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Given a number $N$ with some properties:

  • It is only constructed using these digits: $\{1,3,5,7,9\}$ (e.g. $135$ , $7713$. NOT: $1231$ which includes the digit $2$)
  • It has at least $3$ distinct odd digits. (for example these are not allowed: $7755$, $13$, $11$, $3$ etc..)
  • Its sum is not divisible by $3$ (and thus $3 \nmid N$)

I've written a (very clunky) Python program that takes a random number with these very properties, and checked:

If the number is divisible by $3$ - skip
Else - check each permutation of the number if it is a prime.

For example (for simplicity I've used $5131$ which has only $2$ unique odd digits):

$335511$ - skip, it is divisible by $3$.
$5131$ - check each permutation of this number if it is a prime:
$5131$ - not a prime. $5113$ - not a prime. etc... until we hit a prime: $5113$

The results were quite nice, I've noticed that if the numbers meet those constraints then:

  • At least one permutation of $N$ is a prime

These constraints look very 'harsh' at the beginning, but I don't think it is that trivial - that a number with $3$ or more unique odd digits has at least one permutation that is a prime (If it is not divisible by $3$).

Is there any reason why these numbers behave this way?

UPDATE

Now assume $N$ can be constructed only using $\{1,3,7,9\}$ (no $5$ allowed) and it has at least $3$ unique digits (so maybe it contains $1,3,7$ or $1,3,7,9$ or $3,7,9$ etc..)

Now the number is either divisible by $3$ or has at least one prime permutation.
I could not seem to find any counter-example to this, however it must exist as what @lulu said, it is a matter of probability, but after checking millions and million of numbers - I couldn't ... so such number does exist?

Thank you!

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    $\begingroup$ There are a lot of prime numbers. Just for example, if you stick with $8$ digits or fewer then there are $\pi(10^8)\approx 5.7\times 10^6$ primes. It follows that if you choose a random odd number with $8$ digits there's a greater than $\frac 19$ chance that it is prime. Since it is likely that your number has many more than $9$ permutations, this may just be chance. $\endgroup$ – lulu Jul 16 at 10:51
  • $\begingroup$ To build counterexamples, I'd take a long number with all but two digits equal to $5$ and then two other odd digits. That keeps the number of relevant permutations as low as possible (roughly $2L$ where $L$ is the length). $\endgroup$ – lulu Jul 16 at 11:27
  • $\begingroup$ @lulu Hey thank you for answering! please see my update question it has more constraints $\endgroup$ – CSch of x Jul 16 at 18:04
  • $\begingroup$ Just as a sidelight, you may want to investigate the topic of permutation primes: those primes for which every permutation of their digits produces a prime, e.g. 991, the largest such with three digits. There is a Wiki page on them if you are interested. $\endgroup$ – Chris Leary Jul 16 at 18:13
  • $\begingroup$ @ChrisLeary Thanks for the answer! but I am talking about at least one* prime permutation, not all $\endgroup$ – CSch of x Jul 16 at 18:17
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No permutation of the digits $1,5,5,5,7$ creates a prime. Obviously it cannot end in a $5$, and the others are:

$15557=47\times331$
$51557=11\times43\times109$
$55157=19\times2903$
$55517=7\times7\times11\times103$
$55571=61\times911$
$55751=197\times283$
$57551=13\times19\times233$
$75551=7\times43\times251$

For 10 digits there is also $3555555557$ and $5555555579$.

I found these by computer. There are no other such combinations with 12 digits or fewer.

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  • $\begingroup$ Nice catch! I've actually looked it up beforehand (not the same number but other number with 5 in it) and it seems that this what causes the problem, what about a number that is constructed from $\{1,3,7,9\}$ ? $\endgroup$ – CSch of x Jul 16 at 10:51
  • $\begingroup$ @Remember1312 As lulu points out in their comment, the probability is pretty low. It is also not surprising that in the ones I found, all but two of the digits are $5$, because then there are only $2n-2$ numbers that would have to be composite. WIthout a $5$, the number of permutations that would have to be composite is so much bigger that it is very very unlikely to happen, though not necessarily impossible, and I think with more digits it gets ever more unlikely. $\endgroup$ – Jaap Scherphuis Jul 16 at 11:21
  • $\begingroup$ Thank you sir! Please see my update, I am very curious to look for your answers, thank you again $\endgroup$ – CSch of x Jul 16 at 18:05

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