
17 and 23
A little variation on the problem here:
http://www.flooble.com/perplexus/show.php?pid=35
A sequence of digits is arranged in such a way that the first digit is a 6, and any two consecutive digits make up a number that is divisible by either 17 or 23.
What is the 28th digit in this sequence if the sequence is 100 digits long?
What about if the sequence were 29 digits long?
levik
Wednesday, May 01, 2002
The only pairs you can use for the all but the last 4 digits of a sequence are: 23, 34, 46, 69, and 92. This is because if you ever start down the path of 68, 85, 51, 17, you end up stuck, because there's no twodigit number starting with 7 that's divisible by 17 or 23.
So you repeat a 5digit sequence: 692346923469234 etc.
For the last 4 digits of a sequence, it doesn't matter if you go down the other path because it doesn't matter if you get stuck on the 7 if it's the last digit.
So in a 100number sequence, the 28th digit is going to be a 2 (the sequence restarts on the 26th digit). But in a 29number sequence, it could either be a 2 or a 5. Since the 26th digit is a 6, the next two numbers could either be 92 or 85.
Chris Shieh
Thursday, May 09, 2002
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