number properties problem
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- theboyleman32
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- ceilidh.erickson
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If the product xyz is even, one of them, two of them, or all three of them could be even. If we want to know if z is even, we can test numbers, or think conceptually.
1) x/y = z
Think conceptually:
We can rearrange this to say x = yz. If x is equal to the product of y and z, then xyz = x^2. Since we know that xyz is even, it must also follow that x^2 is even, and thus x is even. (An even squared = even; odd squared = odd).
Since x is even, and it's the product of y and z, it must follow that either y or z is even, or both (because odd*odd = odd, so they can't both be odd). It's possible, though, that y is even and z is odd.
Test numbers:
x=4, y=2, z=2 --> 4/2 = 2
yes, z is even
x=6, y=2, z=3 --> 6/2 = 3
no, z is no even
Insufficient.
2) z = xy
Think conceptually:
If z is equal to the product of x and y, then xyz = z^2. Since we know that xyz is even, it must also follow that z^2 is even, and thus z is even. (An even squared = even; odd squared = odd).
Or thinking about it another way, if x and y were both odd, then z would be odd, and the product xyz would be odd, which can't be true. Thus, either x or y or both must be even, making z even.
Test numbers:
x=2, y=3, z=6 --> 6 = 2*3
There are no numbers that we could test here that would give us an even product of xyz, but not an even product of xy.
Sufficient.
The answer is B.
1) x/y = z
Think conceptually:
We can rearrange this to say x = yz. If x is equal to the product of y and z, then xyz = x^2. Since we know that xyz is even, it must also follow that x^2 is even, and thus x is even. (An even squared = even; odd squared = odd).
Since x is even, and it's the product of y and z, it must follow that either y or z is even, or both (because odd*odd = odd, so they can't both be odd). It's possible, though, that y is even and z is odd.
Test numbers:
x=4, y=2, z=2 --> 4/2 = 2
yes, z is even
x=6, y=2, z=3 --> 6/2 = 3
no, z is no even
Insufficient.
2) z = xy
Think conceptually:
If z is equal to the product of x and y, then xyz = z^2. Since we know that xyz is even, it must also follow that z^2 is even, and thus z is even. (An even squared = even; odd squared = odd).
Or thinking about it another way, if x and y were both odd, then z would be odd, and the product xyz would be odd, which can't be true. Thus, either x or y or both must be even, making z even.
Test numbers:
x=2, y=3, z=6 --> 6 = 2*3
There are no numbers that we could test here that would give us an even product of xyz, but not an even product of xy.
Sufficient.
The answer is B.
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
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- Brent@GMATPrepNow
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The product of integers x, y and z is even. Is z even?
1) x/y = z
2) z = xy
Target question: Is z even?
Given: The product of integers x, y and z is even.
KEY CONCEPT: If xyz is even, then it MUST be the case that at least one of x,y and z is even.. Otherwise, the product xyz would be odd
Statement 1: x/y = z
There are several values of x, y and z that satisfy this condition. Here are two:
Case a: x= 8, y = 2 and z = 3 (notice that xyz = 48, which is even). In this case z IS even
Case b: x= 6, y = 2 and z = 3 (notice that xyz = 36, which is even). In this case z is NOT even
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: z = xy
We've already concluded that at least one of x,y and z is even, so either x is even, y is even, or z is even.
Let's consider all 3 cases:
- if x is even, and if z = xy, then z = (some even number)(y), which means z is even
- if y is even, and if z = xy, then z = (x)(some even number)), which means z is even
- if z is even, then z is even
In all 3 possible cases, z is even, so IT MUST BE THE CASE THAT z is even
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
Answer = B
Cheers,
Brent
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Another approach!
Start with the stem, which tells you that at least one of x, y, and z is even.
S1:: Rearrange this to x = y * z. We know that AT LEAST one of the integers is even. If x is even, then y * z is even (since x = y * z). If y * z is even, then either y or z is even, but we don't know which one. INSUFFICIENT
S2:: As before, we already know that at least one of the three integers is even. If it's z, we're set. Now suppose that it isn't, which means that either x or y is even. But z = x * y, so z = even * x or z = even * y, so z is also even. So in either case, z is even. SUFFICIENT!
Start with the stem, which tells you that at least one of x, y, and z is even.
S1:: Rearrange this to x = y * z. We know that AT LEAST one of the integers is even. If x is even, then y * z is even (since x = y * z). If y * z is even, then either y or z is even, but we don't know which one. INSUFFICIENT
S2:: As before, we already know that at least one of the three integers is even. If it's z, we're set. Now suppose that it isn't, which means that either x or y is even. But z = x * y, so z = even * x or z = even * y, so z is also even. So in either case, z is even. SUFFICIENT!