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Odd Digits 2 Differ Adjacently (Posted on 2008-09-22 )

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.

Submitted by K Sengupta
Rating: 4.0000 (1 votes)
Solution:
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Praneeth has submitted a solution in this location .
We will extend this to n-digit positive decimal (base 10) integers in general having the desired property, and denote:
E_n = the number of such n-digit numbers with last digit 1 or 9, and: :
F_n = the number of such n- digit numbers with last digit 3 or 7 :
G_n = the number of such n-digit numbers with last digit 5
Then, we must have: :
E_(n+1) = F_n; :
F_(n+1) = 2* G_n + E_n, and: :
G_(n+1) = F_n :
Accordingly F_(n+2) = 2*G_(n+1) + E_(n+1) = 2*F_n + F_n = 3*F_n :
Evidently, F_1 = 2(3, 7), and F_2 = 4(13, 53, 57, 97) :
Accordingly, we must have: :
F_(2n) = 4*3^(n-1); F_(2n-1) = 2*3^(n-1). :
Therefore, E_(2n) = G_(2n) = 2*3^(n-1), and consequently: :
E_(2n) + F_(2n) + G_(2n) = 8*3^(n-1) :
Substituting n= 500, it is indeed verified that the required number of 1000-digit positive integers is indeed 8*3^{499} as the desired result.. ---------------------------------------------------
In a similar manner, we will obtain the total number of n-digit positive integers having the desired property as :
14*3^{(n-3)/2} , whenever n is odd, which has been pointed out by Dej Mar . ---------------------------------------------------------------------
Dej Mar has given the required number of n-digit positive integers as:
(8 + 6*(x mod 2)) * 3^{([x/2] - 1)} , such that x > 1, and this will hold regardless of whether n is odd or even.

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