# An Age Conundrum with Men of the Cloth

I came across the following problem in several variants; curiously, it was always dressed up as a conversation between two men of the cloth:

The vicar said to the verger, "How old are your three children?" The Verger replied, "If you add their ages you would get the number on my door. If you multiply their ages together you get 36.

The vicar went away for a while but then came back and said he could not solve the problem. The verger said, "Your son is older than any of my children". Then the vicar told the verger the ages of the verger's children. Find the ages of the kids

Solution

The problem made a frequent appearance in elementary puzzle books. Recently I found that a little more involved variant of the problem served to entertain remarkably mature minds as well. Uri Haber-Schaim, a retired physicist from Jerusalem, Israel, published his recollection in an 2011 issue of Il Nuovo Saggiatore – the bulletin of the Italian Physical Society. In particular, Haber-Schaim recalls a summer school in high-energy physics that took place in Varenna, Italy, in 1954, which was attended by, among others, the Italian particle physicist Enrico Fermi.

During the morning break, one of the participants from France - A. Rogozinsky - posed a mathematical problem concerning a priest and a sexton on a walk who encounter three people coming towards them. The sexton asks the priest how old the three people are and is told that "the product of their ages is 2450 and the sum of their ages is twice your [i.e. the sexton's] age".

The sexton, saying that he needs more information to solve the problem, is then told by the priest that he - the priest - is "older than any of them".

So the question is: what are the ages of the three people, the priest and the sexton?

Haber-Schaim recalls that everyone at the meeting realized that writing down equations would not get them anywhere and that he then suggested to Rogozinksy that he present the problem at lunch so that everyone could tackle it together.

Fermi, however, who was a notoriously good problem solver, proceeded to answer the puzzle within a minute.

Solution

What makes numbers 36 and 2450 special? Are they? May you think up a problem of the same sort but with different numbers?

The vicar said to the verger, "How old are your three children?" The Verger replied, "If you add their ages you would get the number on my door. If you multiply their ages together you get 36.

The vicar went away for a while but then came back and said he could not solve the problem. The verger said, "Your son is older than any of my children". Then the vicar told the verger the ages of the verger's children. Find the ages of the kids

### Solution

The question is to find three (whole) numbers whose product is 36 and whose sum must be in some sense ambiguous to preclude the vicar from solving the problem. 36 could be written as a product of three numbers in several ways. I list all of them below and append in the last column the sum of the three factors:

 1 × 1 × 36 38 1 × 2 × 18 21 1 × 3 × 12 16 1 × 4 × 9 14 1 × 6 × 6 13 2 × 2 × 9 13 2 × 3 × 6 11 3 × 3 × 4 10

The only sum knowing which would not help the vicar to solve the problem is 13, because it can be obtained in two ways:

2 + 2 + 9 = 13, and 1 + 6 + 6 = 13.

If the vicar's son is older than 9 the problem is obviously unsolvable, because both triples satisfy that condition. Yet the vicar was able to solve the problem. Therefore, his son's age is below 9. Hence the solution: 1 6 6.

A priest and a sexton encounter three people coming towards them. The sexton asks the priest how old the three people are and is told that "the product of their ages is 2450 and the sum of their ages is twice your [i.e. the sexton's] age".

The sexton, saying that he needs more information to solve the problem, is then told by the priest that he - the priest - is "older than any of them".

So the question is: what are the ages of the three people, the priest and the sexton?

### Solution

Similar to the solution of the first problem, I list all representations of 2450 as a product of three factors and their half-sums (of course, the variants with nonsensically large factors have been excluded):

 1 × 49 × 50 50 2 × 25 × 49 38 2 × 35 × 35 36 5 × 10 × 49 32 5 × 14 × 35 27 7 × 7 × 50 32 7 × 10 × 35 26 7 × 14 × 25 23

The ambiguous one is now 32 because it's the only one that could be obtained in two ways:

5 + 10 + 49 = 2×32 and 7 + 7 + 50 = 2×32.

For the same reason as in the first problem we choose the one with the lesser largest factor, i.e., 5, 10, 49. The sexton is 32; his son's age is 50.