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If x and y are prime numbers and x ≠ y, how many unique factors does

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Bunuel
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If x and y are prime numbers and x ≠ y, how many unique factors does [#permalink] New post  20 May 2018, 11:46
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If x and y are prime numbers and x ≠ y, how many unique factors does the product x^2*y have?

A. 3
B. 4
C. 6
D. 8
E. 9
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EgmatQuantExpert
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Re: If x and y are prime numbers and x ≠ y, how many unique factors does [#permalink] New post  20 May 2018, 12:27
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Solution



Given:
• x an y are two prime numbers

To find:
• Number of unique factors of the product of \(x^2\) and y

Approach and Working:
• As x and y are prime numbers, individually they don’t have any other factors other than 1 and the number itself
• Now we know that, if a number is expressed in terms of its prime factors \(a^p * b^q * c^r\), where a, b, c are prime factors and p, q, r are the corresponding powers of those prime factors, then
    o Total number of factors = (p + 1) (q + 1) (r + 1)
• Hence, for the number \(x^2 * y\), the total number of factors = (2 + 1) * (1 + 1) = 3 * 2 = 6

Hence, the correct answer is option C.

Answer: C
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ScottTargetTestPrep
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Re: If x and y are prime numbers and x ≠ y, how many unique factors does [#permalink] New post  24 May 2018, 13:28
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Bunuel wrote:
If x and y are prime numbers and x ≠ y, how many unique factors does the product x^2*y have?

A. 3
B. 4
C. 6
D. 8
E. 9


We can determine the number of factors of a number by breaking it to primes and adding 1 to each exponent and then multiplying those results.

So the number of unique factors of x^2*y^1 is (2 + 1)(1 + 1) = 3 x 2 = 6.

Answer: C
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Re: If x and y are prime numbers and x ≠ y, how many unique factors does [#permalink] New post  24 May 2018, 22:04
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For those who don't recall the formulae for the total number of factors \(TNF=(a+1)(b+1)(c+1)...\), we can take the values \(x=2\) and \(y=3\), and check.

\(x^2*y=2^2*3=12\)

\(12\) has the following unique factors: \(1,2,3,4,6,12\).

Answer: C
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Re: If x and y are prime numbers and x ≠ y, how many unique factors does [#permalink] New post  26 May 2018, 00:07
Bunuel wrote:
If x and y are prime numbers and x ≠ y, how many unique factors does the product x^2*y have?

A. 3
B. 4
C. 6
D. 8
E. 9

But how does it work when the numbers are 3 and 5. 45 has got no 6 unique factors :(
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Bunuel
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Re: If x and y are prime numbers and x ≠ y, how many unique factors does [#permalink] New post  26 May 2018, 00:10
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vishakha23 wrote:
Bunuel wrote:
If x and y are prime numbers and x ≠ y, how many unique factors does the product x^2*y have?

A. 3
B. 4
C. 6
D. 8
E. 9

But how does it work when the numbers are 3 and 5. 45 has got no 6 unique factors :(


Yes, it does: 1 | 3 | 5 | 9 | 15 | 45 (6 divisors)
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Re: If x and y are prime numbers and x ≠ y, how many unique factors does [#permalink] New post  26 May 2018, 00:14
Bunuel wrote:
vishakha23 wrote:
Bunuel wrote:
If x and y are prime numbers and x ≠ y, how many unique factors does the product x^2*y have?

A. 3
B. 4
C. 6
D. 8
E. 9

But how does it work when the numbers are 3 and 5. 45 has got no 6 unique factors :(


Yes, it does: 1 | 3 | 5 | 9 | 15 | 45 (6 divisors)

Ohh yes ! Actually prime factorization was all that I did and thought those factors as the only unique factors :(
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Re: If x and y are prime numbers and x ≠ y, how many unique factors does [#permalink] New post  26 May 2018, 06:50
Given: x and y are prime numbers; x = y

To find: Unique factors of (x^2) * y

Total number of factors of a number N = (p+1)(q+1)
Where, N = (a^p)*(b^q) , a and b are prime factors; p and q are powers of the prime factors.

Therefore, total number of factors of (x^2) * y = (2+1)*(1+1) = 3 * 2 = 6

Correct answer= C
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Jayiet89
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Re: If x and y are prime numbers and x ≠ y, how many unique factors does [#permalink] New post  07 Jun 2018, 10:58
Number of Factors of prime number can be expressed as product of a^p*b^q = (P+1)*(Q+1)
Similarly in X^2*y have unique factor which can be expressed as (2+1)*(1+1) =6
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Re: If x and y are prime numbers and x ≠ y, how many unique factors does [#permalink] New post  07 Jun 2018, 11:39
Bunuel wrote:
If x and y are prime numbers and x ≠ y, how many unique factors does the product x^2*y have?

A. 3
B. 4
C. 6
D. 8
E. 9


Given, \(x\) & \(y\) are Prime numbers.

Hence the product \(x^2*y\), will have \((2+1)*(1+1)\) = 6 unique factors.


Answer C.

Thanks,
GyM
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Re: If x and y are prime numbers and x ≠ y, how many unique factors does [#permalink] New post  12 Jun 2018, 06:37
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my approach is the following to find nu of distinct primes x^2*y*2 , if x is prime, it must have 2 factors, when x^2, it will have 3 factors and prime y has 2 factors. in total 2*3=6
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Re: If x and y are prime numbers and x ≠ y, how many unique factors does [#permalink] New post  18 Jun 2020, 03:00
Tulkin987 wrote:
For those who don't recall the formulae for the total number of factors \(TNF=(a+1)(b+1)(c+1)...\), we can take the values \(x=2\) and \(y=3\), and check.

\(x^2*y=2^2*3=12\)

\(12\) has the following unique factors: \(1,2,3,4,6,12\).

Answer: C



No need for such huge calculations.

Tip 01 : Square of any prime number has 3 unique factors so for Xsqaure -----> 3 factors
Tip 02 : Unique factor for any prime number is 2 i.e itself and 1

So finally x2.y = 3*2 =6
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