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Let p be a prime number, here I'm interested with the last p digits of $p^p$ if it forms a prime number. And if I've not mistaken, the smallest p with that property is $433$. Meaning that the last $433$ digits of $433^{433}$ is a prime number. I have checked p up to $1999$ but I couldn't find anymore p with such property. Is there other p with that property ?

Takuya Nazuga
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  • Two consecutive primes are $139$ and $149$. The number $143$ is not prime. – Piquito Dec 24 '16 at 14:27
  • In fact, the smallest $p$ with the desired property is $433$ – Peter May 10 '17 at 18:15
  • Upto $p=21000$ , there are $5$ numbers $p$ which lead to a prime of the desired form : $143,433,1687,3283,14949$. Of those numbers, only $433$ is prime. – Peter May 14 '17 at 08:43

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Of course this problem is base dependent. In binary arithmetic $11$ (meaning three) raised to its own power gives the last three bits $011$ which is a prime number.

Oscar Lanzi
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