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

Question

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

EgmatQuantExpert · Accepted Answer

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

ScottTargetTestPrep · Answer

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

Tulkin987 · Answer

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

vishakha23 · Answer

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

Bunuel · Answer

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

vishakha23 · Answer

Ohh yes ! Actually prime factorization was all that I did and thought those factors as the only unique factors

PP777 · Answer

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

Jayiet89 · Answer

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

GyMrAT · Answer

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

nigina93 · Answer

С
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

Ved22 · Answer

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

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

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