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Consider two cases a) we reverse the digits of a prime number b) we reverse the digits of a composite number. Are we more likely to obtain a prime in case a) or in case b). Since the last digit of primes other than $2$and $5$ end in $1,3,7$ or $9$ hence if a prime or composite number begins in $2,4,5, 6$ or $8$ there is no way its reverse will be a prime. So to make a fair comparison, I only considered those prime and composite numbers whose first and last digits is $1,3,7$ or $9$.

Let $C$ and $P$ be the set of such composites and prime numbers respectively. I looked at the first $10^8$ numbers (in ascending order) in $C$ and observed the density of numbers whose reverse is a prime is roughly $\frac{2.4n}{\log n}$. However in case of the set $P$, the density of roughly $\frac{3.5n}{\log n}$ i.e. about $45\%$ higher which is significant.

Question: Given the set of numbers whose first and last digits is $1,3,7$ or $9$, why is the reverse of a prime about $45\%$ more likely to be a prime than that of a composite?

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    $\begingroup$ If a composite is divisible by $3$, then also its reverse. This could be one reason. But I wonder why this does not cause an even stronger effect. $\endgroup$
    – Peter
    Apr 7, 2021 at 8:10
  • $\begingroup$ @Peter The effect could be slightly bigger. My computing is still running and at the moment, the coefficient is oscillating roughly between $3.45 - 3.55$. But I don't think it would be significantly bigger. $\endgroup$ Apr 7, 2021 at 8:18
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    $\begingroup$ Reverses of multiples of 11 are still multiples of 11. $\endgroup$
    – MJD
    Apr 7, 2021 at 8:39
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    $\begingroup$ I wonder how the result changes when one goes to base 2, etc. That may give some insight on importance of the base $\pm1$, I don't know... $\endgroup$ Apr 7, 2021 at 8:46
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    $\begingroup$ Reverses of multiples of 10001 are all multiples of 10001. I think in other bases, numbers like this might be more common than in base 10. $\endgroup$
    – MJD
    Apr 7, 2021 at 9:18

1 Answer 1

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  1. If a composite number is divisible by 3, the sum of its digits is also divisible by 3.

  2. If a composite number is divisible by 11, the alternating sum of its digits is also divisible by 11.

Just two of those facts will give you a $\frac13+\frac1{11}=43\%$ better chance. You can further increase this number by starting adding up rarer cases. For example, if the number of digits in $p$ is a multiple of 3, then if $p = \overline{a_1a_2a_3a_4a_5a_6\ldots}$ is divisible by 7, then $(a_1+2a_2+a_3)+(a_4+2a_5+a_6)+...$ is also divisible by 7.

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  • $\begingroup$ plus if you keep it odd, your mixture of $\{1,4,7\}$ and $\{2,5,8\}$ the same creates another number with the same remwinder on division by 6 ... $\endgroup$ Apr 7, 2021 at 10:31
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    $\begingroup$ You need $\frac13+\frac1{11} -\frac1{33}$ or you have double-counted the multiples of $33$. $\endgroup$
    – MJD
    Apr 7, 2021 at 15:24
  • $\begingroup$ It took me this long to realize that the reverse of a prime is also never a multiple of $10$, so $\frac13+\frac1{10}+\frac1{11}-\frac1{30}-\frac1{33}-\frac1{110}+\frac1{330}=\frac5{11}=45.45\ldots\%$ $\endgroup$
    – MJD
    Dec 13, 2021 at 13:20

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