Quick reminder of what is a magical operation
A magical operation on a particular number (not ending by 0) is the addition of this number with his symetrical number. For example the magical operation for 2018 would be 2018 + 8102 = 10120.

The puzzle
After working on that Question about magical operation, I found an amazing number with 3 digits abc that gives the exact same number after 5 magical operations with the difference that it was surrounded by 9 and 8 giving the form 9abc8. Can you find it?

  • $\begingroup$ does xxx mean all the digits are the same? if not, you should use abc or xyz. $\endgroup$
    – Quintec
    Feb 20 '18 at 13:39
  • 1
    $\begingroup$ You are right, i have edited $\endgroup$
    – Untitpoi
    Feb 20 '18 at 13:42
  • $\begingroup$ damn I was about to write a script to solve this, but I saw the no-computers tag :( $\endgroup$ Feb 20 '18 at 13:50
  • $\begingroup$ Yeah the script is easy, I think I found the number but I have no clue how to get it without a computer. $\endgroup$
    – Doomenik
    Feb 20 '18 at 13:54
  • $\begingroup$ I can tell you there's a 6 digit number that has the exact same propriety, if you want next level $\endgroup$ Feb 20 '18 at 14:13

This is a very bad solution, but I've done similar problems in the past so I made a few assumptions and used intuition.

I take the first and last digits, a and c. I'm already guessing the number is above 500, and assuming a+c > 10. Then the first and last digits after one magical operation become 1 and a+c-10. After 2 magical operations, the first and last digits both become a+c-9, assuming the numbers are not extremely big(a rather safe assumption). From here it's a lot of casework - I'll only detail the correct path. I'll also refer to the first and last digits only as "the digits". After trying making the digits double the previous digits and failing, I realized there was probably carrying involved so the digits become 2a+2c-18 and 2a+2c-17. Now my intuition tells me the digits probably have gotten big enough to exceed 10 so the next digits will be 1 and 4a+4c-45. Again, after casework we realize there is carrying here and the digits become 4a+4c-44 and 4a+4c-43, which add up to 17. Solving we find a+c=13. Making the assumption that b must be rather big because of the carrying involved, we try several cases and get lucky(yay! tried a=7 c=6 first) to find the answer as $\boxed{776}$.


I used a calculator and defined the magic function - I hope this still counts as no computers as it was only saving me a few keystrokes each time I was trying cases.

  • $\begingroup$ @Flying_whale aahhhhhhh there's more??!!!? :P this took far too much time and bashing to do, i'll pass :P $\endgroup$
    – Quintec
    Feb 20 '18 at 14:13
  • $\begingroup$ haha, just tried for fun with a perl script, rigth now the script is running to find a 8 digit one, it begins to take serious execution time :| $\endgroup$ Feb 20 '18 at 14:16

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