Tom, Dick and Harry each chose two different 2-digit semiprime numbers (a semiprime is the product of exactly two different prime numbers). Any given semiprime may have been used by more than one of these characters, but any one person chose two different ones. In each of the three cases, the sum of the two semiprimes was a 3-digit semiprime.

Also, in each instance, if the units digits of the two semiprimes were swapped, to be paired with the other tens digit, the result was two semiprimes that were different from the original pair. Of course they added up to the same 3-digit semiprime as the first pair. In each of Tom's, Dick's and Harry's three semiprimes, none of the six primes from which they were made were the same: for example, if Tom's semiprimes were 93 and 94 adding up to 187, all six primes (3 and 31 for 93, 2 and 47 for 94 and 11 and 17 for 187) would all be different, satisfying this rule. The same was also true for each of the three people after the swap of units digits between the two 2-digit numbers.

Tom's, Dick's and Harry's 3-digit sums were all different, but Tom's and Harry's 3-digit sums did share a common prime factor, while Dick's did not.

What were Dick's numbers?
What were the other two sets of numbers?

(In reply to re: Problem by brianjn)

No, I see my mistake now.  In calculating two digit semi-primes, I took all multiples of primes between 2 and 47.  Then I used those results to make a list of 3-digit semi-primes.  Big mistake.  So naturally, I did not have 142 and 146 in my list.

Posted by hoodat on 2011-01-21 19:00:49