A tedious task
- Write this number on a piece of paper:
12934996293713980777956211320414507833168958121509352023936597191219256371203395909015136942592299897341739965172537947539094640548058114108344536009517557472657951376125193218497301739825284484200587100843036070437186169123281989270447213468773197364177545732400566354286573812353215825719736176201107229662483254843916868063633231803653049906331265155973799421340095757296913084020666317153543090048344557354229610746075393960639170405640508149789223361366489510025142218946265579225383722572551826544002518120526273969108006400395519943443915451626412808939875355655305864196751610859649120113057173349048339651188978924574050918175504509711167331402101689837630663894139276885179838726146760019028061011507576181410180919283778341140685458575850675934204580486450746220191218883551691042709777420171839895634209399719781235776969190460995625777711826833561305159693107745730395569619432484234085266635968637393382052421052845612509046583452372172633658310561964319100821209704459379034026359063872072140388436561079922479696324277594082319639007737836595198711067763925983345012095305024850873782566817553375119553395393676346463233021149844976494929260364488377790578452096139797843400612320281551607107939626470207832767212256432333567614976276260289205090266789253604531670393182535559755013087796302012372829339471398413066311459138614151872154655628153436832950575663923189543552027655525776192337898763800375948613929701903024012268382303899498661897516192938283503843137624483153789551737489944836344551476954974894767203824679774953849666679904493875312925974368917558431331569938719026264693263855693898709733131818004909093425803007469067016674892746352809728063768655007518332423543
- OK.
- Erase its rightmost digit and add five times that digit to what remains.
- OK, I erase the 3 and add 15. Done!
- Here is the challenge. Repeating this process will you ever reach this number: 131169120753693435909543.
- I need some help.
- The first number is 7^2007 and the last 2007^7.
Problem source: Problem of the Fortnight, San Diego State University.
May 29th, 2009 at 1:57 pm
Hardly, as the number of digits would need to decrease, and the process never does that.
Have you changed time zone, Jan? I’m in Britain, where your problems used to appear at 5pm BST (4pm UTC). I see today’s puzzle has already appeared and it’s not yet 8am here.
May 29th, 2009 at 2:02 pm
Yes, it seems, but only by an hour. I see that my earlier message is timestamped 1:57pm, which is correct for timezone GMT+7.
May 29th, 2009 at 6:54 pm
Looking at the last few digits in each case, we have X=100A+10B+C. Each iteration of this problem we replace 100A+10B+C with 10A+B+5C. Clearly this does in fact reduce its size over time.
The last few digits are:
423543
The next few iterations are:
42369 (42354+15)
4281 (4236+45)
433 (428+5)
58 (43+15)
Can it ever become 131169120753693435909543?
I dunno…. I’d have to ponder that for a while and if I allow myself to dive in, I may miss a few hundred important work e-mails. Still, I wanted to show that this is indeed plausible.
May 29th, 2009 at 7:59 pm
No, unless I missed something in my spreadsheet. I parsed through this by hand and ended up with a final iteration of 49 after 1,696 runs.
49 converts to 4+(9×5)=45 which obviously repeats indefinitely.
To determine this, I just had to do it with brute force. I pulled a standard # of digits off the end (10), verified by spreadsheet that the initial digit of those 10 isn’t a ’0′, ran it through 6 iterations, then concatenated it back onto the end of the initial number.
May 29th, 2009 at 9:54 pm
Richard, no time zone change. I am still in Thailand.
It is a mystery why you could see the post early. Maybe I posted it first with the wrong time stamp.
I post around midnight local time so in the UK, with summer time, you should get them at 6PM your time.
May 29th, 2009 at 10:06 pm
Michael, that it ends at 49 does not imply that it never reached 131169120753693435909543.
May 29th, 2009 at 10:11 pm
Only the last couple of digits change in each iteration. I could check the spreadsheet to verify this but I see no way to have ever hit 1311 as the starting four digits.
May 30th, 2009 at 6:36 am
I agree. My javascript program is at http://easyquestion.net/files/calc.html.
Scroll down to the end and you will see that only two changes were made to the initial digits. What a disappointment!
What, then, is the beauty of this problem? My guess is that the answer can be found by some clever reasoning instead of a lot of calculations. Any suggestions?
May 30th, 2009 at 7:50 am
That was my guess as well… and originally I was trying to work it out in my head but then just resorted to using Numbers to parse through it.
I’m pretty sure any shortcut here uses the fact that every iteration what is added must end with a 5 or 0 but I couldn’t figure out where to go from there.