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June 6th, 2013, 03:01 PM   #1
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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 non-bruteforce procedure to find it.
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June 6th, 2013, 03:50 PM   #2
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Re: new riddle

Does x have to be an integer, because the context of the question makes it seem like it does?
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June 6th, 2013, 04:03 PM   #3
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Re: new riddle

Sorry, yes, x must be an integer. Positive actually.
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June 7th, 2013, 01:40 AM   #4
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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 non-bruteforce procedure to find it.
Extra option:
Prove that such a value for x exists.
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June 7th, 2013, 07:20 AM   #5
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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 fine-tuning 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.
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June 7th, 2013, 04:39 PM   #6
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Re: new riddle

Don't know the answer, but while fiddling with it had some fun with 987654321*9

Ok, back to work!
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June 7th, 2013, 04:43 PM   #7
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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.
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June 7th, 2013, 07:54 PM   #8
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Re: new riddle

Quote:
Originally Posted by johnr
Don't know the answer, but while fiddling with it had some fun with 987654321*9
I like fun too: 123456789*9*8 + 80
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June 8th, 2013, 10:44 AM   #9
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Re: new riddle

I don't count CRG's solution as bruteforce. It's a good solution. You can ask a new one CRGreathouse.
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June 8th, 2013, 11:10 AM   #10
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Re: new riddle

Quote:
Originally Posted by Hoempa
I don't count CRG's solution as bruteforce. It's a good solution. You can ask a new one CRGreathouse.
So what is the numerical value of the exponent? Please post your method if different from [color=#BF0000]CRGreathouse [/color].

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