28 Dec 04
Originally posted by AThousandYoungNo, he was not. He agreed with my general method earlier on. Why then would he explain the LOGIC over again if he thought I knew what I was doing, at least logically? Wouldn't the most likely cause of an error on my part be a programming error? You see, he offered no math to show why (4,61) isn't a solution, so that didn't help me very much.
He was specific.
Now, please do read my post above this--I made one error in thinking that the 7+58 case had a bearing on the solution, and I offered (hopefully a correct!) explanation of the real reason why (4, 61) is now a solution.
I think it's fair that if I go through the trouble of writing a program, you should provide analysis of at least the (4, 61) case before you claim that I am wrong. After all, if you did not perform any analysis, how do you really know I am wrong?
28 Dec 04
I may be missing something here, but in the original problem, Mr. Product did not know either number. That would seem to indicate that both n1 and n2 are prime numbers greater than 1. That limits the candidates to only 100>x>1. In all the solutions above, it was assumed that only 1 of the numbers was prime. Can someone please explain to me what I misunderstood on that point?
(the solution that I came up with was 7 and 41, but I admittedly made some logical "leaps" to get there)
28 Dec 04
1 edit
Originally posted by BLReidP knows the product of n1*n2. This means he knows all of their factors. If n1 and n2 were both prime, then the product would have only two factors - n1 and n2. Since he doesn't know n1 and n2, one of them must not be prime.
I may be missing something here, but in the original problem, Mr. Product did not know either number. That would seem to indicate that both n1 and n2 are prime numbers greater than 1. That limits the candidates to only 100>x>1. In all the so ...[text shortened]... nd 41, but I admittedly made some logical "leaps" to get there)
29 Dec 04
3 edits
Originally posted by AThousandYoungAcceptable sum S (S=n1+n2) must satisfy the two conditions ;
OK. That works. CoolPlayer, what other number combination would have been possible from SUM's point of view as of the end of step 3?
(a) S must be odd.
(b) S -2 must be composite( non-prime).
With (4,61) , from SUM's point of view , after PRODUCT's second assertion, at least two solutions are possible viz (4,61) and (29, 36).
29*36= 12*87= 4*261 = some other products with both even factors.
Obviously here only 65 is acceptable sum. For 12*87 the sum would be 99 which is not acceptable. Similarly with 4*261 the sum is 265 ,which is also not acceptable ,as 263 is prime. With all other factors the sum is even and hence not acceptable. Hence with (4,61) , SUM could not have made the second declaration. Hence (4,61) is not a solution.
Some how, the condition involved in the second statement of SUM has not been properly or fully incorporated in your computer program, Mr. BigDoggProblem.
29 Dec 04
2 edits
Originally posted by BigDoggProblemI have clarified my point in my last post above.
No, he was not. He agreed with my general method earlier on. Why then would he explain the LOGIC over again if he thought I knew what I was doing, at least logically? Wouldn't the most likely cause of an error on my part be a progra ...[text shortened]... u did not perform any analysis, how do you really know I am wrong?
29 Dec 04
1 edit
Originally posted by BLReidThis is the crucial point which everybody is overlooking. PRODUCT while considering the possible admissible sums , will factorise the known value of the product, into only those possible factors n1 and n2 which satisfy the
That limits the candidates to only 100>x>1.
condition 1<n1<100 as well as 1<n2<100. And with such possible factors only he will examine whether the sum is acceptable or not.
29 Dec 04
Originally posted by ranjan sinhaIn this view, for (4,61) there will be many posible and admissible solutions for PRODUCT, in SUM's viewpoint at the end of step 3 , thereby
This is the crucial point which everybody is overlooking. PRODUCT while considering the possible admissible sums , will factorise the known value of the product, into only those possible factors n1 and n2 which satisfy the
condition 1<n1<100 as well as 1<n2<100. And with such possible factors only he will examine whether the sum is acceptable or not.
making (4,61) a non-solution .
29 Dec 04
1 edit
Originally posted by CoolPlayerNow I think I see your error. 29*36 also equals 9*116, which you did not list. 9+116=125, which is an acceptable sum, thus giving two acceptable sums (65 and 125)--thus Mr. SUM knows (29, 36) cannot be a solution.
Acceptable sum S (S=n1+n2) must satisfy the two conditions ;
(a) S must be odd.
(b) S -2 must be composite( non-prime).
With (4,61) , from SUM's point of view , after PRODUCT's second assertion, at least ...[text shortened]... even factors.
Obviously here only 65 is acceptable sum.
To everyone else - please note that I agreed with Plumber that (4, 13) is the only solution to the original problem. This sidebar with CoolPlayer started when I mentioned I had written a program to solve the problem and he asked what would happen if we allow n1 and n2 to go up to (999, 999).
My conclusion is that the problem is no longer sound once you allow n1 and n2 to go up to 437 or higher.
It's a pity that Ranjan Sinha did not explain how he would get the solution to the original problem. I hope there is a more elegant method than using a program.
30 Dec 04
3 edits
Originally posted by BigDoggProblemYes, There indeed is an elegant way of solving this puzzle. This puzzle was posed by our mathematics professor in our class, in Bombay University, way back in the late eighties when i was astudent. It has a unique solution (4,13)
Now I think I see your error. 29*36 also equals 9*116, which you did not list. 9+116=125, which is an acceptable sum, thus giving [b]two acceptable sums (65 and 125)--thus Mr. SUM knows (29, 36) cannot be a solution.
To everyo ...[text shortened]... oblem. I hope there is a more elegant method than using a program.[/b]
The way to solve it involves guided application of logic as well as number theory, and thereby narrowing down the possible candidates for solution. By applying the properties of prime numbers and combining them with the conditions at step no. 1, 2,3, and step 4. you arrive at the unique solution.
Step 1( First statement of PRODUCT) implies that the product P (P=n1*n2) is factorisable in at least two ways.That is P is not a simple product of 2 primes. Step 2( First statement of SUM) implies that the sum S(S=n1+n2) cannot be expressed as a sum of 2 primes. This in turn implies that S will be odd and S-2 must be composite number.(Let us call these twin conditions, the acceptability criterion of the sum.).
Thus one of the two factors is even & the other is odd. Without loss of generality let us take n1 as the even factor and n2, the odd factor.
Step 3(second statement of PRODUCT) implies that of all the different ways of factorisations of P, in only one case the sum of factors satisfies the acceptibility criterion above.
It further implies that n1 is not just even, it will have to be evenly even, that is n1 cannot be oddly even. Thus n1 must be a multiple of 4 i.e.of the type 4k where k is some natural number.
While examining the possible factorisations of the product ( known to PRODUCT), he considers only such factorisations where both factors are below hundred and from among such possible factors, if in only one factorisation, the sum of factors satisfies the accebility criterion of the sum , he knows this is it.
Step 4(second statement of SUM) he examines all products
P1,P2,P3,.... of the type 4*(S-4), 8*(S-8), 12(S-12)....
where Pi = 4i*(S-4i); i = 1,2,3,........
For each such products Pi, he considers all possible factorisations.(Pi is not P*i, it is just sub-cased i th product under examination).
Pi = 4k*(Pi/4k); k= 1,2,3....taking only the possible values of k and also taking care that both factors are <100.
If in only one Pi , only one sum satisfies acceptability criterion and for all other Pi's more than one sum satisfies such acceptibility criterion, then only SUM could have validly made his second statement.
Appying the above , considerably reduces the possible domain of search and it can be done on a spreadsheet.
I didn't give the solution ,earlier, on the thread , hoping that other smart guys might try and come up with some smarter solution. I too hope that some still better and smarter way to solve it can definitely be found.
By the way the logic applied by Plumber is not accurate.
30 Dec 04
Originally posted by BigDoggProblemI concede my mistake. I overlooked the possible factorisation
Now I think I see your error. 29*36 also equals 9*116, which you did not list. 9+116=125, which is an acceptable sum, thus giving [b]two acceptable sums (65 and 125)--thus Mr. SUM knows (29, 36) cannot be a solution.
To everyone else - please note that I agreed with Plumber that (4, 13) is the only solution to the original problem. This sidebar ...[text shortened]... he solution to the original problem. I hope there is a more elegant method than using a program.[/b]
9*116. However if you take into account the logical flaw pointed out by the poser of this puzzle, my contention stands correct, though i did not have that in my mind at that point of time..
30 Dec 04
Originally posted by CoolPlayerLogical flaw? I sure hope you're not talking about limiting n1 and n2 to 99, because it was your suggestion to see what happens when n1 and n2 go up to 999. And now you claim your "contention stands correct". Hmm. That would be downright weasel-like behavior.
However if you take into account the logical flaw pointed out by the poser of this puzzle, my contention stands correct, though i did not have that in my mind at that point of time..
Again, please be specific. Which "logical flaw" are you referring to?
30 Dec 04
1 edit
Originally posted by ranjan sinhaThe problem with programs is that they're only too happy to show you an exception to a rule. It is false to say that there are no even acceptable sums. 174, 182, 184, 188, 190, 192, 196, and 198 are all acceptable. Why? Let's start with 174. The summing combinations can no longer start with something like 2+172, because 172 is too large for n1 or n2. So you have to go all the way up to 75+99, 76+98, etc. until 87+87. The numbers are sufficiently large to avoid a prime+prime combination, thus, 174 is acceptable.
Step 2( First statement of SUM) implies that the sum S(S=n1+n2) cannot be expressed as a sum of 2 primes. This in turn implies that S will be odd and S-2 must be composite number.(Let us call these twin conditions, the acceptability criterion of the sum.).
This seemingly has no bearing on the solution, but it shows how careful you must be in a problem of this complexity.
30 Dec 04
Originally posted by BigDoggProblemI just realized that I have not run the program since I set it to exclude sums with n1 or n2 greater than 100. Now it claims that (96, 99) and (97, 99) are solutions! I don't have time to analyze them by hand right now, but this is an interesting development nonetheless.
This seemingly has no bearing on the solution, but it shows how careful you must be in a problem of this complexity.
Could it be that the problem is unsound??