mathematics - Deducing Two Numbers based on their Difference and Ratio

Yesterday I met the perfect logicians Divvy and Subtra. I told them:

I have chosen two positive integers $x$ and $y$ with $2\le x\lt y\le 100$ and with $x$ a divisor of $y$. I will now whisper the ratio $r=y/x$ into the ear of Divvy and I will whisper the difference $d=y-x$ into the ear of Subtra.

And then I whispered the ratio and the difference into their ears.

Then the following conversation took place:

• Divvy: I don't know the numbers.


• Subtra confirmed: I already knew this.


• Divvy: I already knew that you knew that I don't.


• Subtra: I still can't determine the numbers.


• Divvy: At the very beginning of this conversation, I knew that you did not know the numbers. I forgot to mention this. Does this information change something for you?


• Subtra: Yes, it changes everything! Finally I am able to determine the numbers!

Question: What on earth are these magic numbers $x$ and $y$?

Answer

We are dealing with four integers $x$, $y$, $r$ and $d$ that satisfy $y=rx$ and $d=(r-1)x$. Note that $x

(1) Divvy: I don't know the numbers.

• If $r\ge34$, then $x\ge3$ is impossible (as this would imply $y=rx\ge3\cdot34=102$). Hence this would imply $x=2$ and $y=2r$, so that Divvy knows $x$ and $y$ in contradiction to his statement.


• If $r\le33$, then $x=2$ and $y=2r$ as well as $x=3$ and $y=3r$ are feasible solutions. Divvy is not able to determine $x$ and $y$.

Divvy's first statement is equivalent to $2\le r\le33$.

(2) Subtra confirmed: I already knew this.

As Subtra's difference is $d=(r-1)x$, her statement means that $d$ cannot be written as the product of two factors $r-1$ and $x$, one of which is $\ge33$ and one of which is $\ge2$.

Subtra's statement means that the difference $d$ is either from the range $0\le d\le65$, or that $d$ is odd and from the range $67\le d\le97$. Equivalently, $d\notin\{66,68,70,\ldots,98\}$.

(3) Divvy: I already knew that you knew that I don't.

Divvy's statement means that there does not exist any integer $x$ with $2\le x\le100/r$, so that the corresponding difference $(r-1)x$ is an even integer in the so-called bad range $B=\{66,68,\ldots,98\}$. We make some case distinctions on $2\le r\le33$.

• If $r$ is odd, then $r-1$ is an even integer from the range $2,4,\ldots,32$. As the diameter of the bad range $B$ is $98-66=32\ge r-1$, the bad range contains an (even!) integer multiple $(r-1)x$ of $r-1$. Hence all such values $r$ collide with Divvy's statement.


• If $r=2$, then for any $x$ with $2\le x\le 100/r=50$, the corresponding product $(r-1)x=x\le50$ will not fall into the bad range. This case is possible.



• If $r$ is even with $4\le r\le24$, then $2(r-1)$ is an even integer from the range $6,10,14\ldots,46$. The values $6,10,\ldots,30$ are smaller than the diameter of the bad range, and hence have an integer multiple in $B$. The value $34,38,42,46$ respectively have the multiples $68,76,84,92$ in $B$. For a multiple of the form $2(r-1)k$ in $B$, we may choose $x=2k$. Hence all such values $r$ collide with Divvy's statement.


• If $r\in\{26,28,30,32\}$, then none of the multiples of $r-1\in\{25,27,29,31\}$ fall into $B$. Hence these four cases are possible.

Summarizing, we see that Divvy's second statement is equivalent to $r\in\{2,26,28,30,32\}$.

(4) Subtra: I still can't determine the numbers.

At this point of the conversation, we are left with the following options for the pairs $(x,y)$:

• $r=2$: The pairs $(x,2x)$, where $x\in\{2,3,\ldots,50\}$. Then $d=x$.



• $r=26$: The pairs $(2,52)$ and $(3,78)$. Then $d=50$ or $d=75$.


• $r=28$: The pairs $(2,56)$ and $(3,84)$. Then $d=54$ or $d=81$.


• $r=30$: The pairs $(2,60)$ and $(3,90)$. Then $d=58$ or $d=87$.


• $r=32$: The pairs $(2,64)$ and $(3,96)$. Then $d=62$ or $d=93$.

Since Subtra is not able to determine the numbers $x$ and $y$ from this, her value $d$ must be compatible with two distinct pairs $(x,y)$ in this enumeration. This reduces the possible options for $(x,y)$ to the following short list:

• $r=2$ and $d=50$: The pair $(50,100)$.


• $r=26$ and $d=50$: The pair $(2,52)$.

Subtra's second statement means that $d=50$.

(5) Divvy: At the very beginning of this conversation, I already knew that you did not know the numbers.

We are only left with two possible values $r=2$ and $r=26$ for the ratio.

• If $r=2$, then at the very beginning of the conversation the pair $(x,y)=(2,4)$ with difference $d=2$ would have been a feasible option for Divvy. But then Subtra would have been able to deduce the two numbers right away, as $d=(r-1)x=2$ and $x\ge2$; a contradiction.


• If $r=26$, then at the very beginning of the conversation only the two pairs $(x,y)=(2,52)$ with $d=50$ and $(x,y)=(3,78)$ with $d=75$ would have been feasible options for Divvy. The first difference $d=50=(r-1)x$ is compatible with $(r,x)=(26,2)$, as well as with $(r,x)=(6,10)$. The second difference $d=75=(r-1)x$ is compatible with $(r,x)=(26,3)$, as well as with $(r,x)=(6,15)$. In either case, at the very beginning of the conversation Subtra is not able to deduce the numbers from the difference $d$.

Divvy's third statement implies $r=26$.

(6) Subtra: But now I can determine the numbers.

Subtra now knows that $d=50$ and $r=26$, and deduces the numbers.

Answer: The magic numbers are $x=2$ and $y=52$.