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

We could start by writing formulas for f and g

f(n) = 4(10^2n - 1)/33
g(n) =(10^n - 1)/3

and then we will have the product

f(n)*g(n) = 4/99 (10^3n - 10^2n - 10^n + 1)

Ignoring the 4/99 the powers of 10 part expands nicely (expanded for clarity)

    8    9     1
  9 8   99   0 1
 99 8  999  00 1 
999 8 9999 000 1
The multiplication by 4/99 will produce the results Charlie showed, but I don't have time to explore them further right now...

Edited on September 20, 2012, 10:39 am

Posted by Jer on 2012-09-19 18:34:53