Consider all possible 1000 digit positive base 10 integers all of whose digits are odd. For how many of these integers, do each pair of adjacent digits differ precisely by 2 ?
Note: Try to solve this problem analytically, although computer program/ spreadsheet solutions are welcome.
I came to the same 239 digit number* as did Charile but using a big number calculator to find the answer to the following equation plugging in 1000 for the value of
x such that
x is the number of digits and
n being the total number of distinct numbers of odd digits where adjacent digits have a difference of exactly 2 from their neighboring digits:
n = (8 + 6*(
x mod 2)) * 3
^{([x/2]  1)}, such that x > 1
The above equation was derived by examining the number of new numbers formed with each appendage of a digit, with [
A] as the FLOOR function of
A and (
A mod
B) as
A modulo
B.
(If I did my equation correctly, for 999 digits the solution would be 14*3
^{498}. Perhaps Charlie might be kind enough to confirm this.)
*
8*3^{499} =
9696077812231983157969404554544885098139569340
2670994774256095553548267177557054624529355367
9450367554884792067584102820076158173787015097
1464648244416540208005805666092704920033424756
4175501893260240990907073543274689029307039454
340293336.
Edited on September 23, 2008, 3:10 am
Edited on September 24, 2008, 4:55 am

Posted by Dej Mar
on 20080923 02:58:22 