Let p(n) denote the product of the nonzero digits of n. For example, p(5) = 5, p(37) = 21, and p(604) = 24. Without resorting to a computer, evaluate
p(1) + p(2) + p(3) + ... + p(999999).
(In reply to
re: solution -- I forgot about something by Charlie)
As described in my previous post, we need to add in the values for all the numbers containing zeros (1 through n-1 of them in all the combinations of positions after the first digit). We need the sum of each of these lines:
45
45^2 + 45
45^3 + 2*45^2 + 45
45^4 + 3*45^3 + 3*45^2 + 45
45^5 + 4*45^4 + 6*45^3 + 4*45^2 + 45
45^6 + 5*45^5 + 10*45^4 + 10*45^3 + 5*45^2 + 45
The sum of all these is 9,474,296,895. I did use a computer-based calculator program for this, as I didn't see a shortcut the way I did in the previous "solution" and it's no fun typing these into a calculator rather than pasting.
Added:
I see I could have simplified to
6*45 + 15*45^2 + 20*45^3 + 15*45^4 + 6*45^5 +45^6
which would not have been as horrendous typing into a calculator.
Edited on August 23, 2019, 12:25 pm
|
|
Posted by Charlie
on 2019-08-23 12:18:08 |
revised solution
