Puzzling Stack Exchange is a question and answer site for those who create, solve, and study puzzles. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The teacher tells Summo and Prodo: "I have picked three positive integers $x\le y\le z$. I have whispered the sum $S=x+y+z$ into Summo's ear, and I have whispered the product $P=xyz$ into Prodo's ear." Now the following conversation takes place.

Summo: "I do not know $x,y,z$. But if I knew that your number $P$ is greater than my number $S$, then I would be able to determine $x,y,z$."

Prodo: "Aha! Actually my number $P$ is less than your number $S$. And I am able to determine $x,y,z$."

Question: What are these numbers $x,y,z$?

share|improve this question
up vote 13 down vote accepted

The fact $3z\geq x+y+z>xyz$ implies that $3>xy$, so $x=1$ and $y=1$ or $x=1$ and $y=2$. In the former one, we get $2+z>z$. Otherwise we get $3+z>2z$, so $z<3$ and then in fact $z=2$.

So the possible tuples are:

$(1,1,z)$ and $(1,2,2)$.

However, Summo can only deduce what the tuple is if he knows that $P>S$. It follows that for his sum, there is exactly one possible tuple with $P>S$ and at least one with $P\leq S$.

If $S\geq 8$, we can split it in two triples $(2,3, S-5)$ and $(2,2,S-4)$. For both we have that $P>S$. Hence $S<8$. When $S\leq 5$, there are no tuples with $P>S$. We either have $(1,2,2)$ for a sum of 5 and a product of 4, or $(1,1,k)$ for a sum of $k+2$ and a product of $k$.

If $S=7$, then both $(2,2,3)$ and $(1,2,4)$ were possible.

Hence $S=6$.

Now we can finish it.

The fact that $P<S$ now implies that $x=y=1$ and $z=4$, from the first part of my answer.

share|improve this answer
    
You removed S>= 8 and S=7 from possible answers and then deduce S=6,but what about S=5 ? – Woeitg Feb 21 at 15:40
    
@Woeistg When $S\leq5$, there are not tuples with $P>S$. It is in the answer. – wythagoras Feb 21 at 16:29
    
I'm vaguely dissatisfied with this answer, not because it's not correct, but because it doesn't follow the logic of Summo and Prodo. For example, you start with $3z \geq x+y+z > xyz$, but Summo doesn't know that for his opening statement and presumably Prodo doesn't know that until after Summo's first statement. – Duncan Feb 25 at 21:38

Solution:

$(x,y,z) = (1,1,4)

Here's my reasoning (following the reasoning of Summo and Prodo).

Half of Summo's statement:

Summo can't tell the numbers based on the sum, so that means they can't sum to either $3$ or $4$. Both of those would have unique digits ($1,1,1$ and $1,1,2$ respectively). That means $S \geq 5$.

Rest of Summo's statement:

The rest of Summo's statement tells us that based on his sum, $P$ could be greater than $S$ for exactly one combo of $x,y,z$, while there may be multiple ways for $P$ to be less than $S$. This rules out $S=5$, since for both solutions ($1,2,2$ and $1,1,3$) we get $P$ smaller than $S$. $S$ could be $6$, since we have possible solutions $(1,1,4), (1,2,3) \text{ and } (2,2,2)$, where $1$ of those has product greater than $S$.

I believe this means that $S=6$, since for any larger number there would be multiple products larger that $S$. Take $S=7$, for example. The possible solutions are $(1,1,5),(1,2,4),(1,3,3),(2,2,3)$. A number of those have products larger than $7$, and that will only get worse as $S$ gets higher.

Getting to Prodo:

Prodo now knows that $S=6$. She also says that her product is less than $S$, so $P\leq5$. For $S=6$, the only solution where $P\leq5$ is $(1,1,4)$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.