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Math Help - Number-theoretic functions

  1. #1
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    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.

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

    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?
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  2. #2
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    Quote Originally Posted by glowplug19 View Post
    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

    one-to-one and multiplicative, then your answer would be correct.
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  3. #3
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    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?
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  4. #4
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    Quote Originally Posted by glowplug19 View Post
    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.
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