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Topic: a 101-digit number (Read 580 times) |
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BNC
Uberpuzzler
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a 101-digit number
« on: Feb 23rd, 2003, 11:26am » |
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Consider this 101-digit number:
1. It's leftmost (MSB) digit is 6
2. Every 2 adjacent digits of the number are a 2-digit number that is divisible by either 17 or by 23.
What is the rightmost (LSB) digit?
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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aero_guy
Senior Riddler
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Re: a 101-digit number
« Reply #1 on: Feb 23rd, 2003, 3:53pm » |
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First we write out all the two-digit numbers that have 17 or 23 as a factor.
17
23
34
46
51
68
69
85
92
If we start with a six we have two possibilities,
1) 68517 terminates since there is no number on the list that starts with 7
2) 692346 repeating
So, the second set must be used, at least for the most part. It has five repeating digits, so at 100 we will just have finished the twentieth set. Since we need one last digit, the last pattern must be type 2 as well, which gives us a 6 at the end.
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