# The construction of a number with given digits [closed]

Given a set of decimal digits. And given a set of primes $\mathbb{P}$, find some $p \in \mathbb{P}$ such that $p^n, n \in \mathbb{N}$ contained in itself all the digites from a given set, and it does not matter in what order.

## closed as unclear what you're asking by B. Mehta, Hagen von Eitzen, Delta-u, J.R., jvdhooftMay 25 '18 at 23:51

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• You mean, you are given something like 0,1,4,5,6,7,8 and a possible answer the algotithm should find might be $7^8 = 5764801$? – Hagen von Eitzen May 25 '18 at 13:22
• Yep.............. – Vladislav Kharlamov May 25 '18 at 13:23
• Moreover, an algorithm is desirable that constructs at least a finite set of such examples, not the cardinality of the set 1 – Vladislav Kharlamov May 25 '18 at 13:24
• The longer you explain in reply to comments, the less clear the problem statement gets – Hagen von Eitzen May 25 '18 at 14:36
• The lesson is to think clearly about what you are asking. In the spirit of my answer, any large power of a prime will have lots of digits so will almost certainly include at least one of each (but $2^{99}$ is missing $9$). For any $p$ you have very good odds that $p^{10000}$ will satisfy your need. – Ross Millikan May 25 '18 at 15:04

$2^{100}=1267650600228229401496703205376$ includes all the digits so it will answer your request for any set.

For any given prime it is believed that there is a highest power that does not contain all the digits, so if your list of primes doesn't include $2$ I would just raise one of them to the $1000$ power and expect it to be good. Proving that there is a highest power seems to be hard.

• Is this the first power of $2$ that has all ten digits? – Barry Cipra May 25 '18 at 14:59
• @BarryCipra: I doubt it. I started with powers around $50$ and found at least one digit missing, then jumped to $2^{99}$ which is missing $9$. I was just computing in Alpha and looking. Next I tried $100$ and found them all. – Ross Millikan May 25 '18 at 15:01
• I wonder if there is a largest power of $2$ that doesn't contain each digit at least once.... – Barry Cipra May 25 '18 at 15:11
• @BarryCipra $2^{68}$ is the first power of $2$ that contains all digits, followed by $70,79,82,84,87,88,89,94$ – B. Mehta May 25 '18 at 15:31
• @BarryCipra: I believe that has been asked here. I am sure there is-once you have enough digits you are almost sure to get them all, but I suspect it is very hard to prove. This and this ask it. No solutions are given. – Ross Millikan May 25 '18 at 15:35

Remark: this answer pertains to the initial version of the question. The current edit restricts the search to powers within a given set of primes.

Another theoretically "cheap" way to look for prime powers that contain a given set of digits is to concatenate the $n$ digits, append a $1$ to get an $(n+1)$-digit number $d$ and then apply Dirichlet's theorem on primes in arithmetic progressions to the progression $10^{n+2}m+d$ for $m=1,2,3,\ldots$. (Appending the $1$ is necessary when, for example, the given digits are all even, in order to get a number $d$ that is relatively prime to $10$.) Dirichlet's theorem guarantees you'll find a prime, which can be checked for successive values of $m$ using, say, the AKS primality test if $n$ is large. If $n$ is small, say $n\approx10$, just about any primality test will do. Furthermore, the Prime Number Theorem (for primes in arithmetic progression) suggests you should find a prime relatively quickly.

More is true:

For any finite sequence of decimal digits, there is a power of $2$ whose decimal expansion begins with these sequence.

See here for a proof. For algorithms, see this question.

On the other hand, I've just run a computer search and every allowed subset of digits are the digits of a prime less than 304456880. Not all subsets of digits are allowed, of course: those whose sum is a multiple of $3$ are not. Nor are those solely composed of even digits. I've found $78$ forbidden subsets.

• I know, but here we need an algorithm – Vladislav Kharlamov May 25 '18 at 13:53
• There's an obvious algorithm: start at $1$ and keep multiplying by $2$ until you get a number that contains the digits you're looking for. The theorem guarantees the process will produce an answer. (It doesn't say how quickly, but that's a different question.) – Barry Cipra May 25 '18 at 13:57
• Well, it would be nice to somehow improve, but so well. – Vladislav Kharlamov May 25 '18 at 14:00
• How have you concluded that multiples of $3$ and even digits are disallowed? Recall that OP asks for prime powers, not just primes. In addition, primes like $277$ give the set $\{2,7\}$, which has a sum a multiple of 3. – B. Mehta May 25 '18 at 14:23
• Fair, but this still leaves numbers such as $243 = 3^5$ missing. – B. Mehta May 25 '18 at 14:31