Eighteen answers have been received. One of the writers merely asserts that the first occasion was 12 years ago, that the ages were then 9, 6, and 3; and that on the second occasion they were 14, 11, and 8! As a Roman father, I _ought_ to withhold the name of the rash writer; but respect for age makes me break the rule: it is THREE SCORE AND TEN. JANE E. also asserts that the ages at first were 9, 6, 3: then she calculates the present ages, leaving the _second_ occasion unnoticed. OLD HEN is nearly as bad; she "tried various numbers till I found one that fitted _all_ the conditions"; but merely scratching up the earth, and pecking about, is _not_ the way to solve a problem, oh venerable bird! And close after OLD HEN prowls, with hungry eyes, OLD CAT, who calmly assumes, to begin with, that the son who comes of age is the _eldest_. Eat your bird, Puss, for you will get nothing from me! There are yet two zeroes to dispose of. MINERVA assumes that, on _every_ occasion, a son comes of age; and that it is only such a son who is "tipped with gold." Is it wise thus to interpret "now, my boys, calculate your ages, and you shall have the money"? BRADSHAW OF THE FUTURE says "let" the ages at first be 9, 6, 3, then assumes that the second occasion was 6 years afterwards, and on these baseless assumptions brings out the right answers. Guide _future_ travellers, an thou wilt: thou art no Bradshaw for _this_ Age! Of those who win honours, the merely "honourable" are two. DINAH MITE ascertains (rightly) the relationship between the three ages at first, but then _assumes_ one of them to be "6," thus making the rest of her solution tentative. M. F. C. does the algebra all right up to the conclusion that the present ages are 5_z_, 6_z_, and 7_z_; it then assumes, without giving any reason, that 7_z_ = 21.