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).
Consider the 1digit, 2digit, etc. numbers separately.
The values for the 1digit numbers total 45.
That 45 is, of course, 1+2+3+4+5+6+7+8+9. But these are also the possibilities for the second digit of 2digit numbers. Each is combined with each of the 1digit numbers, and due to the associative property, when you multiply each 2digit numbers' digits and add them, you'll get (1+2+3+4+5+6+7+8+9)*(1+2+3+4+5+6+7+8+9) or 45^2.
Similarly for each number of digits, the new digit is appended to the end of the previous digits, giving 45^n as the total within number of digits.
We need to sum the first six powers of 45. It's like taking a repunit in base 45, seven long, and subtracting 1 (the 45^0 position).
The sum of powers zero through six of 45 is (45^71)/44 = 8,492,487,571; subtracting 1 makes it 8,492,487,570, the answer.

Posted by Charlie
on 20190823 10:54:18 