Math problem: two unknown numbers, known product and sum (resumed closed thread)

[CraigD]

The 3/8/85 thread 1896 closed without a conclusive answer to its starting question.

 

Here’s the problem

There are two unknown whole numbers, m and n, both greater than 1, and less than 100. One mathematician, Mr. Product is given the product of these two numbers, while another mathematician, Mr. Sum is given the sum of these two numbers. The following conversation takes place:

Mr. Product: I do not know the numbers.

Mr. Sum: I knew you didn't know the numbers.

Mr. Product: Now I know the numbers.

Mr. Sum: Now I know the numbers, too.

 

What are the numbers?

It is suggested that a solution of “4 and 13” is correct and unique.

 

My analysis disagrees, both that 4,13 is a correct solution, and that the problem has a unique answer. My analysis produces the following 11 possible solutions:

Product=4558 Sum=139 Numbers: 53,86
Product=4582 Sum=137 Numbers: 58,79
Product=5518 Sum=151 Numbers: 62,89
Product=5494 Sum=149 Numbers: 67,82
Product=6862 Sum=167 Numbers: 73,94
Product=6586 Sum=163 Numbers: 74,89
Product=7426 Sum=173 Numbers: 79,94
Product=7954 Sum=179 Numbers: 82,97
Product=7138 Sum=169 Numbers: 83,86
Product=7654 Sum=175 Numbers: 86,89
Product=9118 Sum=191 Numbers: 94,97

My analysis is based on the following constraints, extracted from the text

Mr. Product: I do not know the numbers.

[1] Product has more than 2 factors

[2] Product isn’t the product of 3 of the same prime

Mr. Sum: I knew you didn't know the numbers.

[3] Sum is odd

[4] Sum not sum of 2 primes

[5] Sum not 3 times a prime

Mr. Product: Now I know the numbers.

[6] Product has 3 factors (one of them 2)

[7] Product uniquely identifies the 2 numbers

Mr. Sum: Now I know the numbers, too.

[8] Sum uniquely identifies the 2 numbers

Here’s the MUMPS code I used for the analysis

S PP="2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97" ;a list of the primes <100
F I=1:1:$L(PP," ") S X=$P(PP," ",I) F J=1:1:$L(PP," ") S Y=$P(PP," ",J) S S(X,Y,3)="" ;[1]
F I=1:1:$L(PP," ") S X=$P(PP," ",I) S S(X,X*X,4)="" ;[2]
F A=1:1:99 F B=1:1:99 s:A+B#2=0 S(A,B,5)="" ;[3]
F I=1:1:$L(PP," ") S X=$P(PP," ",I) F J=1:1:$L(PP," ") S Y=$P(PP," ",J) F A=1:1:99 F B=1:1:99 S:A+B=(X+Y) S(A,B,7)="" ;[4]
F I=1:1:$L(PP," ") S X=$P(PP," ",I) F A=1:1:99 F B=1:1:99 S:A+B=(X*3) S(A,B,8)="" ;[5]
K A F I=1:1:$L(PP," ") F J=1:1:$L(PP," ") S X=$P(PP," ",I),Y=$P(PP," ",J),(A(X*2,Y),A(Y,X*2),A(X,Y*2))="" ;[6]…
F A=2:1:99 F B=2:1:99 S:'$D(A(A,:)) S(A,B,10)="" ;…[6] continued
K C F A=2:1:99 F B=A:1:99 I '$D(S(A,:hihi:) S C(A*:hihi:=$G(C(A*:))+1 ;[7]…
F A=2:1:99 F B=A:1:99 i '$D(S(A,:cup:),C(A*:)>1 s S(A,B,11)="" ;…[7] continued
K C F A=2:1:99 F B=A:1:99 I '$D(S(A,:D) S C(A+B)=$G(C(A+B))+1 ;[8]…
F A=2:1:99 F B=A:1:99 i '$D(S(A,B)),C(A+B)>1 s S(A,B,12)="" ;…[8] continued
F A=2:1:99 F B=A:1:99 I '$D(S(A,B)) W "Product=",A*B," Sum=",A+B," Numbers: ",A,",",B,! ;output results

[8] disqualifies 4,13 as a solution, as it satisfies rules [1]-[7], but is indistinguishable by the sum 17 from 3,14 4,13 6,11 and 7,10, which also satisfy rules [1]-[7]

 

I’m unable to find a flaw in my analysis. Although it’s possible I failed to use a rule that would have allowed a unique solution to be found, that would not explain my rejection of 4,13 as a solution.

 

Can anyone find a flaw in my results? Or is the 4,13 unique solution is a “Math myth”, and my 11 non-unique solutions correct? :)

[snark1100]

suppose the number is 4558

this can be factored into 2*53*43 as a unique set of primes,

so mr product will know the only possibility < 100 for it's factorization

are 86 and 53, and so he does know the numbers, contrary to the assumption,

similarly for your other possible solutions.

 

52 has two possible solutions, (4,13) and (2,26)

and so product cannot find a unique solution,

but if the numbers are (2,26), sum will know

28, which, being even, could be the sum of two primes, 11 and 17,

but then product wouldl know the two numbers, as 11*17 is the only

factorization of 187, so the numbers cannot be (2,26).

now, when sum reports to product that sum knows that product

does not know the answer, mr. product can then rule out (2,26),

and so will know (4,13) are the two numbers. As for uniqueness, well, maybe i'll be back...

 

if the numbers are (3,14), mr. product will know 42, which has 3 possible factorizations-

6*7, 3*14, and 2*21

mr. sum will again know 17, and again know that mr. product cannot know the answer,

but this time, that information does not rule out any of mr. products possibilities for

the sum of each of his factors is odd, so he cannot determine which of his possible choices

is right. This rules out (3,14) as a possibility.

the same argument holds for (6,11) and (7,10) -

mr. product will know 66, allowing (2,33), (3,22) or (6,11) if (6,11) are the two numbers

mr. product will know 70, allowing (2,35), (5,14), (10,7) if (7,10) are the numbers.

I think that does it...(4,13) is the only possibility.