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?

A: No.

B: Aha! **AT LAST** I know how old you are!

Rem: Taking in account the fatherhood limitations, this is uniquely solvable.

Charlie

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
sorry.

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
children?

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?