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?
Thoughts which I think are valid from my reading:
A 2-digit semi-prime will be generated as the product of primes from 2 to 47 when appropriately paired.
Since the maximum sum of two 2-digit numbers is less than 200 then a 3-digit semi-prime will be generated as the product of primes from 2 to 149 when appropriately paired.
One prime is common in the composition of Tom and Harry's 3-digit semi-prime which means that 5 distinct primes feature in Tom, Dick and Harry's 3-digit semi-primes.
While four of Tom's primes create his two original semi-primes there is no stipulation that the digit swapping needs to have any of his primes to be factors of those newly formed.
Now, since a semi-prime is not necessarily unique to one person, the primes forming the original set of semi-primes can range from 8 to 12.
It does not appear to be an issue if any of that latter set are also members of the final set of 5 providing each person can demonstrate having used 6 primes in the final analysis.
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Posted by brianjn
on 2011-01-20 02:21:17 |