Said Albert to Bertrand, "The last time we met, our ages were both prime numbers, and when I was a quarter of the age I am now, you were that age plus half the age your father would have been thirty years previous to when he was six times the age you would have been when I was half your age".
How old were Albert and Bertrand the last time they met?
A quick assessment of the problem reveals that we know nothing about the ages in question, the last time they met, except that both numbers are prime and the difference d in their ages must (in theory, since that too can vary with time of year) be the same as any other time. Since a difference of any even number will yield an infinite number of prime pairs with that difference, a fair assumption would be that the difference is odd, and Albert's age at that time must therefore have been 2. Once we find the difference d, then Bertrand's age at that time is 2+d (which, we would hope, is indeed a prime number).
Looking at the rest of the problem, there are three times mentioned at which we half to consider their ages: let a1 and b1 be their ages now, a2 and b2 be the ages when Albert was one quarter of his age now, and a3 and b3 be their ages when Albert was half of Bertrand's age. We can then say:
b2a2 = (1/2)(6b330) = 3b315 = d
b3 = 2a3
b3a3 = 2a3a3 = a3 = d
b3 = 2d
Substituting:
3(2d)15 = d
Solve:
d=3
Therefore, Bertrand's age was 5, and Albert's (as we already figured) was 2.

Posted by DJ
on 20030303 15:53:24 