19
$\begingroup$

You are given the first 20 digits of π: 31415926535897932384. In each move, you can select a contiguous group of 5 digits and increase/decrease them all by the same integer, provided that each resulting digit stays between 0 and 9 inclusive. For example you can increase the first 5 digits by 2, giving you 53637926535897932384. However you cannot decrease the same 5 digits by 2 as that would result in negative digits. Can you bring every digit to 0?

To allow new users to solve this puzzle and earn reputation points, I encourage all users whose reputation is 200 or more to not post an answer until 48 hours after this question is posted. Thank you!

This is a variant of the question Reducing $\pi$ to zero by Dmitry Kamenetsky.

Improve this question
$\endgroup$
2
  • $\begingroup$ Is there any practical usefulness in solving such a problem? $\endgroup$
    – web adventurer
    Sep 26 at 5:27
  • 1
    $\begingroup$ @webadventurer I enjoy puzzles. I don’t care if they have any real world application. I would characterize myself as a pure mathematician not an applied mathematician. I do math because I enjoy its intrinsic beauty. $\endgroup$
    – Will Octagon Gibson
    Sep 26 at 8:01

1 Answer 1

Reset to default
38
$\begingroup$

At the time of writing, I am currently at 196 reputation, so I am just under the limit.

I believe that you cannot reduce the first 20 digits of $\pi$ to 0.

We start by adding all the digits in the question together, to get the number $97$. This is our starting sum $S$. We also notice that each time we increase a contiguous group of 5 digits by a number $k$, we increase $S$ by $5k$, where $k$ is some integer. (Reducing a group of 5 digits is the same as having a negative $k$).

Therefore, if we are only able to increase or reduce $S$ by multiples of 5 at each step, the last digit of $S$ can only ever be $7$ or $2$. However, for $\pi$ to be reduced to 0, $S$ must equal $0$. Therefore, it is impossible to reduce $\pi$ to 0 by only increasing/decreasing groups of 5 digits.

Improve this answer
$\endgroup$
5
  • 5
    $\begingroup$ Your answer is correct even if the 5 digits are not contiguous. $\endgroup$
    – Will Octagon Gibson
    Sep 23 at 21:43
  • 5
    $\begingroup$ Yes, I was just following the wording of the question $\endgroup$
    – SquareFinder
    Sep 23 at 23:43
  • $\begingroup$ It is common to use spoilers to give others a chance at completing it themselves. I believe you can add them with >! in front of your text. $\endgroup$
    – infinitezero
    Sep 25 at 6:15
  • $\begingroup$ @infinitezero I have now spoilered the explanation, do you think I should spoiler the answer as well? $\endgroup$
    – SquareFinder
    Sep 25 at 6:28
  • $\begingroup$ I think it's good this way :) $\endgroup$
    – infinitezero
    Sep 25 at 7:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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