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If y, z, and b are positive integers greater than 1, what is the value 02 May 2016, 22:30
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41% (02:29) correct 59% (02:52) wrong based on 98 sessions

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If y, z, and b are positive integers greater than 1, what is the value of z?

(1) \(y^3*z^b\) is a perfect square less than 100.
(2) \(y^3*z^b = x^3\), where x is an integer.
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If y, z, and b are positive integers greater than 1, what is the value 03 May 2016, 08:20
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y > 1; z > 1; b > 1; z = ?

St1: y^3*z^b is a perfect square less than 100
Possible Squares: 4, 9, 16, 25, 36, 49, 64, 81
Possible Cubes: 8, 27, 81

Cube * z^b = Square
Only 8 * 8 = 64 is a valid value
8 = 2^3; z^b = 2^3 --> z = 2

St2: y^3*z^b = x^3
8 * 2^3 = 4^3 --> z = 2
8 * 8^2 = 8^3 --> z = 8
Not Sufficient

Answer: A
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If y, z, and b are positive integers greater than 1, what is the value 22 Sep 2018, 23:18
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Hi chetan2u Bunuel PKN gmatbusters amanvermagmat Abhishek009 VeritasKarishma

I could not understand above solution. Could you elaborate?
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If y, z, and b are positive integers greater than 1, what is the value 23 Sep 2018, 19:19
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adkikani
Hi chetan2u Bunuel PKN gmatbusters amanvermagmat Abhishek009 VeritasKarishma

I could not understand above solution. Could you elaborate?
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Hi adkikani,

Question stem:- We need to find out an unique value of z.

Given, y,z, and b are positive integers greater than 1.

St1:- \(y^3=\)A perfect cube*\(z^b<\) perfect square less than 100( \(2^2, 3^2, 4^2,5^2, 6^2, 7^2, 8^2, 9^2\))
Since y is an integer greater than 1, hence the possible values of \(y^3\)\(* z^b\):
a) \(2^3*8=8^2=64<100\) (A PERFECT SQUARE). KEEP
b) \(3^3*3^1=9^2\)<100( (A PERFECT SQUARE), However, b>1. DISCARD this case.

So, we have an unique value of z from (a).
Sufficient.

St2:- \(y^3*z^b=x^3\), where x is an integer.
Here, there are no restrictions on the values of the variables x,y,z, and b.
Hence, we can have numerous values of z.

Insufficient.

Ans. (A)

P.S:- You may raise specific queries(if any)
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If y, z, and b are positive integers greater than 1, what is the value 05 Jun 2021, 06:10
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If z and b are integers greater than 1, then the value of z^b is at least 4. Statement 1 tells us:

y^3 * z^b < 100

and since z^b is 4 or greater, y^3 must be less than 25, and if y is an integer greater than 1, y can only equal 2. So we know (2^3)(z^b) = 8(z^b) is a square less than 100, but is at least 32 (since z^b is at least 4), and the only positive square between 32 and 100 that is divisible by 8 is 64. So 8(z^b) = 64, and z^b = 8, and z = 2 and b = 3 is the only possibility if z and b are integers greater than 1.

Statement 2 is not sufficient, because it's possible, say, that b = 3. Then (y^3)(z^b) = (y^3)(z^3) = (yz)^3, and yz is automatically the cube of some integer no matter what z is equal to, and z can have any value at all.

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