Question

If f(a^b) = b + 1 and f(a^b*c^d) = (b+1)(d+1), can we conclude that f(a^b*c^d*e^f) = (b+1)(d+1)(f+1)?
(assuming a, c, and e aren't all equal)

Answer 1

maybe I should explain what I mean a little better...

You can take a number, say 16, and break it down into its prime factors.
For 16 this would be: 2 * 2 * 2 * 2
If the the factors can be expressed as a^b (2^4 for 16), then the number of factors that number has is b + 1. The factors of 16 are 1, 2, 4, 8, and 16. As you can see, there are 5 or (4 + 1) of them, so the rule checks out.

I've seen a proof of this for numbers that can be expressed as a^b and a^b*c^d, but i'm not sure how to prove that you could extend it

Answer 2

I see! That makes the problem more clear. Let me think about it for a bit more.

Answer 3

here's a link to an explanation of the proof of the first two cases http://mathschallenge.net/library/number/number_of_divisors

Answer 4

I see...well here's the first bit of news: if just given your first post. The answer is no you can't necessarily conclude that. Here's why:
First here are our three equations (I will refer to them by their designated number):
\[1:f(a^{b}) = b+1\]
\[2:f(a^{b} \times c^{d}) = (b+1)(d+1)\]
\[3:f(a^{b} \times c^{d} \times e^{f}) = (b+1)(d+1)(f+1)\]
For us to conclude that f(a^b*c^d*e^f) = (b+1)(d+1)(f+1), based on what we are given, we must assume that a,b,c,d,e,f can be anything aka their equations hold true for all real numbers, but a,c,e aren’t equal to one another. Does this make sense?

And so since that must be true, if we can prove it to have one instance of being false, we can prove that we can’t conclusively state that f(a^b*c^d*e^f) = (b+1)(d+1)(f+1).

To do so, we can take advantage of the fact that we can actually turn the second equation into the last, by taking advantage of the property if c*e (from the third equation) equals c from the second. This would mean d and f(from the third equation) would be the same as d from the second.

This means that \[(b+1)(d+1) = (b+1)(d+1)(d+1)

This is not true for all values of a,b,c,d,e,f. In fact only if d = f = 0 will this work out which is essentially the same as the first one.

In order to prove your question we actually must take into account that we are talking about factors and counting them. Which honestly is a much harder question, and I am unsure whether I’ll be able to solve it.

Answer 5

*by first one I mean the first equation.

Answer 6

actually no it's actually quite easy now that I think about it to prove it, if you are talking about the number of factors. It's really just a probability problem.

Answer 7

if we assume that n = a^b*c^d*e^f then to find the number of divisors what we are really saying is how many ways to put "b" number of a's, "d" number of c's, and "f" number of e's into a box, where you can leave things out of the box. Since that's what finding divisors of something really is right? And so thinking of it that way, it becomes clear that it's really just (b+1)(d+1)(f+1). Does this make sense?

Answer 8

@freckles since you are looking at this, does this make sense?

Answer 9

I think landon mean to have another condition...
because \[factors_+(4^3)=\left\{ 1,2,4,8,16,32,64\right\} \\ \text{ so } |factors_+(4^3)|=7=6+1 \neq 4=3+1\]

Answer 10

Yes I think he meant to say that a,c,e must be prime.