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I had asked for a hint here USAMO problem hint. i had tried induction once but i thought it won't work so i left it, but after seeing @lulu 's comment, i decided to give it a go again. Please see if my solution is correct.

[USAMO 2003] Prove that for every positive integer n there exists an n-digit number divisible by $5^n$ all of whose digits are odd.

MY SOLUTION: So first thing, i checked some small cases and figured we could generate the number with (n+1) digits satisfying the property by adding a number to it's front, ie. adding b 10$^n$ to the number with n digits.

We'll proceed with induction, let P(n) mean there exists an n-digit number divisible by $5^n$ all of whose digits are odd.

P(1) is true as 5|5.

Let P(k) be true, ie. let 5$^k$ | $a_ka_{k-1}...a_1$ with $a_i$ $\neq$ 2l for i $\in$ {1,2...k} .

I'll attempt to prove that by adding $ b \cdot 10^k $ with $ b \in {1,3,5,7,9} $ . we can have a number that is divisible by $5^{k+1}$.

So we want 5$^{k+1}$ $|$ $ b \cdot 10^k $ + $a_ka_{k-1}...a_1$ . --> eq.1

Let $a_ka_{k-1}...a_1$ = $5^km $

So from eq. 1, inputting $a_ka_{k-1}...a_1$ = $5^km $ , we will get

5$^{k+1}$ $|$ $ b \cdot 10^k $ + $5^k$m, then dividing throught by $5^k$ , we need

5 $|$ $2^k \cdot b + m$

as $ b \in {1,3,5,7,9} $ , $\equiv$ 0,1,2,3,4 (mod 5)

So now m $\equiv$ 0,1,2,3,4 (mod 5) , let m $\equiv$ r (mod 5),

We need $2^k \cdot b + r =0 (mod 5)$

now,$2^k \equiv$ 1,2,3,4 (mod 5)

so painstakingly going through each and every case of possible values of $2^k$ and m (mod 5) (there are 16 cases), we prove that we can find a $ b \in {1,3,5,7,9} $ such that 5 $|$ $2^k \cdot b + m$ .

This is the first time i have written so much in latex, so i am sorry if there is any mistake.

If you were a grader, out of 7, how many points would you give me?

Bill Dubuque
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Aditya_math
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2 Answers2

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I don't think math.se answers can address how they'd mark it, but I can advise on a cleaner way to write answers, because your ideas are right but they could do with algebraic clarity, and clarity regarding modulo arithmetic. (If you find yourself claiming that if we do something we eventually get a particular outcome, try to state this as an existence theorem that's either obvious, well-known or proven within your work.) To wit:

We claim some sequence $a_n$ of $n$-digit numbers in base $10$, all digits odd, satisfies $5^n|a_n,\,10^n|a_{n+1}-a_n$. In particular write $a_n=5^nb_n,\,a_{n+1}=a_n+10^nc_n$, so $b_1=1$ (because $a_1=5$) and$$5^{n+1}b_{n+1}=a_{n+1}=c_n10^n+5^nb_n\iff5b_{n+1}=c_n2^n+b_n,$$so it suffices to choose $c_n\in\{1,\,3,\,5,\,7,\,9\}$ with $5|c_n2^n+b_n$. This choice is possible because these $5$ choices of the $c_n$ each achieve a different residue class modulo $5$ (because $5\nmid k2^n$ for $k\in\{2,\,4,\,6,\,8\}$), and exactly one achieves $5|c_n2^n+b_n$.

J.G.
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  • thanks, so should i work more on my proof writing skills, like is it necessary to write a really well polished solution such as yours. Actually i dont have much experience with proof writing, so is there a specific way we need to write our proofs in, and if there is, do you know where i can learn that, thank you – Aditya_math Nov 22 '20 at 19:57
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    @Aditya_math I wish I could remember which chapter of which book I wanted to recommend. You can start with guides like these, but be warned some of their advice is about typesetting, instead of what will matter in your handwritten exploits. What may help the most is to read a lot of proofs to see how they articulate an idea clearly. – J.G. Nov 22 '20 at 20:06
  • @Aditya The argument after "this choice is possible" is a special case of a general criterion for solvability of linear congreunces. You should invoke this criterion instead of repeating its proof inline as above - see my answer. This criterion (and related) results are fundamental so you should be sure to know them well. – Bill Dubuque Nov 23 '20 at 11:00
  • @BillDubuque are you reffering to the last three lines? – Aditya_math Nov 23 '20 at 14:15
  • @Aditya_math Yes. – Bill Dubuque Nov 23 '20 at 14:53
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There is a $\,x\in\Bbb Z\,$ with $\,5\mid 2^k x - m\!\iff\! \bmod 5\!:\ 2^k x \equiv m\,$ has a root $\,x.\,$ By general theory

$$\begin{align} \color{#c00}c\ x &\equiv \, d\!\!\pmod{\!\color{#0a0}n}\ \ \text{has a root}\ x\!\iff\! \gcd(\color{#c00}c,\,\color{#0a0}n)\mid d\\[.3em] {\rm thus}\ \ \color{#c00}{2^k} x&\equiv m\!\!\!\pmod{\!\color{#0a0}5}\ \ \text{has a root}\ x,\ \, {\rm by}\ \ \gcd(\color{#c00}{2^k},\color{#0a0}5)\!=\!1\end{align}\qquad$$

and we can choose a root $\,x\in \{1, 3, 5, 7, 9\}\,$ since it is a complete residue system $\!\bmod 5;\,$ alternatively: $ $ if $\,0\le x < 5\,$ is even then $\,x':= x\!+\!5\,$ is odd $< 10,\,$ and $\,x'$ remains a root by $\,x'\equiv x\pmod{\!5}$.

Bill Dubuque
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  • Thanks, i knew what this says but i never knew it had a name. So it is essentially like finding an inverse modulo $2^k$ . Also, i have one question, if sometimes, in a contest i don't remember the name of a theorem/ can't frame it, would it be ok to do the case bash? although unpleasent, does it work? – Aditya_math Nov 23 '20 at 14:24
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    @Aditya If you know modular inverses then we can write $\bmod 5!:\ x\equiv m(2^{-k}) \equiv m(3^k),$ by $,2^{-1}\equiv 3.,$ But this problem only requires the existence of a solution, not its value. Of course it is always correct to repeat the proof of a theorem in a special case application, even though that is not optimal. – Bill Dubuque Nov 23 '20 at 14:57