Let us denote f(n) = 12...12 (repeated n times) and g(n) = 3...3 (repeated n times). For example: f(3) = 121212 and, g(3) = 333.

(i) Determine the distinct digits in the base ten expansion of f(n)*g(n) whenever n ≥ 3

(ii) Express s.o.d (f(n)*g(n)) in terms of n

*** s.o.d (x) denotes the sum of digits in the base ten expansion of x

Looking at Charlie's output:

i. the DISTINCT digits appearing: 0,3,4,5,6 and 9

ii. even n......the sod is 18*n :  odd  9/2*(3n-1)

All numbers (n>2) start with a  40.. (repeated) and that is easily explained by evaluating the limit:

1/3*12/99=4/99= .040404040...

 

Edited on September 20, 2012, 12:07 am

Posted by Ady TZIDON on 2012-09-19 23:12:41