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Problem: Prove that for every odd integer $n$, exist a multiple $m$ of $n$ whose decimal representation entirely consists of odd digits.

My work:

+) For all $n:(n,5)=1$, $10^{\varphi (n)}-1$ works

+) If we can prove the statement for all $5^k$,assume we have $s5^k$ have m digits which are all odd

Then $n=5^kt$,

$\overline{(s5^k)(s5^k)(s5^k)...(s5^k)(s5^k)} \times \frac{10^{\, mn\varphi (n)}\, \, \, \, \, \, \, -1}{10^{\, m\varphi (n)}\,\,\,\, \, \, -1}$ ,($\varphi(n)$ time $s5^k$)

=$\overline{(s5^k)(s5^k)(s5^k)...(s5^k)(s5^k)} \times \overline{1000..0100..0100...001}$, (the 0's has $m\varphi(n)-1$ digit 0)

=$\overline{(s5^k)(s5^k)(s5^k)...(s5^k)(s5^k)}$ works

Because :

+)$\frac{10^{\,\,ab}\,\,\,\,-1}{10^{\,\,b}\,\,\,\,-1}=\overline{100..0100..0100...0100..01}\,\,$ have a digits 1 and the 0's has b-1 digits 0

+) (Every number t has b digit) $\times \frac{10^{\,\,ab}\,\,-1}{10^{\,\,b}-1}\,$= $t \times \overline{100..0100..0100...0100..01}=$ $\overline{ttt...t}$

+)For every $p|n, p\neq 5,$follow LTE:

$v_{p}(\frac{10^{\,\, mn\varphi (n)}\,\,\,\,\,\,\,\, -1}{10^{\,\, m\varphi (n)} \,\,\,\,\,\,-1}\,\,)=v_p(10^{\varphi(n)}-1)+v_p(mn)-v_p(10^{\varphi(n)}-1)-v_p(m)=v_p(n)$

So if we can do with $5^k$,we are done

Someone help me ? Thanks :)

Bill Dubuque
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hoang
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    I do not understand the argument that we would be done if we could show it for $5^k$ : The product of numbers with only odd digts need not be a number with only odd digits, so how do you justify this implication ? – Peter Nov 29 '21 at 10:04
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    For every number upto $10^7$, such a multiple exists. I checked this simply with brute force using PARI/GP, but I found no useful pattern allowing a general proof. – Peter Nov 29 '21 at 10:55
  • @Peter You are correct : I tried, but I can't figure out how the second + point works. – Sarvesh Ravichandran Iyer Nov 29 '21 at 17:04
  • Yes, there is a typo, it's my bad. i've fixed, can you check it ? – hoang Nov 30 '21 at 01:38

1 Answers1

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First I claim there is an $n$ digit number whose digits are odd and which is divisible by $5^n$. It’s true for $n=1$. Assume it’s true for $n-1$. That number is $u \cdot 5^{n-1}$ for some integer $u \equiv 0,1,2,3,4 \bmod 5$. Now $10^{n-1}$ is also $v \cdot 5^{n-1}$ for some $v \equiv 1,2,3,4 \bmod 5,$ so by adding a first digit which is $1,3,5,7,9$ the new number is the same power of $5$ times

$$u+v, u+3v, u+5v, u +7v, u+9v \bmod 5,$$

And exactly one of these will be divisible by $5$, and so one is done.

This produces a number $m$ with $n$ digits divisible by $5^n$. The number $mmmmmmmmm\ldots$ Is then

$$m \cdot (10^{kn}-1)/(10^n-1),$$

And then taking $kn$ divisible by $\varphi((10^n-1)r)$ we can for $(r,10)=1$ we can get divisiblity by $5^n \cdot r$ by a number with only odd digits.

user989894
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