# A magical operation #2

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?

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

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 $\boxed{776}$.