25
$\begingroup$

Peter, Paul and Mary have between them 282 marbles.

  • Though of the three friends, it is Peter who has the most marbles, he is the one with the fewest silver marbles, just the square root of the number of all his marbles.

  • On the other hand, Mary, who has the fewest marbles, is the one with the largest number of silver marbles, precisely one seventh of her marbles.

How many silver marbles does Paul have?

$\endgroup$
  • 2
    $\begingroup$ Great puzzle. I managed to solve it without looking at the posted solution. There was a lot of constraint juggling. I admire people who create such puzzles. $\endgroup$ – Kornelije Petak May 12 '17 at 6:55
27
$\begingroup$

Paul has

11 Silver Marbles

The exact numbers are

Peter has 100 marbles, 10 silver, Paul has 98 marbles, 11 silver, and Mary has 84 marbles, 12 silver.

This is the only solution. (See bottom for why)


Note/warning: no spoiler tags from here on in

So how did you work that out?

Well to start, there are three people, with 282 marbles. So each person has an average of 94 marbles, because the average is sum over quantity, which here is $282/3$, which is indeed 94.

We know Peter Rabbit here has a number of marbles that is a square number. We also know that he has the most marbles, so his number must be $>94$. The first square number that fits is $100$, so let's try that out.

If Peter Piper has 100 marbles, and the amount of silver marbles is the square root of the total then we get the nice easy sum of $\sqrt100$ which is 10. So Peter has the least amount of silver marbles which is 10.

Okay, but what about the others?

Our Mary Poppins has the most silver marbles. So the amount of silver marbles, must be $>10$, which is the least. Because there are three people, and she has the most, not the joint most, then she has to have an amount of silver marbles $≥12$, so there is at least one number in between her number and Peter's number for Paul the Dickie Bird to have which isn't equal to the most or least.

Mary and her little lamb can therefore have a minimum of $12*7$ marbles which is $84$, as a seventh of her marbles are silver.

I see, but what about Paul?

We already know that Paul has 11 silver marbles if the most is 12 and the least is 10, but how many does he have total?

Well Peter the Pumpkin Eater and Mary quite Contrary have $100+84$ marbles between them, then Paul must have $282-(100+84)$ marbles, which means he has 98 marbles.


This is one answer that works, but are there others? After all you only said that 100 is the first square number that fitted for Peter, what about the second?

The second square number which is $>94$ is $121$ ($11^2$). That would mean that Peter had 11 silver marbles, the least.

This would mean Mary would have to have at least 13 silver marbles (the most), and that would mean she would have at least 91 marbles ($13*7$).

So Paul would have 12 silver marbles. That fits doesn't it?

Well, Peter and Mary have $121+91$ marbles between them, which is 212. That means Paul would have $282-212$ marbles which is only 70. Mary has to have the least amount of marbles, but 70 is less than 91 so this doesn't work. If you keep increasing Peter's amount, then Paul's will keep going down, so 100 is the only square that works.

Okay that doesn't work, but you also said that Mary had to have at least 12 silver marbles. What if she had more?

Well the next number she could have is 13. As I already showed, that would mean Mary would have 91 marbles.

Mary and Peter would then have 191 marbles between them ($100+91$). Paul would then have $282-191$ which is equal to 91.

Well, that's not less than Mary's amount, it's the same! There are multiple solutions then.

Actually no. Mary has the least, not the joint least. So if she has the same as Paul, she doesn't have the least. And by increasing Mary's amount of marbles, you are therefore decreasing Paul's amount, so if Mary had 14 marbles, Paul would have less than Mary. So Mary has to have 12 silver marbles.

Thanks.

Hope this helps. If you have any other questions on things you want clarifications on, ask in the comments.

$\endgroup$
  • 4
    $\begingroup$ No spoiler tags? $\endgroup$ – user5971 May 8 '17 at 6:54
  • $\begingroup$ @Evert do you think this needs them? I'll spoiler the first section at the top and leave the rest, but tell me if you think I should also spoiler the rest $\endgroup$ – Beastly Gerbil May 8 '17 at 12:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.