wu :: forums « wu :: forums - String of Nines » Welcome, Guest. Please Login or Register. Jan 11th, 2025, 10:30am

RIDDLES SITE WRITE MATH! Home Help Search Members Login Register

wu :: forums    riddles    easy (Moderators: towr, ThudnBlunder, Icarus, SMQ, Grimbal, Eigenray, william wu)    String of Nines « Previous topic | Next topic » Pages: 1  Reply Notify of replies Send Topic Print    Author  Topic: String of Nines  (Read 438 times) Sir Col Uberpuzzler impudens simia et macrologus profundus fabulae     Gender: Posts: 1825 String of Nines   « on: Jun 20th, 2003, 12:30pm » Quote Modify Given that k is a positive odd integer, prove that (i) 9k ends in 9. (ii) 99k ends in 99. (iii) 999k ends in 999.   Generalise. « Last Edit: Jun 20th, 2003, 12:42pm by Sir Col » IP Logged mathschallenge.net / projecteuler.net NickH Senior Riddler     Gender: Posts: 341 Re: String of Nines   « Reply #1 on: Jun 20th, 2003, 1:20pm » Quote Modify :: It follows by considering the results, respectively, mod 10, mod 100, mod 1000.  For example, for (iii), (-1)k = -1 (mod 1000) if k is odd.  (And equals 1 if k is even, of course.)  Generally, (-1)k = -1 (mod ar) if k is odd, so the number represented in base a by a string of r (a-1)s will end in the same string of (a-1)s when raised to an odd power.  (If a is an integer > 1.)   Could also be proved using the binomial theorem. :: « Last Edit: Jun 20th, 2003, 1:22pm by NickH » IP Logged Nick's Mathematical Puzzles Sir Col Uberpuzzler impudens simia et macrologus profundus fabulae     Gender: Posts: 1825 Re: String of Nines   « Reply #2 on: Jun 20th, 2003, 1:42pm » Quote Modify Nice proof, NickH.     :: Of course, your conguence base 10n method is a simplification of the binomial theorem approach anyway.   For odd k, (10n–1)k=(10n)k–k(10n)k–1+...+k(10n)–1=(10 n)Q–1==–1 mod (10n). :: IP Logged mathschallenge.net / projecteuler.net Sir Col Uberpuzzler impudens simia et macrologus profundus fabulae     Gender: Posts: 1825 Re: String of Nines   « Reply #3 on: Jun 20th, 2003, 6:19pm » Quote Modify It can be verified that 99 is a 9 digit number, but 9949 has 98 digits and 9950 has 100 digits.   Can you find a value for k, where k is a natural number, such that 999k contains 999 digits?   What about (10m–1)k, containing 10m–1 digits? « Last Edit: Jun 20th, 2003, 6:34pm by Sir Col » IP Logged mathschallenge.net / projecteuler.net LZJ Junior Member     Gender: Posts: 82 Re: String of Nines   « Reply #4 on: Jun 21st, 2003, 8:37am » Quote Modify For 9993, I suspect that k = 999/3. Furthermore, for (10m - 1)k, k is only a natural number if (10m - 1)/m is a natural number (ie k = (10m - 1)/m) ...but I still need to work on it to be sure... IP Logged NickH Senior Riddler     Gender: Posts: 341 Re: String of Nines   « Reply #5 on: Jun 22nd, 2003, 5:46am » Quote Modify I think LZJ's result is correct... :: We require 1010^m-2 <= (10m - 1)k < 1010^m-1 Using the binomial theorem, (10m - 1)k = 10mk(1 - C(k,1)/10m + C(k,2)/102m - ...) If k = (10m - 1)/m, then C(k,1) < 10m, and it can easily be shown that C(k,i+1)/C(k,i) < 10m, for 1 < i < k. With a bit of hand waving (regarding the alternating signs, for one), the result follows. :: « Last Edit: Jun 22nd, 2003, 5:52am by NickH » IP Logged Nick's Mathematical Puzzles Sir Col Uberpuzzler impudens simia et macrologus profundus fabulae     Gender: Posts: 1825 Re: String of Nines   « Reply #6 on: Jun 22nd, 2003, 4:52pm » Quote Modify Good spot, LZJ! It's a really nice result and surprising, isn't it?   I like your method, NickH, although a deductive proof that leads to the solution would be nice. Perhaps someone can tidy up my 'proof' (I don't the hand waving that I do towards the end):   (10m–1)k lies between two consecutive perfect powers of 10. 10a < (10m–1)k < 10a+1   Solving 10b = (10m–1)k   b=k log(10m–1)   As a<b<a+1, it follows that a=[b] (the integer part function).   Therefore a=[k log(10m–1)].   As m–1<log(10m–1)<m, and is incredibly close to m...   *waves magic wand* [k log(10m–1)]=km–1 (I know, it's shameful, but it is true).   As 10a contains a+1 digits and we wish our number to contain 10m–1 digits, a=10m–2.   Therefore a=km–1=10m–2.   Hence k=(10m–1)/m. IP Logged mathschallenge.net / projecteuler.net Pages: 1  Reply Notify of replies Send Topic Print Forum Jump: ----------------------------- riddles -----------------------------=> easy   - medium   - hard   - what am i   - what happened   - microsoft   - cs   - putnam exam (pure math)   - suggestions, help, and FAQ   - general problem-solving / chatting / whatever ----------------------------- general -----------------------------  - guestbook   - truth   - complex analysis   - wanted   - psychology   - chinese « Previous topic | Next topic »

Powered by YaBB 1 Gold - SP 1.4!
Forum software copyright © 2000-2004 Yet another Bulletin Board