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# For all positive integers x, y, and z, x@y^(z+1) means that y^(z+1) an

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For all positive integers x, y, and z, x@y^(z+1) means that y^(z+1) an  [#permalink]

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Updated on: 27 Mar 2019, 04:52
00:00

Difficulty:

55% (hard)

Question Stats:

48% (02:18) correct 52% (02:22) wrong based on 27 sessions

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For all positive integers x, y, and z, $$x@y^{z+1}$$ means that $$y^{z+1}$$ and $$y^z$$ are divisors of x. If $$162@3^{z+1}$$, then what does z equal?

a. -3 < z < -1

b. -1 < z < 4

c. -2 < z < 5

d. 0 < z < 6

e. 1 < z < 4

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Originally posted by AkshdeepS on 27 Mar 2019, 03:23.
Last edited by Bunuel on 27 Mar 2019, 04:52, edited 1 time in total.
Renamed the topic.
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Re: For all positive integers x, y, and z, x@y^(z+1) means that y^(z+1) an  [#permalink]

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27 Mar 2019, 06:57
2
162@3^z+1 where 3^z+1 and 3^z are divisors of 162..

162 = 3^4 * 2
Let's focus on 3^z+1..

The highest value of 3^z+1 can be 3^4..
Z+1= 4
Z= 3..

The lowest value of 3^z+1 = 3^1 = z+1= 1
Z= 0

Therefore 0<= z <= 3

The only inequality that matches with our answer is -1<z<4... Hence b is the right answer..

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Re: For all positive integers x, y, and z, x@y^(z+1) means that y^(z+1) an   [#permalink] 27 Mar 2019, 06:57
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