A total of 12 digits are placed in a sequence such that the 11 pairs of subsequent digits form 11 different prime numbers.

What is the second digit in this sequence, and what is the last?

There are a total of 21 two-digit primes. These are: 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97.

As no prime (but 2) ends with an even digit and as no prime (but 5) ends with a 5, in order for one of these primes that begins with an even digit or 5 to be included in the sequence would have to be in the initial position. Ignoring these primes in our list leaves ten primes. As the sequence involves eleven pairs of primes, the sequence must begin with one of these primes.

Of the digits of the ten primes, there are four primes that begin with a 1 and three that end with a 1 (4+3 is odd); two that begin with a 3 and two that end with a 3 (2+2 is even); three that begin with a 7 and three that end with a 7 (3+3 is even); one that begins with a 9 and two that end with a 9 (1+2 is odd).

As the primes are PAIRS of primes, the primes with the odd number must be at the beginning and ending of the sequence. As there are more primes that begin with a 1 than end with a 1, the 1 must be at the beginning of the sequence of the ten, and as there are more primes that end with 9 than begin with a 9, the 9 must be at the end of the sequence of ten.

Thus, the second digit is a 1 and the final digit is a 9.
As the sequence of ten must begin with a 1, the initial prime of the sequence of eleven must then be either 41 or 61, and as the sequence ends in a 9, the final prime must be a 19 or 79.

Edited on September 21, 2008, 9:15 pm

Posted by Dej Mar on 2008-09-21 12:15:16