Attempted spoiler: The first 26 primes go up to 101, so it seems like a good bet that this number would start with 101. There are 25! choices for the remaining bunch of digits and it sure seems likely that one of them would be prime. Hm, unless there's something like the digit sum divisible by 3 that makes it impossible. Let's see. 2+3+5+7+11+13+17+19+23+29+ 31+37+41+43+47+53+59+61+67+71+ 73+79+83+89+97+101 = 1161 so it is indeed divisible by 3. OK, then, how about 103 in place of the 101? (It seems clear that we want to keep the number of digits smaller by having only one three-digit prime and using all the first 25 primes). I'll conjecture 103, and not feed the 25! possibilities into a probable-prime test to see which one(s) are prime. I'm betting a lot of them are and I could probably have a solution to this within a short bit of computer time. --Joshua On Tue, Aug 3, 2010 at 9:37 AM, 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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