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\centerline{\smalltt INTEGERS: 
 \smallrm ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 5 
(2005), \#A10} 
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{\bf A NEGATIVE ANSWER TO TWO QUESTIONS ABOUT THE SMALLEST PRIME NUMBERS
 HAVING GIVEN DIGITAL SUMS}
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{\bf Tom M\"uller}\\
{\smallit Institut f\"ur Cusanus-Forschung an der Universit\"at und der Theologischen Fakult\"at Trier,\\
Domfreihof 3, 54290 Trier, Germany}\\
{\tt muel4503@uni-trier.de}\\
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\centerline{\smallit Received: 10/28/04, Revised: 5/22/05,
 Accepted: 5/23/05, Published: 6/20/05}
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\centerline{\bf Abstract}

\noindent
Denote by $\rho(k)$ the smallest prime number with digital sum $k$ (not a multiple of 3).
Richard K. Guy asked whether the congruences $\rho(k)\equiv 99$ (mod 100) and $\rho(k)\equiv
999$ (mod 1000) hold for all $k>38$, respectively $k>59$. Counterexamples to this are given,
inter alia, for $k=86$ and $k=104$. Moreover, several open problems and conjectures about
$\rho(k)$ are discussed.

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\markright{\smalltt INTEGERS: \smallrm ELECTRONIC
 JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 5 (2005), \#A10\hfill}

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\section*{\normalsize 1. Introduction}
In his famous book about unsolved problems in Number Theory, Richard K. Guy~\cite{guy} asks in section A3 the following three questions:
\begin{quote}
``If $\rho(k)$ is the smallest prime with digital sum $k$, is $\rho(k)\equiv 9$ (mod 10) for $k>25$?
Is it $\equiv 99$ (mod 100) for $k>38$? And $\equiv 999$ (mod 1000) for $k>59$?"
\end{quote}
We intend to show that the two latter congruences do not hold by constructing a counterexample in the following paragraph. The final section is concerned with some open problems and conjectures concerning the prime numbers $\rho(k)$ for $k$ not a multiple of 3.
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\section*{\normalsize 2. Counterexamples}
A first counterexample to two of the claims above is given for $k=86$. Using some simple combinatorial methods, we find that the smallest
number with digital sum $86$ is
\begin{eqnarray}
5\, 999\, 999\, 999
\end{eqnarray}
followed by the numbers
\begin{eqnarray}
6\, 899\, 999\, 999,\\
6\, 989\, 999\, 999,\\
6\, 998\, 999\, 999,\\
6\, 999\, 899\, 999,\\
6\, 999\, 989\, 999,\\
6\, 999\, 998\, 999,\\
6\, 999\, 999\, 899,\\
6\, 999\, 999\, 989.
\end{eqnarray}
The integers (1) to (8) turn out to be composites, as they are divisible respectively by 7, 6089, 827, 19, 61, 31, 7 and 3011.\\
The number (9) is prime. So $\rho(86)=6 999 999 989 \equiv 989$ (mod 1000) $\equiv 89$ (mod 100). Therefore the two latter questions above have to be answered with ``no", while the first one remains open. It might moreover be interesting to note that $k=86$ is not the only known case for which the said congruences do not hold. Other examples are $\rho(104) = 699999999989$, $\rho(137) = 4\cdot 10^{15}-11$, $\rho(317) = 4\cdot 10^{35}-11$, $\rho(400) = 6\cdot 10^{44}-11$ and $\rho(580) = 6\cdot 10^{64}-11$.\\
Perhaps there are many other primes of the form $a\cdot10^n-11$ with $a\in\{1,3,4,6,7,9\}$ (the parameters $a\in\{2,5,8\}$ give only multiples of $3$), which lead to further examples. Lists of such primes can be found in the Online Encyclopedia of
Integer Sequences~\cite{sloane} (the reference codes are respectively A092767, A102737, A102738, A102739, A102740 and A102741). Additionally, we have $\rho(260) = 10^{29}-101$.
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\section*{\normalsize 3. Open problems and conjectures}
A number of open problems dealing with $\rho(k)$ are waiting to be solved. Is there a $k>25$ with $\rho(k)\equiv 79$, or $\equiv 97$ (mod 100)? Moreover, one might ask whether $\rho(k)$ exists for every $k$ not a multiple of $3$. The answer to this is very probably positive, and therefore the above question could be changed into the following one (Question 1):
\begin{quote}
Does an integer $m$ exist, such that the congruences $\rho(k)\equiv 99$ (mod 100) and $\rho(k)\equiv 999$ (mod 1000) hold for all $k>m$?
\end{quote}
Let $S(k)$ be the set consisting of the $k$ smallest integers
 with digital sum $k$, and let $N(k)$ be the number of
 integers in $S(k)$ that end in ``999".  We know that
 as $k\rightarrow \infty$, $\frac{N(k)}{k} \rightarrow 1$;
 however, as $N(k)<k$ for all $k>10$, we conjecture that there
 exist infinitely many values of $k$ 
for which $\rho(k)$ does not satisfy the congruences 
in Question 1. If this is true, the integer $m$ does not exist.
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\section*{\normalsize Acknowledgements} Thanks are due to the friendly referee who pointed out a known sequence in the Online Encyclopedia of Integer Sequences. The author is grateful to the referee, Daniel O'Connell and Prof. Dr. J\"urgen M\"uller for their help in finalizing this paper.
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\begin{thebibliography}{2}

\footnotesize
\bibitem{guy}
Richard K. Guy, \textit{Unsolved problems in number theory}, Springer Verlag, New York 2004 (3th edition), 15

\bibitem{sloane}
Neil J. A. Sloane, \textit{Online Encyclopedia of Integer Sequences}, published electronically at \texttt{http://www.research.att.com/njas/sequences/}

\end{thebibliography}

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