In particular, $f$ is defined such that $f(mn)=f(m)+f(n)$, where $m$ and $n$ are positive integers. Also $f(n)\ge 0$ for all $n$, $f(10)=0$, and $f(n)=0$ if $n$ ends in a $3$. Thank you.
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Not true, define $f(n)=\sigma_2(n)-\sigma_5(n)$ where $\sigma_p(n)$ denotes the highest power of $p$ that divides $n$. – Did Oct 02 '16 at 17:11
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Please would you clarify your response? Thank you – John Smith Oct 02 '16 at 17:17
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Did's function has all the properties required in your question: f(mn)=f(m)+f(n) because the number of 2s (or 5s) dividing mn is the number of 2s (or 5s) dividing m plus the number of 2s (or 5s) dividing n. Also f(10)=1-1=0 and if n ends in a 3 then it is not divisible by 2 or by 5, hence f(n)=0-0=0. – Embarassed Guy Oct 02 '16 at 17:23
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Oh, now the question has been changed by someone. The requirement $f(n)\geq 0$ wasn't there before. – Embarassed Guy Oct 02 '16 at 17:24
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@EmbarassedGuy Oops, that embarasses me - it means I need to corect my proof ... – Hagen von Eitzen Oct 02 '16 at 17:27
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Thank you. Did "not true" mean that actually f(n) wasn't always equal to 0? – John Smith Oct 02 '16 at 17:31
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@JohnSmith: Exactly, For Did's example one has e.g. f(2)=1 and f(5)=-1. – Embarassed Guy Oct 02 '16 at 17:33
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My apologies, thank you everyone for your help. – John Smith Oct 02 '16 at 17:34
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OP: Next time you modify crucially the text of a question while people are discussing it, at least mention the modification through a comment... – Did Oct 02 '16 at 17:38
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- If $n\equiv 1\pmod{10}$ then $3n\equiv 3\pmod{10}$ and so $f(n)=f(3n)-f(3)=0$.
- If $n\equiv 7\pmod{10}$ then $3n\equiv 1\pmod{10}$ and so $f(n)=f(3n)-f(3)=0$.
- If $n\equiv 9\pmod{10}$ then $3n\equiv 7\pmod{10}$ and so $f(n)=f(3n)-f(3)=0$.
From $f(2)\ge 0$ and $f(5)\ge 0$ and $f(2)+f(5)=f(10)=0$, we obtain $f(2)=f(5)=0$.
Now let $n\in\Bbb N$ be arbitrary. Write $n=2^r5^su$ with $\gcd(u,10)=1$. Then $u\equiv 1,3,7,9\pmod{10}$ and hence $f(n)=rf(2)+sf(5)+f(u)=0$.
Hagen von Eitzen
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