Define a function $P_b(x)$ as $$P_b(x)=\text{"the product of digits of x in base b"}$$
Is it true that for any $b\in\mathbb{N}$, except the powers of $3$, there exists at least one positive integer $n$ that satisfy $P_b(n)=n/3$?
For example, in the case of $b=16$, we have $n=24,54,2970$ because $$\begin{array}{l}&P_{16}(24)=1\cdot 8=24/3\\&P_{16}(54)=3\cdot6=54/3\\&P_{16}(2970)=11\cdot9\cdot10=2970/3\end{array}$$
I tried to find solutions for $b\leq 100000$. Most of them were found, but I haven't yet found for $b=16707,94779$; no solution found for $b=16707,n\leq16707^6$ and $b=94779,n\leq94779^5$.
What makes the problem difficult is that there are only a finite number of $n$ that can satisfy this. If $n$ is a $k$ digit number, $P_b(n)\leq(b-1)^k$ and $n\geq b^{k-1}$. So, $P_b(n)=n/3\implies(b-1)^k\geq\frac{b^{k-1}}{3}\implies k\leq\frac{\log{3}}{\log{\frac{b}{b-1}}}$.
My attempts:
- For $b\in 2\mathbb{N}$, we have two digit number $n=1\cdot b+b/2$, so this is true for any even $b$.
- For $b=3m-1,m\in\mathbb{N}$, we have two digit number $n=m\cdot b+x$.
- For $b=6m+1,m\in\mathbb{N}$, we have three digit number $n=1\cdot b^2+(4m+1)\cdot b+(5m+1)$.
- In the case of $b=3^k$, $P_b(n)=n/3\implies 3\mid n\implies$the last digit of $n$ in base $b$ is divisible by $3$$\implies 3\mid P_b(n)\implies9\mid n\implies$the last digit of $n$ is divisible by $9$$\implies\ldots\implies 3^k\mid n$. However, this implies that the last digit of $n$ is $0$, and $P_b(n)=0$.
This is my result of $b\leq100000$ whose digit is greater than $3$. $$\begin{array}{c|r|r} b&\text{the smallest }n&n\text{ in base }b\\\hline 111&61315056&44\ 92\ 51\ 99\\ 171&9989426304&11\ 116\ 136\ 164\ 117\\ 291&277687433520&38\ 210\ 229\ 201\ 252\\ 489&42981939000&367\ 286\ 364\ 375\\ 999&728487513000&730\ 676\ 675\ 729\\ 1371&15297465600&5\ 1283\ 690\ 1152\\ 2271&2148217774020&183\ 934\ 1913\ 2190\\ 2619&67678340184&3\ 2009\ 2236\ 1674\\ 2721&1807944879312000&32\ 2670\ 2145\ 1923\ 1710\\ 2979&184425289344&6\ 2907\ 1856\ 1899\\ 3279&63027219024&1\ 2582\ 3256\ 2499\\ 3891&579834488880&9\ 3279\ 1882\ 3480\\ 4887&758438820144&6\ 2434\ 3602\ 4806\\ 5211&3497208707631600&4\ 3870\ 4533\ 4733\ 3510\\ 5787&519014171052&2\ 3923\ 5083\ 4338\\ 7299&1494480881088&3\ 6154\ 7207\ 3744\\ 7731&6204334299715300896&1736\ 6267\ 6702\ 7114\ 3987\\ 11079&7359977315880&5\ 4566\ 9917\ 10836\\ 16707&>16707^6\text{ or not exist}\\ 16911&505775286533088&104\ 9812\ 14443\ 11439\\ 21249&197497042814880&20\ 12424\ 18273\ 14499\\ 23889&923571500757948&67\ 17794\ 22398\ 11529\\ 25059&27526222159200&1\ 18775\ 20322\ 24048\\ 25227&124239347654688&7\ 18632\ 16044\ 19791\\ 27891&258476067299736&11\ 25469\ 22074\ 13932\\ 30159&6650570175588900&242\ 13333\ 24826\ 27675\\ 32679&201107903798340&5\ 24922\ 29694\ 18117\\ 32871&10369855581776460&291\ 31790\ 16121\ 23178\\ 35031&341247650220540&7\ 32859\ 24494\ 20190\\ 35481&389195145360000&8\ 25306\ 21250\ 30156\\ 37299&707049339082596000&13625\ 25328\ 18982\ 35979\\ 37911&1030916303835264&18\ 34888\ 35892\ 15246\\ 42969&136292456717568&1\ 30848\ 36046\ 40857\\ 45219&356537315481336&3\ 38709\ 23194\ 44124\\ 47451&191176066968896711988&37\ 33671\ 34281\ 39943\ 37356\\ 52011&365305056780924&2\ 31018\ 46041\ 42633\\ 52167&4061220222560256&28\ 31653\ 42496\ 35943\\ 62649&18606111483304800&75\ 41859\ 45578\ 43344\\ 63243&180596621753119728&713\ 60541\ 61588\ 22644\\ 66699&1150322810219388&3\ 58474\ 50482\ 43299\\ 67311&6999295321933200&22\ 63993\ 31720\ 52245\\ 69039&87847167941186472&266\ 66196\ 40709\ 40851\\ 69177&633981494732400&1\ 63303\ 59380\ 56220\\ 70689&1311868845107948520&3713\ 66460\ 32009\ 55362\\ 75879&703506233044944&1\ 46307\ 70546\ 71784\\ 77391&1416831300621642240&3056\ 50878\ 46296\ 65610\\ 77547&34711476672464640&74\ 33744\ 63146\ 73380\\ 78759&27319621329237600&55\ 72528\ 46203\ 49410\\ 79491&6516703861972992&12\ 77424\ 58404\ 40032\\ 81819&21796802712158292&39\ 65059\ 38454\ 74466\\ 94779&>94779^5\text{ or not exist}\\ 98199&43456370315630400&45\ 87544\ 40435\ 90936\\ 98409&46188177128578368&48\ 45739\ 84977\ 82524\\ 98451&12091447235911680&12\ 66080\ 71344\ 71244\\ 99111&3438347292408211920&3531\ 69668\ 82205\ 56676\\ \end{array}$$ I would appreciate any help. Thank you.
Not a full solution, but I’ll add that
For $b = (3i - 1)j$ with $i, j \ge 1$ we have a two-digit number $n = (i)b + (ij)$.
This narrows the potential counterexamples $b$ to those that are a product of primes of the form $3i + 1$ or $3$. This set has zero asymptotic density.
For $b = 6i + 1$ with $i \ge 1$ we have a second three-digit number $n = (2i + 1)b^2 + (3i + 1)b + (4i + 1)$.
And here are some two-parameter three-digit families. Since $b$ is linear in $j$, these solve an entire congruence class for each fixed $i$:
For $b = (2i - 1)(3(3i - 1)j + 1)$ with $i, j \ge 1$, we have a three-digit number $n = (1)b^2 + ((3i - 1)^2 j + i)b + (2i - 1)((9i - 4)j + 1)$.
(Solves $b = 6j + 1, 45j + 3, 120j + 5, 231j + 7, \ldots$.)
For $b = (3i - 2)((3i - 1)j - 2i + 1)$ with $i \ge 2, j \ge 1$, we have a three-digit number $n = (1)b^2 + ((3i - 2)^2 j - 3(i - 1)(2i - 1))b + (2i - 1)((3i - 1)j - 2i + 1)$.
(Solves $b = 20j - 12, 56j - 35, 110j - 70, 182j - 117, \ldots$. Too bad the third digit is too big for $i = 1$—otherwise we’d get $b = 2j - 1$ and we’d be done!)
For $b = i((9i - 3)j - 6i + 1)$ with $i, j \ge 1$ (except $i = j = 1$) we have a three-digit number $n = (1)b^2 + ((3i - 1)^2 j - 3i(2i - 1))b + i((6i - 1)j - 4i)$.
(Solves $b = 6j - 5, 30j - 22, 72j - 51, 132j - 92, \ldots$.)