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 oddWe 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 even1. 2*3*4 = 24 even
2. 1*2*3 = 6 even
3. -1*0*-1 = 0 even
Then I is
OKII. x + y + z is even1. 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 OKIII. (x + y)(y + z) is odd1. (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
OKSo I,III is OK then we choose E.More tricksWe 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
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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.
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same type (even-even, odd-odd) that is + or - will have eveneven + even = even
odd + odd = even
but different type will have oddeven + 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.