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If x, y, z and n are positive integers and t = x^2y^2z^n, how many di

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Bunuel
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If x, y, z and n are positive integers and t = x^2y^2z^n, how many di [#permalink] New post   15 Jan 2021, 01:15
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If x, y, z and n are positive integers and \(t = x^{2}y^{2}z^{n}\), then t hs how many different positive factors?

(1) x, y and z are prime numbers

(2) n = 2



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E
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Re: If x, y, z and n are positive integers and t = x^2y^2z^n, how many di [#permalink] New post   15 Jan 2021, 03:47
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Given x, y, z and n are positive integers and we need find the total number of factors for t which is x^2*y^2*z^n.

Total Number of factors for any number expressed as powers of its distinct prime factors m,n and p as m^a*n^b*p^c is given by (a+1)*(b+1)*(c+1)


Statement 1: x,y and z are prime numbers. We have no info whether x,y and z are distinct prime numbers and value of n is not known. INSUFFICIENT

Statement 2: n=2

No info on x, y and z. INSUFFICIENT

Combining 1 & 2

Still no info whether x,y and z are distinct. INSUFFICIENT

ANS: E

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Re: If x, y, z and n are positive integers and t = x^2y^2z^n, how many di [#permalink] New post   15 Jan 2021, 21:47
If x, y, z and n are positive integers and t=x^2 *y^2 *z^n

Stat1: x, y and z are prime numbers
Total nos. of factors = (2+1)*(2+1)*(n+1), we don't know value of n. Not sufficient.

Stat2: n = 2
we don't know, if x, y and z are prime or not. We need to break them in prime numbers then, total nos. of factors can be calculated. Not sufficient.

Combining both stats,
if x=y=z then, total nos. of factors = 6+1=7, if x, y and z are different, then total nos. of factors = 3*3*3=27; Not sufficient

So, I think E. :)
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Re: If x, y, z and n are positive integers and t = x^2y^2z^n, how many di [#permalink] New post   16 Jan 2021, 02:17
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If x, y, z and n are positive integers and t=x^2*y^2*z^n, then t has how many different positive factors?

(1) x, y and z are prime numbers
--> we have no information about n. Insufficient.

(2) n = 2
--> No information about x,y,z. Insufficient.

Combining both the statements:-

if x=y=z=2, No. of distinct positive factors of 2^2*2^2*2^2 => (6+1) = 7.
If x,y,z are distinct prime numbers, No. of distinct positive factors of 2^2*3^2*5^2 => 27

No definite answer. Hence, Insufficient.

IMO E.
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Re: If x, y, z and n are positive integers and t = x^2y^2z^n, how many di [#permalink] New post   Updated on: 16 Jan 2021, 11:34
When any positive integer "N" is written in the form of \( N = a^pb^qc^rd^s....\) (where a, b, c, d,.... are distinct prime numbers), then the total no. of factors of N is given by the expression (p+1)(q+1)(r+1)(s+1).....
Keeping the above in mind,

Statement 1: x, y, z are prime numbers. Hence, Total no. of factors of t = (2+1)(2+1)(n+1). Since we don't know the value of n, we don't know the answer. Insufficient.

Statement 2: n = 2. But now we don't know whether x, y, z are prime numbers. Insufficient.

Combining Statements 1 & 2:
We know that x, y, z are prime numbers and n = 2. However, we still don't know whether x, y, z are distinct prime numbers.
Insufficient.

Ans is E IMO

Originally posted by Vivor on 16 Jan 2021, 07:52.
Last edited by Vivor on 16 Jan 2021, 11:34, edited 2 times in total.
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Re: If x, y, z and n are positive integers and t = x^2y^2z^n, how many di [#permalink] New post   16 Jan 2021, 09:08
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Asked: If x, y, z and n are positive integers and \(t = x^{2}y^{2}z^{n}\), then t hs how many different positive factors?

(1) x, y and z are prime numbers
n is unknown
NOT SUFFICIENT

(2) n = 2
It is unknown whether x,y & z are prime numbers or not.
NOT SUFFICIENT

(1)+(2)
Still it is unknown whether x,y&z are distinct or not
NOT SUFFICIENT

IMO E




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Re: If x, y, z and n are positive integers and t = x^2y^2z^n, how many di [#permalink] New post   17 Jan 2021, 00:23
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If x, y, z and n are positive integers and x^2*y^2*z^2, then t hs how many different positive factors?

(1) x, y and z are prime numbers
n is not known.
Not sufficient

(2) n = 2
No info. About x, y, and z
Not sufficient

(1) + (2)

For x=y=z=2, x^2*y^2*z^2 = 2^2*2^2*2^2 = 2^6
Number of factors = 6+1 = 7

For x=y=3 and z=2, x^2*y^2*z^2 = 3^2*3^2*2^2 = 3^4*2^2
Number of factors = 5*3 = 15

Not sufficient

Option E

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Re: If x, y, z and n are positive integers and t = x^2y^2z^n, how many di [#permalink] New post   17 Jan 2021, 00:39
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Bunuel wrote:
If x, y, z and n are positive integers and \(t = x^{2}y^{2}z^{n}\), then t hs how many different positive factors?

(1) x, y and z are prime numbers

(2) n = 2



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The number of distinct positive factors of t will depend on.
a) Whether x, y and z are DISTINCT prime numbers. If not, then which of them are same.
b) the value of n.

(1) x, y and z are prime numbers
We do not know whether they are distinct and n is not known.

(2) n = 2
Point (a) not known

Combined
We still do not know whether x, y and z are distinct

E
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Re: If x, y, z and n are positive integers and t = x^2y^2z^n, how many di [#permalink] New post   17 Jan 2021, 01:20
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great question and you got me.......


The "C-Trap" at its greatest.

S1 and S2 are not sufficient alone:

Together:


Case 1: x = y = z = same prime number

t = (3)^2 * (3)^2 * (3)^2

t = (3)^8

No. of Unique Positive Factors = (8 + 1) = 9


Case 2: X = 2 ---- Y = 3 ----- Z = 5

t = (2)^2 * (3)^2 * (5)^2

No. of Unique Positive Factors = (2 + 1) * (2 + 1) * (2 + 1) = 3 * 3 * 3 = 27



Different Values,

E
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Re: If x, y, z and n are positive integers and t = x^2y^2z^n, how many di [#permalink] New post   27 Sep 2023, 13:29
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Re: If x, y, z and n are positive integers and t = x^2y^2z^n, how many di [#permalink]
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