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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 $x,y,z$?

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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.

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  • $\begingroup$ You removed S>= 8 and S=7 from possible answers and then deduce S=6,but what about S=5 ? $\endgroup$ – Woeitg Feb 21 '16 at 15:40
  • $\begingroup$ @Woeistg When $S\leq5$, there are not tuples with $P>S$. It is in the answer. $\endgroup$ – wythagoras Feb 21 '16 at 16:29
  • $\begingroup$ 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. $\endgroup$ – Duncan Feb 25 '16 at 21:38
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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 $(1,1,4)$.

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