A puzzle by Princeton mathematician John Horton Conway:
Last night I sat behind two wizards on a bus, and overheard the following:
A: I have a positive integral number of children, whose ages are positive integers, the sum of which is the number of this bus, while the product is my own age.
B: How interesting! Perhaps if you told me your age and the number of your children, I could work out their individual ages?
B: Aha! AT LAST I know how old you are!
Rem: Taking in account the fatherhood limitations, this is uniquely solvable.
Thank you for introducing to the extensive world of essays
triggered by the original puzzle.
I have solved this problem long time ago, read some article
quoting (48,4) as the only possible ambiguous
set(age, #of children) corresponding to bus #12.
At that time I realized that there is something wrong about
the uniqueness of the set (a,b,c),
provided by you and
other mathematicians i.e. (48, 12, 4).
I wrongly remembered 36 as a possible age, and
found support in Steve Herman's comment.
Did not check too much and left it to you. MEA CULPA, really
However , the question regarding duplicity of ages as created
by “propagation” , prevails.
What’s wrong with (48, 13, 5) i.e. the guys are on
bus number 13 and A, aged 48 has 5
I do not introduce quintuplets 1,1,1,1,1 with quadruplets
2,2,2,2 for a 16 yrs old father of 9
children riding on the same bus.
The ages are funny, but possible:
13 48 5 1 1 3 4 4
13 48 5 1 2 2 2 6 (from your listing)
If I err, tell me where.
If not, how did the others get away with it?