We can't use 2, 3, ..., 101 because it'd be divisible by 3. So we use 2, 3, ..., 97 and 103. The first three digits are very likely 103 because we can expect that some permutation of the remaining primes will make it prime. Let x be the concatenation of 103, 11, 13, 17, 19, 2, 23, 29,31, 3, 37, 41, 43, 47, 53, 5, 59, 61, 67, 71, 73, 7. The smallest candidates and their smallest divisors are: x79838997, 19 x79839789, 17 x79898397, 29 x79899783, 21773 x79978389, 43 x79978983, 53 x83798997, 1664337338519 x83799789, 149 x83897997, 479 x83899779, 36767 x83977989, 7 x83978979, 97 x89798397, 59 x89799783, 17 x89837997, 7 x89839779, 7057327 x89977983, 7 x89978379, 7283 x97798389, 23 x97798983 is prime!!! We can prove the primality of 1031113171922329313374143475355961677173797798983 by using the fact that 5 is its (smallest) primitive root and 1031113171922329313374143475355961677173797798983 - 1 = 2 * 48866600953 * 74195778089023 * 142195225375699804440389. Hope this answer would make me pass Number Theory 101. :) Warut On Tue, Aug 3, 2010 at 11:37 PM, Hans Havermann <pxp@rogers.com> wrote:
The smallest decimal-digits prime composed of the concatenation of 7 distinct decimal-digit primes is 1113257317, composed of 11, 13, 2, 5, 7, 3, and 17. What are the first three digits of the smallest decimal-digits prime composed of the concatenation of 26 distinct decimal-digit primes? (Bonus marks: Allowing probable primes, what is the entire number?)
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