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For nonnegative integers x, y, and z, is x^z odd? [#permalink]
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29 Sep 2015, 05:32
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Re: For nonnegative integers x, y, and z, is x^z odd? [#permalink]
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29 Sep 2015, 05:46
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Bunuel wrote: For nonnegative integers x, y, and z, is x^z odd?
(1) The product xz is odd. (2) x = 2^y
Kudos for a correct solution. (1) The product xz is odd. => x, z are odd => x^z odd => SUF (2) x = 2^y => x is even => x^z even => SUF Answer: D



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Re: For nonnegative integers x, y, and z, is x^z odd? [#permalink]
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29 Sep 2015, 06:32
For nonnegative integers x, y, and z, is x^z odd? > this expression is odd only if x is odd (z is integer and doesn't matter in this case) (1) The product xz is odd. > if the product of 2 numbers is odd then we have Odd*Odd. SUFFICIENT (2) x = 2^y > x=even, means x^z is also even SUFFICIENT Answer D
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Re: For nonnegative integers x, y, and z, is x^z odd? [#permalink]
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29 Sep 2015, 06:39
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Statement 1 : The product xz is odd implies that x is odd and z is odd, hence x^z is odd. Sufficient
Statement 2 : x = 2^y and we are asked to check whether 2^(yz) is odd.
Here y can be 0 and thus X^Z = 1(odd) or Y and Z can be any other positive integer and then X^Z is even.( Insuff)
So, Answer (A)



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Re: For nonnegative integers x, y, and z, is x^z odd? [#permalink]
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29 Sep 2015, 06:48
Actually Issaml is right... just missed the point about non negative can be 0 (it's not said that it's positive)
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Re: For nonnegative integers x, y, and z, is x^z odd? [#permalink]
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29 Sep 2015, 08:47
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Bunuel wrote: For nonnegative integers x, y, and z, is x^z odd?
(1) The product xz is odd. (2) x = 2^y
Kudos for a correct solution. A stmt 1 says that both x and z are odd. suff stmt 2 x could be even or odd depending on if y>0. insuff
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Re: For nonnegative integers x, y, and z, is x^z odd? [#permalink]
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01 Oct 2015, 03:46
I marked A but when I saw the first 2 comments, I was like WHAT???



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Re: For nonnegative integers x, y, and z, is x^z odd? [#permalink]
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05 Oct 2015, 06:11
Bunuel wrote: For nonnegative integers x, y, and z, is x^z odd?
(1) The product xz is odd. (2) x = 2^y
Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION:In this problem, you hopefully found statement (1) to be sufficient, without too much effort. If the product xz is odd, then both integers must be odd. And if x, the base of x^z, is odd, then when raised to any integer it will be odd. Statement (1) guarantees that xz is odd. But statement (2) seems to guarantee that x is even, as 2 to any exponent is even. Right? There’s one glaring exception: 2^0 = 1 — an important property for many exponent problems. So while most potential values of x, given statement (2), are even, one does exist where it’s odd. Statement (2) is not sufficient, and the correct answer is A. Here’s where you can use some higherlevel strategy. If you were certain (as you should be) that statement (1) is sufficient with the answer “odd,” then as soon as you see a potential “even” with statement (2) you know that statement (2) is not sufficient! Why? Because the rules of the game dictate that if statement (1) gives you the answer “x must be odd,” then statement (2) cannot say “x must be even.” The only options are “x must be odd” or “x could be either odd or even.” Once you’ve found that “even” answer, then you know the “maybe” part must be coming, based on the rules of the game.
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Re: For nonnegative integers x, y, and z, is x^z odd? [#permalink]
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21 Nov 2016, 02:50
Bunuel wrote: For nonnegative integers x, y, and z, is x^z odd?
(1) The product xz is odd. (2) x = 2^y
Kudos for a correct solution. Great Question It really Tests our Knowledge on the concepts on evens/odds Here we are given three non negative integers x,y,z and we need to check if x^z is odd or not There are two ways in which x^z can be odd here Case 1=> If z=0, Then the value of x won't even matter as anything^0=1 Case 2=> If x=odd number. Then the value of z wont matter as odd^any nonnegative integer is odd Statement 1 Here as xz=odd so x and z both must be odd hence sufficient to say that x^z all be odd Statement 2 x=2^y so x can be either 1 (for y=0) or an even number If x=1 => x^z will be odd but if x=even , then unless z is zero => x^z will be even And we are given no info on z hence this statement is insufficient Hence A
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