Thank you, Bunuel for the clarifications. Just one question: The formula to find the number of factors returns only to positive factors, right? If negative prime factors would be allowed, then we would need to take them into account as well, wouldn't we? Is there a formula to find both positive and negative factors at the same time?
shalva wrote:
IMO it's (A)
From Statement 1:
n = a * a * a * a * b * b * b
n has 13 factors: 12 different combinations of a & b + 1.
Statement 2 tells nothing: n could have only 2 prime factors but what about non-prime factors?! we should consider them too. f.e. 35, 25, 49 and so on.
The answer is (A), but n has (4+1)(3+1)=20 factors, including 1 and n itself, not 13.
Finding the Number of Factors of an IntegerFirst 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.
In original question \(n =a^4*b^3\), so number of factors =(4+1)(3+1)=20.