June 6th, 2013, 03:01 PM  #1 
Math Team Joined: Apr 2010 Posts: 2,778 Thanks: 361  new riddle
Hard to think of something for you guys. I have one, but some of you may already have seen it. Find x such that 3^x starts with 8 leading 6's, so 3^x = 66666666..... Or give a nonbruteforce procedure to find it. 
June 6th, 2013, 03:50 PM  #2 
Senior Member Joined: Sep 2012 From: British Columbia, Canada Posts: 764 Thanks: 53  Re: new riddle
Does x have to be an integer, because the context of the question makes it seem like it does?

June 6th, 2013, 04:03 PM  #3 
Math Team Joined: Apr 2010 Posts: 2,778 Thanks: 361  Re: new riddle
Sorry, yes, x must be an integer. Positive actually.

June 7th, 2013, 01:40 AM  #4 
Math Team Joined: Apr 2010 Posts: 2,778 Thanks: 361  Re: new riddle
Ok, the question needs more specification. Find a positive integer x such that 3^x has at least 8 leading 6's. I.e. 3^x = 66666666.... or give a nonbruteforce procedure to find it. Extra option: Prove that such a value for x exists. 
June 7th, 2013, 07:20 AM  #5 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: new riddle
OK, so log_{10} 3 is about 0.477. Twice this is about 0.954, close to 1 since 9 is a decent approximation of 10. We want to work mod 1 here, so this is really 0.0457. Now let's find a decent positive approximation. Probably we need an odd number so we can start with .477 and reduce it by .0457 a set number of times. .477/.0457 is about 10, so let's try 2*10+1 = 21: log_{10} (3^21) = 10.0195... or 0.0195 for our purposes. Looks like we need about one of the former and two of the latter, so 21*2+2 = 44 is a good exponent for approximating a power of 10, with value 0.00666. Three of those with one of 21 gives 3*44+21 = 153 which yields 0.000448 mod 1. We need a decent positive approximation, so poking around I found 109 which is over by 0.00621. OK, so that should make finetuning better when I find something close to log_{10} (20/3) = 0.8239...k, but I haven't found that yet. 2 is vaguely decent, being too large by 0.130. 8 is better, with 0.00693 excess. Now I can use the approximations from the first paragraph to tweak this, since 0.00621  0.00693 is small (in magnitude), so 109+8 looks promising. Well, I'm not going to take this to a full solution but it should make the approach clear. At each step you can improve the approximation of 10^n, coming up with one power of 3 which is just below 10^n and another which is just above 10^m (for some n and m which don't matter). Once these are precise enough you can use them to get better and better approximations of 2/3 * 10^n. 
June 7th, 2013, 04:39 PM  #6 
Math Team Joined: Apr 2012 Posts: 1,579 Thanks: 22  Re: new riddle
Don't know the answer, but while fiddling with it had some fun with 987654321*9 Ok, back to work! 
June 7th, 2013, 04:43 PM  #7 
Senior Member Joined: Sep 2012 From: British Columbia, Canada Posts: 764 Thanks: 53  Re: new riddle
Does CRG's solution count as brute force? If so, then no point in trying it for me. If not, then the question can pretty much be answered in a matter of minutes.

June 7th, 2013, 07:54 PM  #8  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,631 Thanks: 735  Re: new riddle Quote:
 
June 8th, 2013, 10:44 AM  #9 
Math Team Joined: Apr 2010 Posts: 2,778 Thanks: 361  Re: new riddle
I don't count CRG's solution as bruteforce. It's a good solution. You can ask a new one CRGreathouse.

June 8th, 2013, 11:10 AM  #10  
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Re: new riddle Quote:
 

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