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
    Commented Feb 20, 2018 at 13:39
  • 1
    $\begingroup$ You are right, i have edited $\endgroup$
    – Untitpoi
    Commented Feb 20, 2018 at 13:42
  • $\begingroup$ damn I was about to write a script to solve this, but I saw the no-computers tag :( $\endgroup$ Commented Feb 20, 2018 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
    Commented Feb 20, 2018 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 $\endgroup$ Commented Feb 20, 2018 at 14:13

1 Answer 1


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
    Commented Feb 20, 2018 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 :| $\endgroup$ Commented Feb 20, 2018 at 14:16

Your Answer

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

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