# Solução do Enigma da Edição de Março – Dois Matemáticos encontram-se numa estrada

Na última edição de Março partilhámos um enigma com a comunidade IST e pedimos respostas. O enigma era o seguinte:

<<Um sabe o produto de dois números inteiros que estão entre 2 e 50. O outro conhece a soma desses dois números. Ambos desconhecem que o limite superior é 50. Começam, entretanto, a falar:

– O que sabe o PRODUTO diz: “Não sei quais são os dois números.” – O outro diz: “Já sabia isso”
– O primeiro responde: “Agora já sei os dois números.”
– O segundo diz: “Também já sei os dois números” >>

“4 e 7”, “4 e 13” – estas foram algumas das respostas que recebemos e, pasmem-se, as duas podem estar corretas. É que, na verdade, este enigma é IMPOSSÍVEL. Ou, por outra, tem uma solução indeterminada.
De seguida, reproduzimos uma explicação disponível no fórum stackexchange, que, embora se refira a um enigma ligeiramente diferente, acreditamos que essas diferenças não introduzem consequências significativas na sua solução, ora vejamos:

<<
Someone imagined two positive whole numbers. Both numbers are greater than 11, and less than 2121. That person tells the sum of those two numbers to mathematician A, and the product of those two numbers to mathematician B. Couple of days later, A and B talk to each other:

A: There is no way for you to find the sum.
B: But I know the sum now!
A: And now I know the product.

Which two numbers have person imagined?

In fact, this problem is impossible.

The reason it is impossible is that there are multiple solutions and so the mathematicians would not be able to declare that they had discovered each over’s numbers.

The conversation gives us additional constraints on the solutions. In order for mathematician A to declare that there is no way that B could determine his sum that means he has ascertained that:

1. His sum is NOT the sum of two primes.
2. His sum can never be created using two numbers whose product has a unique factorisation in the interval [2, 20] i.e. 20+20=40, => 20*20=400 (I believe the upshot of this is that the sum must be less than 13 as 13=2+11 and 11*2 = 22 > 20)

(Please note I did the following on paper and may have made a mistake somewhere.)

From the bounds of the problem we know that the possible sums range from 4 to 40 inclusive. Using the first constrain we can eliminate many of these. If take all possible combinations of adding two primes (in the interval [2, 20]) and eliminate those sums we are left with the following as possible sums:

11, 17, 23, 25, 27, 29, 31, 33, 35, 37, 39, 40

Applying constraint 2 we find that the sum must be 11. This allows mathematician B to declare “But I know the sum now!”

In order for there to no unique factorisation we must be able to create the sum (11) using a non-prime in the interval [2, 20]. To find the solutions we must subtract each non-prime from the sum. For each valid solution the result will be in the interval [2, 20]. This leads to the following set of four solutions.

1. (4, 7) Product: 28 = 2 * 14.
2. (5, 6) Product: 30 = 3 * 10 = 2 * 15
3. (3, 8) Product: 24 = 4 * 6 = 2 * 12
4. (2, 9) Product: 18 = 3 * 6

>>

Texto: Afonso Anjos