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When I acquired my Mercedes-Benz car in Germany, the first thing I had to do was to get a license plate. The plate I got had a peculiar number on it. It consisted of 5 different numbers and by mistake when I fixed it upside down the number could be still read, but the value had increased by 78633.
What was my actual license number?

Solve it fast - really fast!

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    $\begingroup$ Solved fast, really fast, as requested :-) $\endgroup$ Commented May 30, 2015 at 21:23
  • $\begingroup$ And that's why number plates are alphanumeric. :p $\endgroup$
    – Eli
    Commented May 30, 2015 at 23:35
  • $\begingroup$ I wrote up a brute force program which solves it: www.pastebin.com/59004630 (Lua) $\endgroup$
    – warspyking
    Commented May 31, 2015 at 13:45
  • $\begingroup$ @warspyking good one buddy :) $\endgroup$
    – jatin2302
    Commented May 31, 2015 at 20:05
  • $\begingroup$ @jatin ty ${}{}$ $\endgroup$
    – warspyking
    Commented May 31, 2015 at 21:52

1 Answer 1

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It was originally

10968.

The upside-down value is

10968+78633=89601.

Proof of uniqueness

The upside-down number must be at least 10000 bigger than 78633, so its first digit must be 8 or 9 while the first digit of the original number must be 1 or 2. If the first digit or the upside-down number was 9, then the last digit of the original number would have to be 6, so the last digit of the upside-down number would have to be 3+6=9, so the first digit of the original number would have to be 6, which is too large. So the first digit of the upside-down number, and hence the last digit of the original number, are 8, which means the last digit of the upside-down number, and hence the first digit of the original number, are 1.

Now the original number must be less than 90000-78633=11367, so its second digit must be 0 or 1. But we know it has five distinct digits, so the second digit of the original number, and hence the fourth digit of the upside-down number, are 0. By considering the fourth and fifth columns of summation, we now see that the fourth digit of the original number is 6 and hence the second digit of the upside-down number is 9.

Now we have 10?68+78633=89?01, where the two ?'s are each other upside down and therefore must be both 2, both 5, or 9 and 6 respectively (bearing in mind that each number has 5 distinct digits). We can check these 3 cases by hand and verify that 9 and 6 is the only one that works.

QED.

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    $\begingroup$ In afdition to your proof I wrote up a brute force program which proves it: pastebin.com/59004630 $\endgroup$
    – warspyking
    Commented May 31, 2015 at 3:47
  • $\begingroup$ It's wrote in Lua btw $\endgroup$
    – warspyking
    Commented May 31, 2015 at 3:48

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