# There is an ambiguity in wording that I do not understand for a factor problem.

The problem is as follows

If $A = 9n$ and $B = 8n$ where $n$ is a positive integer, which one has a greater number of distinct prime factors ?

According to the answer, we cannot tell. However, I am not totally convinced because I don't know exactly how many distinct prime factors there are in the first place.

For $A$, I am not sure if I am supposed to think 3 as one prime factor or $3*3$ as two factors. I want to say that there is only one prime factor that has to do with $9$.

I understand that $n$ can have multiple prime factors, but I don't think that matters because $B$ also has $n$ in it, so I disregarded it (I am suspecting that this is the reason I got it wrong).

For $B$, if $8 = 2*2*2$ considered as having one distinct prime factor, then the number of distinct prime factor would be one, thus I concluded that $A$ and $B$ has the same number of distinct prime factors...

What am I missing here ?

• "For $A,$ I am not sure if I am supposed to think 3 as one prime factor or 3∗3 as two factors." The word "distinct" means that 3 is counted only once as a factor. For example, the distinct prime factors of 9*60 are 2, 3, and 5. The distinct prime factors of 8*60 are also 2, 3, and 5. The distinct prime factors of 9*21 are 3 and 7, while the distinct prime factors of 8*21 are 2, 3, and 7. – Will Orrick Aug 19 '13 at 5:13
• Thanks for the clarification. – hyg17 Aug 19 '13 at 15:08

If $n$ has a factor of $2$ but no factor of $3$, then $8n$ has the same number of distinct prime factors as $n$, and $9n$ has one more, namely, $3$. If $n$ has a factor of $3$ but no factor of $2$, the situation is exactly reversed. And if $n$ has both a factor of $2$ and a factor of $3$, or if it has neither $2$ nor $3$ as a prime factor, then $9n$ and $8n$ have the same number of distinct prime factors.
In a little more detail, suppose that $n$ has $k$ distinct prime factors. If $2\mid n$, then $8n$ also has $k$ distinct prime factors, but if $2\nmid n$, then $8n$ has $k+1$ distinct prime factors: it has $2$ as a prime factor in addition to the ones that $n$ already had. Similarly, if $3\mid n$, then $9n$ has $k$ distinct prime factors, but if $3\nmid n$, then $9n$ has $k+1$ distinct prime factors. Thus,
• $8n$ and $9n$ both have $k$ distinct prime factors if $2\mid n$ and $3\mid n$;
• $8n$ and $9n$ both have $k+1$ distinct prime factors if $2\nmid n$ and $3\nmid n$;
• $8n$ has $k$ distinct prime factors and $9n$ has $k+1$ if $2\mid n$ and $3\nmid n$; and
• $8n$ has $k+1$ distinct prime factors and $9n$ has $k$ if $2\nmid n$ and $3\mid n$.