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Bellringers Problem – Solution

March 3, 2014

In my last post, I gave you the problem of The Vicar, The Bishop And The Bellringers.

This problem really is very neat – an extremely clever interweaving of particular numbers and their factors. It needs only arithmetic as applied to prime factors.

I restate the problem here, at least in part to give you thinking time so you’re not looking straight at the solution…

The Problem

A bishop and her friend the vicar were both amateur but well-practised mathematicians, and often set each other problems to solve.

One day, the bishop visited the vicar. While the two were walking in the grounds of the vicarage, three bellringers passed them by.

“Why, they’re a fine bunch of bellringers!” said the bishop. “How old are they?”

“Well, I can tell you two things…” said the vicar.
“Firstly, if you multiply their ages, you get 2450.”
“Secondly, if you add their ages together, you get twice your age, Bishop.”

The bishop had come prepared for such a challenge. She took pen and paper from her pocket and scribbled down a few numbers, but then looked up and said, “You haven’t given me enough information.”

The vicar responded, “Oh, yes, I see. Well, I’m older than any of you.”

The bishop looked down at her scribblings, then said, “Ah! Now I know the bellringers’ ages.”

The question is: How old is the vicar?


(1) A solution does exist and is unique for the conversation as described.
(2) This is not a trick problem. For example, no one was born on February 29.
(3) Neither has any tricky wording has been used. For example, the vicar saying that he is “older than any of you” means that the vicar is the eldest of the five people in the story.
(4) You can assume from the course of the conversation that we are dealing with whole number ages.
(5) You can also assume that the bishop was correct about not having enough information to solve her problem.
(6) Lastly, you can assume that the vicar was correct in using his age to differentiate between the possible values for the bellringers’ ages.

Over time, I’ve developed a shorthand version of the conversation, as follows:

B: a, b, c = ?
V: a x b x c = 2450
      a + b + c = 2 x B
B: Not enough info
V: V > a, b, c, B
B: a, b, c = !

This version is very useful if writing it on a restaurant serviette.

The Solution

Let’s look at the conversation by parts. We use the variable names in the serviette version above.

B: a, b, c = ?
V: a x b x c = 2450
     a + b + c = 2 x B

The vicar’s information allowed the bishop to make something like the following table. As per note (4) above, the four ages (a, b, c and B) are all whole numbers. Since 2450 = 2x5x5x7x7, ab and c must be products of these prime factors. For each combination of ab and c, the age of the bishop then follows immediately.









































I have excluded extreme combinations such as 2x5x245, although including them would make no difference to the problem solution. You can add them to the table yourself, if you like. I have also added the bishop’s age for each combination of the bellringers’ ages.

B: Not enough info

The bishop knows her own age. So, if the vicar has not provided enough information, her age must be not unique in the table above. This is the distinguishing feature which allows you to ignore combinations such as 2x5x245.

So the bishop must be 32.

Then the bellringers’ ages are one of the following combinations: 5, 10, 49 or 7, 7, 50.

V: V > a, b, c, B
B: a, b, c = !

The extra fact that the vicar is the eldest of the five allowed the bishop to differentiate between  the above combinations of bellringers’ ages. (Of course, this also means that the bishop knew the vicar’s age.)

If the vicar was 51 or older, neither option would be excluded.
If the vicar was 49 or younger, both options would be excluded.
So, the vicar is 50.

As I said, it’s the nicest problem I know…


I started wondering just how good this problem is, compared to other similar problems.

I discuss below some possible variations to the problem, and comment about the quality of those variations. This section is both more mathematical and more esoteric, so feel free to skip it…

‘Realistic’ bishop’s ages

The youngest Anglican bishop of the 20th century was Bishop Anthony Burton’s in the Diocese of Saskatchewan. He was 34 when consecrated in 1993. Further, in  1932, Raymond Augustine Kearney became the Catholic bishop of Brooklyn New York, at the age of 32 – the youngest Catholic bishop in the last century. Today a Catholic bishop must be at least 35 years old. I did not look at other faiths, but I imagine the picture is similar. The number of female bishops (Catholic, Anglican or otherwise) is still so small that there is no list that I could find of youngest female bishops. For example, Penny Jamieson was 47 when ordained a bishop in 1989.

(Please tell me of other 32 year old – or younger – bishops or equivalent if I am wrong.)

So, if someone wants to get picky with me, the only bishop I could find of age 32 – at least in the last 100 years or so – was male (not female, as above).

Human lifespans

We should at least maintain some semblance of reality and restrict ourselves to documented human lifespans.

The person with the oldest documented age, Jeanne Calment, lived to the age of 122. On this basis, all counts given below assume ages not more than 122. The current oldest person (as at 3 March 2014), Misao Okawa, has her 116th birthday on 5 March 2014, and no one else has been documented as living over 119, so we are safe, for a while at least, with such an age restriction.

Variation 1: Other values of P

On the other hand, let’s ignore the facts that all five people in this story should be neither too young nor too old to perform their respective roles. We may, of course, have to retell the story with different roles for the people in it.

Then there are a surprisingly large list of numbers P which can replace 2450 above, but where each still produce a problem with a unique solution. In the analysis above, one of the nice things about the problem was having to consider what the bishop knew. In what follows, we are the puzzle maker, so we must also consider what the puzzle solver knows (or doesn’t know).

Let’s start with the fewest restrictions and progressively filter out potential solutions according to the problem parameters.

B: a, b, c = ?
V: a x b x c = P

The number of different combinations of three ages up to and including n is n(n+1)(n+2)/6, which for n=122 means 310,124 combinations for a,b and c. Try listing the combinations for small values of n and you will see the pattern. Quite a few share the same value of P, of which more below.  

V: a + b + c = 2 x B

The bellringers ages must add to twice the bishop’s age – an even number. Exactly half of the above combinations have a+b+c=2B (at least for bellringers’ ages up to an even n; for odd n, there are an extra (n+1)/4 combinations). This leaves 155,061 combinations.

B: Not enough info

The bishop knows her (his?) own age, so there must be at least two solutions for a valid age of the bishop. So we want solutions only if they share both B and P with at least one other solution. This reduces the number of combinations to 11,019, spread across 5,394 combinations of B and P. This latter figure is the number of potential solutions.

V: V > a, b, c, B
B: a, b, c = !

When the vicar announces that he is the oldest of the five people in the story, we want his age to be enough information for the bishop to solve her (his?) problem, and for the vicar’s age to be unique in doing so.

Now, the bishop has two or more possible solutions to consider, each of which has an oldest member of a, b, c and B.

For the vicar’s age to uniquely distinguish between the possible solutions, the oldest members of each must differ in age by one year.

There are 201 pairs of possible solutions conforming to such a restriction. Note that seven of these solutions have a third set of abc (for a given B and P) but the maximum age there is larger than the other two, and would also be excluded by the vicar’s statement.

Variation 2: I’m older than any of them

Sometimes small changes to the wording of a problem can change things significantly.

For example, if the vicar had said “Oh, yes, I see. Well, I’m older than any of them“, meaning the eldest of a, b and c only, then there are only 82 pairs of solutions.

If we also try to make the original problem a little more realistic by saying the bishop must be between, say, 50 and 75 (which excludes our original problem), and that the others must also be younger than 75, then there are only 12 pairs of solutions (a, b, c and a’, b’, c’), as follows:

a b c a’ b’ c’ B P
56 27 19 57 24 21 51 28728
57 29 18 58 27 19 52 29754
68 23 15 69 20 17 53 23460
54 30 22 55 27 24 53 35640
69 25 14 70 23 15 54 24150
55 32 21 56 30 22 54 36960
62 33 21 63 31 22 58 42966
69 30 21 70 27 23 60 43470
80 27 15 81 25 16 61 32400
65 33 24 66 30 26 61 51480
62 35 27 63 31 30 62 58590
69 35 22 70 33 23 63 53130

Making a pleasing, solvable problem

The numbers just above are really quite large. Working with them would be well beyond the arithmetic abilities of most of us.

Being realistic is all very well, but not if it makes solving the problem unrealistic.

For this problem, I suggest some alternative criteria for a ‘good’ version:

  • Size of prime factors of P
    The prime factors of P should not be too large, otherwise the arithmetic gets too hard. We want the problem to be about logic as much as possible. So let’s use only prime factors less than 12 (i.e. 2, 3, 5, 7 and 11). Then solvers may be able to do the arithmetic in their heads, using their times tables from school.
  • Length of bellringer age combinations
    The bishop’s list of combinations of possible bellringer ages (ab and c) should not be too long. A good rule of thumb is that they could be written on the back of a used envelope. This list should also not be too small, so the fact that two combinations have the same age for the bishop is visible, but not immediately obvious. Between 5 and 20 combinations feels about right. See the ‘abc list length‘ column in the table below.
  • Ages of story characters
    Let’s say that all people in the puzzle could reasonably said to be able to walk and talk. I’m probably still being too inclusive here, but let’s say everyone is at least two years old. Remember also the upper age limit of 122 for all players.
  • ‘Older than them’ is better
    I think the problem feels better if it conforms to the ‘older than them’ option. The bishop’s age is best not involved beyond the equation with it being half the sum of the bellringers’ ages. Remember that this choice is taken by us as problem makers, not by the problem solver – for them it just happens one way or the other.

Applying these criteria, we get the following five possibilities, plus a few extras:

a b c a’ b’ c’ B P abc list length
9 5 2 10 3 3 8 90 4
49 10 5 50 7 7 32 2450 9
21 11 6 22 9 7 19 1386 14
20 7 3 21 5 4 15 420 16
14 9 5 15 7 6 14 630 16
27 12 7 28 9 9 23 2268 25
35 9 4 36 7 5 24 1260 34
44 15 9 45 12 11 34 5940 40
44 18 10 45 16 11 36 7920 65
35 16 9 36 14 10 30 5040 75

Of the first five above, only our original one (with P=2450) has a bishop’s age that is even half way reasonable. It is also the only P with two factors of 5, making multiplication easier. The latter five have far too many combinations for a problem that can be pleasantly solved.

I’ve known for years that there were numerous alternatives to 2450. Still, all of the above analysis suggests that it is actually the best option.

I wonder if I’ve actually reverse engineered the thinking of the first person to think this problem up.

  1. Gösta Ekspong permalink

    My original version of the problem with the age of the Bishop is unique.
    No numerical input information is given, but a large amount of output results, which seems to come from nowhere.

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