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James: These two cards have two numbers $X$ and $Y$, which are both one-digit numbers and are different. Can you find the sum?

Jack: No, that's impossible. Can I get a hint?

James: The last digit of $X$ and $Y$'s product in base 10 is the same as your pen's price.

Jack: I know, it's...

Find the sum only with the help of the dialog. The characters in the dialog were honest.

This is was inspired from this video: https://www.youtube.com/watch?v=8EITLJf1nug&ab_channel=MindYourDecisions. The video showed an adapted version of a 2015 Singapore Olympiad Challenge from Primary 5. The video is wrong; hence the post. The dialog is different here so that all copyright stuff is avoided. I have also made it significantly more mathematical so that the hint is contrasted and easier to notice.

So, can you outsmart a real mathematician? The title is also a hint.

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1 Answer 1

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Neither of $X,Y$ can be zero, because

$0\times Y=0$ for any $Y$, so the last digit of the product is always zero and could never have allowed Jack to determine the non-zero number. One could also argue that Jack's pen would not cost zero dollars.

So we can assume $X,Y$ are both non-zero digits. The following is true:

$X\times Y$ has the same last digit as $(10-X)\times (10-Y)$
For example $3\times4=12$ has the same last digit as $7\times6=42$.

The above relation therefore shows that there will remain at least two possibilities for $X,Y$ unless

$\{X,Y\}$ is actually the same pair of numbers as $\{10-X,10-Y\}$.

From this it follows that the sum of $X$ and $Y$ is

$X+Y=10$

This doesn't actually show that the situation described in the question can actually happen, so I should verify that there are actual values for which it could work:

$1\times9=9$, $2\times8=16$, $3\times7=21$, $4\times6=24$.
Obviously $1\times4$ and $1\times6$ are alternative ways to get a final digit of $4$ or $6$, so knowing that the product has last digit $4$ or $6$ does not provide Jack with a unique set of values for $X,Y$. Maybe a last digit of $1$ or $9$ will.
The only products of single digits ending in $1$ are $1\times1$, $3\times7$, and $9\times9$. Since the factors need to be different, $\{3,7\}$ would work as a unique answer when the price of Jack's pen is 1 dollar.
The only products of single digits ending in $9$ are $1\times9$, $3\times3$, and $7\times7$. Since the factors need to be different, $\{1,9\}$ would also work as a unique answer when the price of Jack's pen is 9 dollars.
We don't know which of the two possibilities is correct, but the sum is $10$ regardless.

Edit:
The above assumes that the numbers involved are not negative. There are single-digit negative numbers too of course. However, if negative numbers are allowed then the dialogue described in the question could not have occurred, as for every pair of single-digit numbers $X,Y$ there is a different pair of single-digit numbers $-X,-Y$ that give the same product. For the dialog to have occurred, the participants (or at least Jack) must have assumed or understood that negative numbers are not allowed, so we can make the same assumption when solving the puzzle.

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  • $\begingroup$ Sorry, I made a mistake in the question. Your answer is correct but the question is itself wrong, so not quite. $\endgroup$
    – Shambhav
    Jan 4, 2022 at 13:32
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    $\begingroup$ four times six isn't usually 36 though. $\endgroup$
    – Bass
    Jan 4, 2022 at 16:13
  • $\begingroup$ @Bass Oops! Fixed it now, thanks. $\endgroup$ Jan 4, 2022 at 16:19
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    $\begingroup$ @ShambhavGautam How is that wrong? I've already excluded 0, and for any non-zero pair of digits $X,Y$ whose product's last digit matches the price of the pen, $10-X,10-Y$ is also a non-zero pair of digits whose product's last digit matches the price of the pen. Those are two different solutions unless $X+Y=10$. $\endgroup$ Jan 5, 2022 at 12:49
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    $\begingroup$ @ShambhavGautam Now that you've changed the question to be about written numbers on cards instead of volumes, negative numbers are possible too, but that was not the case when I wrote my answer. Presumably that could not have been the reason why you thought the video was wrong, otherwise you would not have framed it using volumes. It is very bad form to change a question to invalidate already given answers. $\endgroup$ Jan 5, 2022 at 13:15

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