makaren@alberta.UUCP (11/09/83)
f the
two integers and gives it to mathematician "P".
He then instructs both mathematicians to determine both the
sum and product of the two integers.
After considerable thought mathematician "S" says to mathematician "P"
"There is no way that you can determine the sum"
Mathematician "P" replies "Then I know the sum"
Mathematician "S" responds "Then I know the product"
YOUR PROBLEM: What were the two integers?
note: We assume that mathematicians do not make mistakes in matters of logic.
I will post the solution at a later date if nobody solves it.makaren@alberta (11/14/83)
I have received some mail about this puzzle having been garbled at some
sites and so I am posting it a third time. Appologies included.
Maybe if I include enough
lines of text at the
beginning of the message
then the puzzle will not
end up garbled.
I have also removed the
first blank line in the hopes that this will
have some effect.
Anyway here is the puzzle again. ----
There are two mathematicians which we will call "S" and "P".
They are in a contest.
The judge of the contest chooses two integers tells the mathematicians
"Both integers are greater than one"
"The sum of the two integers is less than forty"
"The product of the two integers is less than four hundred"
The judge then writes down the sum of the two integers and gives
it to mathematician "S". He then writes down the product of the
two integers and gives it to mathematician "P".
He then instructs both mathematicians to determine both the
sum and product of the two integers.
After considerable thought mathematician "S" says to mathematician "P"
"There is no way that you can determine the sum"
Mathematician "P" replies "Then I know the sum"
Mathematician "S" responds "Then I know the product"
YOUR PROBLEM: What were the two integers?
note: We assume that mathematicians do not make mistakes in matters of logic.
I will post the solution at a later date if nobody solves it.Pucc-H:Pucc-I:Pucc-K:ags@CS-Mordred.UUCP (11/17/83)
The S and P puzzle as stated has no solution. In fact, the terms are
contradictory. I will point out the contradiction and state a revised
version of the puzzle which DOES have a solution. I will post an answer
to the revised puzzle in a few days if no one gets it.
Here are the conditions on S and P as stated in the original puzzle:
The judge of the contest chooses two integers tells the mathematicians
"Both integers are greater than one"
"The sum of the two integers is less than forty"
"The product of the two integers is less than four hundred"
First of all, the third condition adds no new information. If the sum
of the two integers is less than 40 and both integers are greater than one,
we can already deduce that the product is not greater than 20*19 = 380.
Here is my revised version of the judge's instructions:
"Both integers are greater than one"
"Neither is greater than one hundred"
As before, the judge gives the sum to S and the product to P and asks them
to deduce both sum and product.
(1) S announces that P does not know the sum.
(2) P says he now knows the sum.
(3) S says he now knows the product.
There is enough information for the reader to determine a unique answer to the
revised problem. What is more surprising is that the answer to the revised
puzzle also satisfies the conditions of the original (namely, the sum is less
than forty). From this it does not follow that the same solution works for the
original problem. Here is why:
(From this point on I assume the original judge's instructions were given)
(1) When S announces that P can not determine the sum, we may deduce that
the sum is either 11 or 17. The products corresponding to 11 are
18, 24, 28 and 30. The products corresponding to 17 are 30, 42, 52,
60, 66, 70 and 72. Each of these products can be factored in more than
one way, yielding different sums. For all other sums between 4 and 39,
there is at least one product that can be factored uniquely.
(2) P has likewise determined that the sum is either 11 or 17. He has a
product that can correspond to only one of these, and therefore deduces
the sum. From this we can determine that the product is not 30, since
30 can be factored as 6 * 5 or as 15 * 2, yielding either 11 or 17 for
the sum. We are left with the following possibilities:
sum = 11, product in {18, 24, 28}
sum = 17, product in {42, 52, 60, 66, 70, 72}
(3) S now claims he knows the product. He must have miscalculated. Whether
his sum is 11 or 17, he cannot know the product.
If you have managed to read this far, your challenge now is to determine
why this reasoning no longer applies if the judge's original instructions
are changed as I have indicated.
Dave Seaman
..!pur-ee!pucc-k!ags