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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: (Hide) 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*3499 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.

 Subject Author Date Solution Praneeth 2008-10-07 07:17:48 re: Solution Charlie 2008-09-23 10:33:15 Solution Dej Mar 2008-09-23 02:58:22 re: computer-aided solution Charlie 2008-09-22 17:47:13 computer-aided solution Charlie 2008-09-22 15:00:34

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