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Blog

Quiz question of the week

By Matin Durrani

fermistamp.jpg

I always find it interesting when little-known anecdotes about some of the greatest figures in physics come to light.

So here’s one that I thought I’d share with you, courtesy of Uri Haber-Schaim, a retired physicist now living in Jerusalem.

Writing in the latest issue of Il Nuovo Saggiatore – the bulletin of the Italian Physical Society – 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.

So over to you, physicsworld.com readers. Can you solve the problem or – even better – beat Fermi and get the answer in under a minute?

For the record, I still haven’t figured it out.

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21 comments

  1. The people are aged 35, 7 and 10. The sexton is therefore 26 and the priest must be at least 36.
    It took me about 8 minutes and I don’t know if there are more possible answers ;-)

  2. Cartik Sharma

    Individual ages of 3 people, 49,10,5.
    Age of sexton:32
    Age of priest: 50
    satisfies sum (64 = 2n)
    xyz = 2450
    or
    35,7,10, sexton: 26, priest:36.

  3. Paul Coxon

    to keep the numbers simple, assume one of the people is only aged 1, so you’re only looking for a pair of numbers that mutliply to make 2450.
    Now 2500 = 50 x 50, so if A x B x C = 2450, with A = 1 then B = 50 and C = 49.
    A + B + C = 100, therefore the sexton is also 50 and the priest is at least 51.

  4. There are 12 solutions that don’t involve people having ages in excess of 150, all of which leave the priest’s age undetermined. If the priest is under the age of 70, we could have
    1,49,50 s=50
    2,25,49 s=38
    2,35,35 s=36
    5,10,49 s=32
    5,14,35 s=54
    7,7,50 s=32
    7,10,35 s=26
    7,14,25 s=23
    The Sexton presumably knew his own age, yet still required further information. So he must have been 32. If the priest’s comment about his age was enough to settle the matter for the sexton (assuming that the sexton knew the age of the priest), then the priest must have been 50.
    The ages are then 5,10,49, s=32, p=50.

  5. Satoru Inoue

    The article is missing the crucial part which makes this solvable. The problem should say that, with the information that the priest is older than the other 4, the sexton was able to figure out everyone’s age. This is a meta-puzzle.

  6. Christine

    The priest never says us, he says them. Their ages are 7, 14, & 25. The sexton is 23 and the priest is 26+

  7. Luis Mochán

    There seem to be many solutions such as 49, 10, 5; 35,, 35, 2; 35, 10, 7; as 2450=7*7*5*5*2. However, the only solution that becomes unique by putting an upper bound to all ages is 25,14, 7. Thus, the priest is aged between 25 and 35 and the sexton 23

  8. Luis Mochán

    I guess I was mistaken. I thought the solution was 25,14,7, as it is has the smallest bound on the largest age so it would be unique if the priest’s age were between 35 and 25, but I didn’t account for the Sexton’s knowledge of his own age, which makes Bob D’s the correct answer.

  9. Pat Bushong

    a + b + c = 2c
    a + b = c therefore.
    abc = 2450
    substituting
    ab(a + b) = 2450
    aab + bba = 2450
    b > a and b > c
    b > a + b
    ?????

  10. The priest is 50 and the sexton does not need to know this in advance.
    First you must factor 2450. It breaks down into 1x2x5x5x7x7. Combining these numbers to make 3 ages reveals 20 different combinations: (1 1 2450), (1 2 1225), …, (7 10 35), and (7 14 25). Now the sexton obviously knows his own age but can’t figure out the ages of the three people. This implies that multiple sets of factors add up to the same value. The sets (5 10 49) and (7 7 50) both have a sum of 64.
    Then the priest reveals that he is older than all of the students, and now the sexton can determine the ages of the three people. If the priest was 51 or older we would not be able to figure out which set of factors is correct, so he must be 50 years old and the three people’s ages must be 5, 10, and 49. Hence the sexton is 32.

  11. Brian P

    Bob D’s solution is the same as mine. You need to make reasonable assumptions about max possible ages, and of course, the key to understanding the solution is that the information provided by the priest was inadequate and then adequate for the sexton. Those facts about adequacy provide the needed info for us, the solvers, as there is only one solution which would have lead to that exact conversation. For all others, the conversation would have been different.

  12. George

    Bob D has it right. The extra information hidden in the riddle is that the sexton required extra information – thus there must be two sets of 3 numbers that add up to the same total. That happens for 49,10,5 and 50,7,7. Since the priest’s age could settle the question, he must be 50, since that eliminates one but not both answers. Thus: 49,10,5 for the three people (a parent and two kids?), 32 for the sexton, and 50 for the priest. Took me about 14 minutes, but initially I was crunching numbers in my head, before I realized that I would have to list them.

  13. Vesselin Gueorguiev

    The obvious equations are: a*b*c=2450=2*5*5*7*7; a+b+c=2*d;
    assume specific choice of order for a, b, and c such that c>=b>=a;
    The age of the priest ‘p’ should relate to a, b, c, and d so that there is a unique solution;
    This is only possible if p is bound from above.
    Thus: d>=p>c is an important constrain that needs to be included.
    I found one solutions and was going to run a code to confirm its uniqueness, but then I realized that Bob D’s list is probably complete and thus the only solution is: 2,35,35,36,36.

  14. VGG

    The obvious equations are: a*b*c=2450=2*5*5*7*7; a+b+c=2*d;
    assume specific choice of order for a, b, and c such that c>=b>=a;
    The age of the priest ‘p’ should relate to a, b, c, and d so that there is a unique solution;
    This is only possible if p is bound from above.
    Thus: d>=p>c is an important constrain that needs to be included.
    I found one solutions and was going to run a code to confirm its uniqueness, but then I realized that Bob D’s list is probably complete and thus the only solution is: 2,35,35,36,36.

  15. Satya

    ages of 3 people: 49,10,5
    sexton’s age: 32
    priest’s age: 50.
    Took me about 40 seconds.

  16. Satya

    ages of 3 people: 49,10,5
    sexton’s age: 32
    priest’s age: 50.
    Took me about 40 seconds.

  17. Jim

    Most of us could figure this out in (much) more than a minute which makes the problem unremarkable. What we’ve learned is that Fermi was exceptional at crunching numbers in his head and keeping track of details (e.g., that the Sexton needed more information) which are both the same mental ability: a high functioning working memory.

  18. Jack

    I think you have mis-stated the problem. There is no way to tell the priest’s age. The last bit of info should say that he,THE SEXTON, is older than any of the three. the ages are then 2,35,35,36(Sexton).

  19. Tachylon

    I guess the bottom line is that fermi is quite good at calculating and storing numbers in the brain.obviously, anyone would have figured it out:BUT UNDER ONE MINUTE,I DOUBT!

  20. Perhaps a more interesting problem is a mathematical proof there are (or not) an infinite number of solutions to the constraints, i.e.
    a * b * c = 2450
    a + b + c = d
    where a,b,c,d:
    - may or may not be equal
    - may or may not be whole numbers
    - may or may not be ‘sensible’ values for human age
    - are non-negative
    Al

  21. Jim

    I divided 2450 by 2,3,5 and 7 to obtain the prime factors 2,5 and 7. All the other numbers are multiples.I multiplied 2x5x7which equals 70. I then divided 2450 by 70 which equals 35. Now I need another number, so considering the sexton’s age range, I factored 70 and got 10×7. ==So 10x7x35=2450 and 10+7+35 =52/2=26. Voila

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