M31-32

Question

How many distinct prime factors does the positive integer  n  possess?

(1)  44 < n^2 < 99 
 
(2)  8n^2  has twelve factors

Bunuel · Accepted Answer

Official Solution:

How many distinct prime factors does the positive integer  n  possess? 
 
(1)  44 < n^2 < 99 . 
 
This statement implies that  n  can be 7, 8, or 9. Each of these numbers has 1 prime factor: 7, 2, and 3, respectively. Sufficient.
 
(2)  8n^2  has twelve factors. 
 
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  11+1 = 12 . In both cases ( n = prime  other than 2 or  n = 2^4 ),  n  has one prime factor. Sufficient.

Answer: D

anders112 · Answer

Hi,

I don't understand the explanation for why (2) is sufficient. Why must "n" be a prime, and how do you get "(3+1)(2+1)=12"?

Bunuel · Answer

Finding the Number of Factors of an Integer :

First make prime factorization of an integer  n=a^p*b^q*c^r , where  a ,  b , and  c  are prime factors of  n  and  p ,  q , and  r  are their powers.
 
The number of factors of  n  will be expressed by the formula  (p+1)(q+1)(r+1) .  NOTE:  this will include 1 and n itself.

Example:  Finding the number of all factors of 450:  450=2^1*3^2*5^2

Total number of factors of 450 including 1 and 450 itself is  (1+1)*(2+1)*(2+1)=2*3*3=18  factors.

ruggerkaz · Answer

Why must N be a prime number? I understand that N greater than anything but 3 will result in more than 12 factors. But, just wondering if there is a trick in the number of factors which tells you that N must be a prime number?

Thanks in advance!

Thanks in advance!

Bunuel · Answer

If n is not a prime then 2^3*n^2 will have more than 12 factors. For example, if n is say 15, then the number of factors of 2^3*n^2 = 2^3*(3*5) will be (3+1)(1+1)(1+1)=16. Only if n is prime 2^3*n^2 will have 12 factors. Check for more here: m31-199943.html#p1599885

ruggerkaz · Answer

Is it fair to say the square of all prime numbers have 3 factors?

17- It has 2 factors: 1 & 17
(17)^2- 3 factors since (2+1)=3

Bunuel · Answer

Yes, the square of a prime has 3 factors: (prime)^2 has (2+1)=3 factors. Check here: m31-199943.html#p1599885

Chets25 · Answer

I think this is a  high-quality  question and the explanation isn't clear enough, please elaborate. Can you please explain, how we got the conclusion that 'n' will be a prime number if the number of factors is 12.

Bunuel · Answer

If n is not a prime then 2^3*n^2 will have more than 12 factors. For example, if n is say 15, then the number of factors of 2^3*n^2 = 2^3*(3*5) will be (3+1)(1+1)(1+1)=16. Only if n is prime 2^3*n^2 will have 12 factors.

dhruv solanki · Answer

hi,
the question ask for prime factors so for 7 it will be 1 for 8 it will be 3 as in (2^3) and 9 it will be (3^2).
i am not able to comprehend properly as the question asks number of prime factors and not the prime numbers it is constituted of..please explain..

Bunuel · Answer

7 has only one prime factor, which is 7.
8 has only one prime factor, which is 2.
9 has only one prime factor, which is 3.

But for example, 36 has two primes: 2, and 3.

danielarasan · Answer

Beautiful Question!

Manku · Answer

Bunuel

I have a confusion: The question categorically asks the number of P. F and not total number of factors.
If 8n^2 has 12 factors then we can have following two possibilities:

Case 1- When n is a Prime number other than 2 - 2^3 *n^2, so we have (3+1)*(2+1)= 12 factors , in all n has 2 P.F

Case 2 - When n is not Prime and equals 16 or 2^4, so we have 2^3*2^8= 2^11= 12 factors total, in all n has 1 PF and i.e. 2 , so here we get two different answer so this statement should be insufficient

Bunuel · Answer

In both cases n has only 1 prime. 
In case 1: n = prime other than 2 --> n has 1 prime factor.
In case 1: n = 2^4 --> n has 1 prime factor.

coreyander · Answer

I think this is a  high-quality  question and I agree with explanation.

akt715 · Answer

44<n2<99 on taking square root on both sides will result into -7 < n < 9 (approx) . Since n is positive 
0<n<9. Correct me if I am wrong??

Bunuel · Answer

No.

Taking the square root from  44<n^2<99  gives:  \sqrt{44}<|n|<\sqrt{99}  (recall that  \sqrt{x^2}=|x| ), which gives:  -9.9<n<-6.6  or  6.6<n<9.9 . Since n is positive, then we'd get  6.6<n<9.9  and since n is an integer we get n = 7, 8, or 9.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

Ontherox · Answer

In statement 2: 
If (2^3)*(n^2) - and n is a prime number, it has 2 distinct prime numbers.
If 2^11 - there's only one distinct prime number.

How can we conclusively determine the number of distinct prime numbers?

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