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
(In reply to
A start by Jer)
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
Multiplying these by 4/9 gives a similar new pattern
3 9 6
4 3 99 5 6
44 3 999 55 6
444 3 9999 555 6
(n1)4's 3 (n)9's (n1)5's 6
The only thing left is to divide by 11. The complicating factor here is these repdigits leave different remainders on division by 11 depending if there is an even or odd number of repeated digits.
Again, I'll finish when I have time...

Posted by Jer
on 20120920 14:00:47 