# Math Help - Number-theoretic functions

1. ## Number-theoretic functions

If F(n) is a number-theoretic function, and therefore multiplicative (?), with
$F(35)=161$ and $F(77)=437$, find $F(55)$.

This was a test question, and I don't know if thats all the info. We have not gone over this much, so a good explanation of how to evaluate this would be great.

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I tried reducing everything to prime factors:

Because F(n) is multiplicative, $F(35)=F(5\cdot7)=F(5)\cdot F(7)=161$, where $161=7\cdot23$.
Also by the same argument, $F(77)=F(7\cdot11)=F(7)\cdot F(11)=437$, where $437=19\cdot23$.

Just looking at it, i.e. ruling out the common factor 7, I got:

$F(7)=23, F(5)=7, F(11)=19$ $\Longrightarrow F(55)=F(5\cdot11)=F(5)\cdot F(11)=(7)\cdot(19)=133$.

Is this right? I relied more on intuition than real number theory, so if it is right, how?

2. Originally Posted by glowplug19
If F(n) is a number-theoretic function, and therefore multiplicative (?), with
$F(35)=161$ and $F(77)=437$, find $F(55)$.

This was a test question, and I don't know if thats all the info. We have not gone over this much, so a good explanation of how to evaluate this would be great.

=================

I tried reducing everything to prime factors:

Because F(n) is multiplicative, $F(35)=F(5\cdot7)=F(5)\cdot F(7)=161$, where $161=7\cdot23$.
Also by the same argument, $F(77)=F(7\cdot11)=F(7)\cdot F(11)=437$, where $437=19\cdot23$.

Just looking at it, i.e. ruling out the common factor 7, I got:

$F(7)=23, F(5)=7, F(11)=19$ $\Longrightarrow F(55)=F(5\cdot11)=F(5)\cdot F(11)=(7)\cdot(19)=133$.

Is this right? I relied more on intuition than real number theory, so if it is right, how?
the question hasn't given us enough information. for example, a number theoretic function is not necessarily multiplicative. also, the range of a number theoretic function doesn't even have to

be integers, it could be real or complex numbers. another thing is that a number theoretic function doesn't have to be one-to-one! so if we assume in the above problem that $F:\mathbb{N} \longrightarrow \mathbb{N}$ is

3. Thanks for the help.

Let us assume it is multiplicative, and the range is $\mathbb{N}$.

How does the one-to-one come into play for this problem?

If it is correct, following these assumptions, can someone prove it?

4. Originally Posted by glowplug19
Thanks for the help.

Let us assume it is multiplicative, and the range is $\mathbb{N}$.

How does the one-to-one come into play for this problem?

If it is correct, following these assumptions, can someone prove it?
if $F$ is multiplicative, then $F(1)=1.$ so if $F$ is one-to-one, then $F(7), F(5),$ and $F(11)$ are all $\neq 1.$ therefore from $F(5)F(7)=7 \times 23,$ we get $F(7) \in \{7,23 \}$ and from $F(7)F(11)=19 \times 23,$

we get $F(7) \in \{19,23 \}.$ thus $F(7)=23.$ the rest is clear.