The following logic puzzle has reportedly appeared in Omni Magazine. It is known as "the impossible puzzle", which was first published in 1969 by Hans Freudenthal, and the name impossible was coined by Martin Gardner. Gadner writes, " This beautiful problem, which I call 'impossible' because it seems to lack sufficient information for a solution...." Interestingly enough, the wording of the problem varies; in some cases, a slight rewording of the puzzle makes it ambiguous or insoluble. The puzzle presented here is indeed solvable, although it may not be easy to solve, and has been reworded to match Fredenthal's original problem constraints!
The Impossible Logic Puzzle
There are four extremely intelligent people*, Alice, Bob, Claire, and Daniel, having lunch in a restaurant. They enjoy playing logic games. One of them, Alice, informs the other three that she has thought of two integer numbers \(a\) and \(b\) greater than 1, with \(a\gt b\). Their sum is no greater than \(100\). Then she says, "The sum of the numbers is..." and she whispers the sum to Bob. Subsequently, Alice says, "The product of the numbers is..." and she whispers the product to Claire. Afterwards, the following conversation takes place:
Bob: Listen, Alice, I don't think that we know the two numbers.
Claire: Aha! Now I know the two numbers.
Bob: Oh, now I also know the two numbers.
Daniel: Well, now I too know the two numbers.
What are the numbers?
*Assume Alice, Bob, Claire, and Daniel are perfect logicians and have perfect mathematical knowledge.
Let \(S\) be the sum of the two integers \(a\) and \(b\). It is trivial that the sum \(S\in \{ n\in \mathbb{Z^{+}}|5\le n\le 100 \}\). From Bob's first statement, we may deduce that the \(a\) and \(b\) are not both prime numbers. Otherwise, Claire, knowing their product, by the unique prime factorization theorem, could've told them immediately apart. Nevertheless, Claire didn't speak, and Bob confirmed their initial ignorance, thereby eliminating every sum \(S\) that can be written as a sum of two prime numbers, leaving us with the following subset of possible values for the sum \(S\),\(^{1}\)
$$\{ 11, 17, 23, 27, 29, 35, 37, 41, 47, 51, 53, 57, 59, 65, 67, 71, 77, 79, 83, 87, 89, 93, 95, 97 \}$$
Claire knows the product \(P\). When she hears Bob's statement, she realizes that the sum \(S\) must be in the set above. Since every sum is odd, we now know that \(a\) must be odd and \(b\) even, or vice versa, by the properties of sums of even and odd integers. Of the possible values for the sum, \(23\) may be written in at least two different ways as \(2^{q}+p\), where \(q\ge 1\) and \(p\) is either a prime or a product of any number of odd primes. For instance, note that \(29=2+27=16+13\).
If one of these quantities is the sum, then Claire can now determine the two numbers because the product will be \(2^{q}\cdot p\), which can uniquely be broken one way into an odd and an even factor. However, if one of these quantities is, in fact, the sum, Bob can never know which representation of \(2^{q}+p\) corresponds to the values of \(a\) and \(b\). But, Bod does determine the values, thus the only possible value for \(S\) is \(17\) !
Now, we need to consider all the possible partitions of \(17\) into \(a+b\):
- \(17=2+15 \to P=30=2\times15=6\times5\)
- \(17=3+14 \to P=42=14\times3=2\times21\)
- \(17=4+13 \to P=52=4\times13\)
- \(17=5+12 \to P=60=12\times5=2\times30\)
- \(17=6+11 \to P=66=6\times11=2\times33\)
- \(17=7+10 \to P=70=10\times7=2\times35\)
- \(17=8+9 \to P=72=8\times9=2\times36\)
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