and z are consecutive integers and x < y < z [#permalink]
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Given that x < y < z, where x, y, and z are consecutive integers.
In the first case, z =4, y=3, x = 2

I) xyz = 2*3*4 =even
II) x+y+z = 2 + 3 +4 = Odd
III) (x+y)(y+z) = (2 +3) * (3+4) = 5*7 = Odd

In the second case, z =5, y=4, x = 3
I) xyz = 3*4*5 =even
II) x+y+z = 3 +4+ 5 = Even
III) (x+y)(y+z) = (3 +4) * (4+5) = 7*9 = Odd


In both cases, (I), and (III) are true.(E)
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Re: If x, y, and z are consecutive integers and x < y < z, which of the fo [#permalink]
If x, y, and z are consecutive integers and x < y < z, which of the following must be true?

I. xyz is even
II. x + y + z is even
III. (x + y)(y + z) is odd



We know that x,y,z are integers that are consecutive.
What we should know next is that there are 3 cases to consider in this question. (because there is a possibility that the middle number would be even, odd, or even zero)

1. even, odd, even
2. odd, even, odd
3. -1,0,1

For me, I believe that input the number is the easiest way to solve this question. This is because we can see that the question I,II,III are all about +,-,*,/ which are not that complicated, so input number could be the right way to do. My input number will be
1. even, odd, even = 2,3,4
2. odd, even, odd = 1,2,3
3. -1,0,1

I. xyz is even
1. 2*3*4 = 24 even
2. 1*2*3 = 6 even
3. -1*0*-1 = 0 even
Then I is OK

II. x + y + z is even
1. 2+3+4 = 9 odd
The question ask about "MUST be true" which means that every cases must suffice. So we dont have to waste time calculate other options
Then II is NOT OK

III. (x + y)(y + z) is odd
1. (2+3)(3+4) = 5*7 = 35 odd
2. (1+2)(2+3) = 3*5 = 15 odd
3. (-1+0)(0+1) = -1*1 = -1 odd
Then III is OK

So I,III is OK then we choose E.

More tricks
We may be able to calculate more 20+ secs faster if we can remember these patterns since we may not waste time to calculate or input simple numbers
- even * any integers = even (because there is gonna be 2 as factors for sure so that the product will be even) This could help us be able to skip question "I" since we know that there is even integer for sure in the answer.
- same type (even-even, odd-odd) that is + or - will have even
even + even = even
odd + odd = even
but different type will have odd
even + odd = odd
odd - even = odd
- But in the end, to input could be a very effective strategy in these kind of questions and not waste that much time.
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Re: If x, y, and z are consecutive integers and x < y < z, which of the fo [#permalink]
I don't get one thing. The question is saying consecutive integers, Not just even consecutive integers. So we can select 1,3 and 5. Then xyz=15 which is odd.
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Re: If x, y, and z are consecutive integers and x < y < z, which of the fo [#permalink]
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Hea234ven wrote:
I don't get one thing. The question is saying consecutive integers, Not just even consecutive integers. So we can select 1,3 and 5. Then xyz=15 which is odd.


"Consecutive integers" ALWAYS mean integers that follow each other in order with common difference of 1: ... x-3, x-2, x-1, x, x+1, x+2, ....

For example:

-7, -6, -5 are consecutive integers.

2, 4, 6 ARE NOT consecutive integers, they are consecutive even integers.

1, 3, 5 ARE NOT consecutive integers, they are consecutive odd integers.
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Re: If x, y, and z are consecutive integers and x < y < z, which of the fo [#permalink]
Hea234ven wrote:
I don't get one thing. The question is saying consecutive integers, Not just even consecutive integers. So we can select 1,3 and 5. Then xyz=15 which is odd.

1. Consecutive integers = ...,-3,-2,-1,0,1,2,3,...
2. Consecutive even integers = ...,-4,-2,0,2,4,...
3. Consecutive odd integers = ...-5,-3,-1,1,3,5,...

So the question is saying "Consecutive integers" which means that it could be ...,-3,-2,-1,0,1,2,3,...
we CAN NOT select 1,3,5 since 1,3,5 are "Consecutive ODD integers"

Not sure if its help?
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Re: If x, y, and z are consecutive integers and x < y < z, which of the fo [#permalink]
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Re: If x, y, and z are consecutive integers and x < y < z, which of the fo [#permalink]
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