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
solution by Charlie)
I forgot that the numbers with zero in them still count, by ignoring the zero digits, and adding in the product of the rest of the digits. So, for example, 4digit numbers include those with a single zero, in any of three positions, thus counting 3*45^3, that is three times the value previously (previous comment) calculated for 3digit numbers; they also include those with two zeros in any of the C(3,2)=3 positions, this time adding in 3*45^2; plus the 45 for the threezero case.
I have to think about this more, or work out the total.

Posted by Charlie
on 20190823 11:11:35 