If 8n^2 has twelve positive factors, how many distinct prime factors

Question

If  8n^2  has twelve positive factors, how many distinct prime factors does the positive integer  n  have?

A. None
B. 1
C. 2
D. 3
E. Cannot be determined from the given information

Bunuel · Accepted Answer

Official Solution:

If  8n^2  has twelve positive factors, how many distinct prime factors does the positive integer  n  have?

A. None
B. 1
C. 2
D. 3
E. Cannot be determined from the given information

For  8n^2 = 2^3n^2  to have twelve factors,  n  must be a prime other than 2:  8n^2 = 2^3 * (prime)^2 . In this case, the number of factors can be calculated as  (3+1)(2+1) = 12 .

Another possibility is when  n = 2^4 :  8n^2 = 2^3 * (2^4)^2 = 2^{11} . In this case, the number of factors is also  11+1 = 12 .

In both cases ( n = prime  other than 2 or  n = 2^4 ),  n  has one prime factor.

Answer: B

gmatophobia · Answer

We can represent 12 in the following ways:

12 = 12 * 1 ; 3 * 4 ; 4 * 3

Case 1  : n = 2

8n^2 = 2^3 * n^2 = 2 ^{11}

The number of prime factors of n= 1

Case 2  : n = any other prime number

8n^2 = 2^3 * n^2

The number of prime factors of n = 1

Option B

Nidzo · Answer

To find the number of positive factors of a number: one needs to first prime factorise the number, then add one to every exponent followed by multiplying these values with one another.

We are told that  8n^2  has 12 factors. Although we cannot prime factorise  n^2 ,  8  can be written as  2^3 . Making use of the above mentioned rule to find the number of positive factors, we take the exponent of  2^3  and add 1 to the exponent:  (3+1)= 4 .

The factor pairs for 12 are as follows:  (12*1) ,  (6*2)  and  (4*3) .

Using these pairs with the  4  we already have from  2^3 :

(6*2)  : This option cannot work. It requires  n^2  to have the prime factorisation of  2^2  and a single prime number as then one would have  2^5 * x^1  (where x is an prime number). However, given that we have  n^2  all prime factors will have even numbered exponents.

(4*3) :  This option works as it requires  n^2  to the square of any single prime number. Let  n = 3 , which would then make  8n^2 = 2^3 * 3^2 . In this case the amount of positive factors will be  (3+1)(2+1) = 12 .

In this case  3  will have a single prime factor.

(12*1) :  In this case, we have  n^2 = 2^8 . Plugging that into  8n^2 :  2^3*2^8 = 2^{11} . Which means that we have  (11+1) = 12  positive factors for  8n^2  and therefore,  n  has a single prime factor.

ANSWER B

Rohstar750 · Answer

I added 2 as well in the count and selected 2. Silly mistake

abhilashsingh · Answer

for 8n2 = 2^3 * n^2

total factors => 12 = (3+1) * ((prime factor of n) + 1) => n must be a prime number for the equation to hold

hence B

ribbons · Answer

I have this doubt only, why should we not count the 2 ?

Bunuel · Answer

If n = 2, then  8n^2 = 8*2^2 = 2^3*2^2=2^5 , which has 6 factors not 12.

ribbons · Answer

No i understand why n isnt 2, my question is in the below, here n is assumed to be some other prime factor (not 2) then doesnt it have 2 distinct prime factors 2 and n ?
Case 2: n = any other prime number

8n2=23∗n28𝑛2=23∗𝑛2

The number of prime factors of n = 1

Bunuel · Answer

The question asks how many distinct prime factors does n have, not 8n^2. So, if n = any other prime number, then n will have only that prime, not that prime and 2.

Does this make sense?

bumpbot · Answer

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If 8n^2 has twelve positive factors, how many distinct prime factors

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If 8n^2 has twelve positive factors, how many distinct prime factors

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