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In a shop there are infinitely many cards with exactly one natural number written on each of them. Suppose for any $n \geq 1$ there are exactly $n$ cards hosting a number $d$ such that it's a divisor of $n$. For example if $n=6$ there are exactly six cards with number belonging to the set ${1,2,3,6}$. Prove that every natural number occurs at least once among the cards.

I found a solution here which proposes the formula that $g(ip)=g(i) \times g(p)$ where $i$ is an integer, $p$ is a prime, and $i,p$ are coprime. Here $g(k)=\text{number of cards with k written on it}$. They provided no proof of it. I tried to prove it with induction but failed. I know that for any prime $p$, $g(p)=p-1$, how do I prove thus formula?

Bill Dubuque
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math_learner
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    "Suppose for any $n \geq 1$ there are exactly one card hosting a number $d$ such that it's a divisor of $n$. For example if $n=6$ there are exactly six cards with number belonging to the set ${1,2,3,6}$."—This seems self-contradictory. Is there one such card or six? – Greg Martin Jan 29 '24 at 20:43
  • I don't see how the condition is possible. Given that $n$ could be $1$, there must be a card which shows $1$. But $1$ divides everything...so what could be on any of the others? – lulu Jan 29 '24 at 20:45
  • @GregMartin I'm very sorry for such a careless mistake, it should have been "there are exactly $n$ cards". Let me edit the post. Thank you! – math_learner Jan 29 '24 at 20:46
  • @BrianTung Good catch...that's very differemt. – lulu Jan 29 '24 at 20:46
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    To me the most natural solution (at least after some experimentation and understanding-gathering) is to prove that $g(k)=\phi(k)$ (the Euler phi function) for all $k$. – Greg Martin Jan 29 '24 at 21:19
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    @GregMartin I'm able to notice the similarity, so I was trying to create a bijection between the numbers less than $k$ that are coprime to $k$ and all the cases where $k$ would appear on a card, but haven't been successful yet. – math_learner Jan 29 '24 at 21:50
  • Am I missing something, or isn't it true that every natural number is a divisor of some $n \ge 1$, and hence if the constraint is true for all $n$, then all the natural numbers will be represented? I feel like I'm missing something. – Eric Snyder Jan 29 '24 at 23:47
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    I would prove $g(k)=\phi(k)$ by strong induction on $k$. The identity $\sum_{d\mid n} \phi(d)=n$ will come into play. – Greg Martin Jan 30 '24 at 03:22
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    @EricSnyder - there is no requirement that every divisor of $n$ must be among the $n$ cards with divisors of $n$. For example, the cards dividing $4$ could be a $1$-card and three $4$-cards, leaving $2$ out. (Except the cases for $n=1$ and $n=2$ force a $2$-card to exist.) – Paul Sinclair Jan 30 '24 at 21:23
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    Reading some proofs about the $\sum_{d\mid n}\phi(d)$ identity, the set $\left{\frac1n,\frac2n,\ldots, \frac nn\right}$ may help to create a bijection with the $n$ cards. – peterwhy Jan 30 '24 at 21:43
  • @PaulSinclair Ah thank you for the explanation. And now I see how the $n = 1,2$ cases and a bit more examination suggests (but doesn't prove) that $g(k) = \phi(k)$ may be what the question is seeking. – Eric Snyder Jan 31 '24 at 02:29

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