PRIZE PUZZLE FOR NOVEMBER 2006
In mathematical notation find the least prime, P, such that
Method. Since we are only dealing with odd primes, one smaller and one larger will both be even. So the two numbers that are adjacent to the prime number must be divisible by 26 and 14 and they must have a difference of two. Once we have found two numbers with these properties we can test that the number between them is prime. If not, add 7*13*2=182 (the extra '2' because the two numbers must remain even) to both of them as this must be the next higher pair of numbers with such a property. Continue to do this until the odd number between them is prime.
So what are the first two such numbers? The easiest way of solving this is by trial an error. If the numbers were much larger you would have to use number theory methods which I will not go into here.
Looking at the multiples of 14 we have 14, 28, 42.. etc which we can test to find the first which has a number two more that is divisible by 26. These are the numbers 16, 30, 44.. etc. The first number in this sequence that is divisible by 26 is 156 with 154 being divisible by 14. So is 155 prime? Clearly not, since it ends in 5. So add 182 to both numbers and we have 338 and 336 with the centre number 337. To test for primality we only need test for division using the prime numbers less than the square root, which are 3, 5, 7, 11, 13 and 17. Using a calculator we quickly find that none of these numbers divides 337 with an integer result so 337 must be prime. So 337 is the answer.
I received only two entries, both correct. They were from John Stafford and Clem Robertson. A draw was not held as there were less than three entries. Not to worry. Some puzzles are easy, some are not. The December puzzle is much easier and a good puzzle for children over Christmas and the New Year holiday.