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TomasB Inner circle Sweden 1144 Posts |
I found the following fun sounding puzzle at http://www.fountain.btinternet.co.uk/nicola/puzzmore.html
Take two integers n and m, selected from the range of integers 2 thru 100. Give the SUM of the two (n+m) to person S, and the PRODUCT of the two (n*m) to person P. Neither S or P know the values of m or n. S says to P: "There is no way you can tell what my sum is" P says to S: "Then, I know your sum" S says to P: "Then, I know your product" What are the values of n and m such that the above statements are true? The appeal of this puzzle is, of course, that it appears to have insufficient information for the solution. Also, an intriguing point is that the first statement is true -- UNTIL S articulates it. Then it becomes false. |
Bong780 Regular user Toronto, Canada 172 Posts |
From the 1st statement:
S knows the sum is NOT a combination of 2 prime numbers. So at least one of m, n is non-prime. Then P couldn't get a definite answer. 2nd statement: P get the info from 1st statement, there must be a 2 combination of (m,n). From one set of (m,n), P can construct a list of sum S(m,n),(m+1,n-1),(m+2,n-2). One of the set would have a sum combination by 2 prime numbers (m+x=prime, n-x=prime where x<n-1) so P can eliminate this set of (m,n), which leave another set that the sum combination don't have both prime numbers. So P knows m and n now. 3rd statement: S now knows P have the answer. So he knows that the product of m,n has only 2 sets of answer. S checks each sum combination (m,n)(m+1,n-1)... There should be only one combination gives 2 sets of product. The way to have a 2-set product is like this, p*p*q, where p,q is prime factor of product, so m=p*p n=q OR m=p n=p*q only 2 sets) Then S checks the sum combination, there should be only one that fits in (p*p*q). By trial and error... I got m=4, n=13 Its a little messy so anyone can understand maybe able to clear up a bit. There should be a more logical ways to solve this, and that's the only way I can think of. I didn't do math for years. I can't think of a better solution. |
TomasB Inner circle Sweden 1144 Posts |
I think you did really well. I see no other option than to write the full tables and look through them to find the desired numbers. Maybe if someone here has studied Abstract Algebra they can come up with a neater method.
/Tomas |
Nir Dahan Inner circle Munich, Germany 1390 Posts |
I think I read someplace (possibly in a Martin Gardner column) that what is strange is that limit on the integers need not be 100. Even if the limit is 1 million, you still get the same answer. I am not sure if it was the same puzzle and I am currently at work and can’t search for it.
I will try to post later. |
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