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 20120919 18:34:53 