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For all positive integers x, y, and z, x@y^(z+1) means that y^(z+1) an 30 Nov 2021, 06:45
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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50% (02:09) correct 50% (02:17) wrong based on 148 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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B
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For all positive integers x, y, and z, x@y^(z+1) means that y^(z+1) an 18 Dec 2021, 06:35
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162 is equal to 2*3^4
Now 3^z+1 and 3^z are divisor of 162
Hence max power of 3 can be 4 and lowest is zero

Hence z+1 less than equal to 4
Z less than equal to 3
Z+1 > equal to 0
Z > eq to -1
Only b satisfies this

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For all positive integers x, y, and z, x@y^(z+1) means that y^(z+1) an 18 Dec 2021, 08:56
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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
B. -2 < z < 5
D. 0 < z < 6
E. 1 < z < 4
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HI I thinks there is some problem with OA,

we have to express function with base form of 3, 162/3^z+1 and 162/3^z now z can be -1 for 162/3^z+1 such that denominator becomes 3^0=1 and for 162/3^z ,z = 0
solutions are -1 and 0 hence the OA should be C
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GMAT Focus 1: 735 Q90 V89 DI81
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For all positive integers x, y, and z, x@y^(z+1) means that y^(z+1) an 18 Dec 2021, 09:57
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Bunuel
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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162 = \(2*3^4\)
So, \(3^0, 3^1,3^2,3^3,3^4\) are factors.
Therefore, \(0\leq z+1\leq 4\)
\(-1\leq z\leq 3\)

Now, z is positive integer, so minimum value of z is 1.
Thus, \(0<z<4\)

Only B matches with the given range as -1<z<4 would also mean 0<z<4 as z>0.
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For all positive integers x, y, and z, x@y^(z+1) means that y^(z+1) an 29 Mar 2026, 05:27
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if you take option B, 4 is not possible as 3^4+1 = 3^5 and both conditions must satisfy.
vipulgoel



HI I thinks there is some problem with OA,

we have to express function with base form of 3, 162/3^z+1 and 162/3^z now z can be -1 for 162/3^z+1 such that denominator becomes 3^0=1 and for 162/3^z ,z = 0
solutions are -1 and 0 hence the OA should be C
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For all positive integers x, y, and z, x@y^(z+1) means that y^(z+1) an 29 Mar 2026, 22:45
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Hi vipulgoel,

The issue is right in the first line of the problem: 'For all positive integers x, y, and z.' This means z must be a positive integer — so z = 1, 2, 3, 4, ... Only. Values like z = -1 or z = 0 are not allowed because they are not positive integers.

Now let's find which positive integer values of z work.

162 = 2 × 3^4, so the highest power of 3 that divides 162 is 3^4.

The operation 162@3^(z+1) requires:
- 3^(z+1) divides 162 → z + 1 ≤ 4 → z ≤ 3
- 3^z divides 162 → z ≤ 4

The tighter constraint is z ≤ 3.

Since z must be a positive integer, valid values are z = 1, 2, or 3.

Now check the answer choices — which range contains exactly [b]1, 2, and 3?[/b]

Answer B: -1 < z < 4 — this captures 1, 2, 3. Correct!
Answer C: -2 < z < 5 — this would also include z = 4, but 3^(4+1) = 3^5 = 243, which does NOT divide 162. So C is too wide.

Key Insight: Your mistake was allowing z = -1 and z = 0, which the problem rules out by saying z is a positive integer. Always check the constraints given at the start of custom operator problems — they're easy to overlook but critical.

Answer: B
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