Concyclicboy wrote: Let $ P$ be the product of all non-zero digits of the positive integer $ n$. For example, $ P(4) = 4$, $ P(50) = 5$, $ P(123) = 6$, $ P(2009) = 18$. Find the value of the sum: P(1) + P(2) + ... + P(2008) + P(2009). $ \sum_{i=1}^9P(i)=45$ For any $ n\in\{1,2,3,4,5,6,7,8,9\}$ : $ \sum_{i=10n}^{10n+9}P(i)=n+45n=46n$ So $ \sum_{i=1}^{99}P(i)=45+46(1+2+3+4+5+6+7+8+9)=45\cdot 47=2115$ For any $ n\in\{1,2,3,4,5,6,7,8,9\}$ : $ \sum_{i=100n}^{100n+99}P(i)=n+2115n=2116n$ So $ \sum_{i=1}^{999}P(i)=2115+2116(1+2+3+4+5+6+7+8+9)=45\cdot 47=97335$ So $ \sum_{i=1}^{2009}P(i)=\sum_{i=1}^{999}P(i)+\sum_{i=1000}^{1999}P(i)+\sum_{i=2000}^{2009}P(i)$ $ =97335+1+97335+2+2(1+2+3+4+5+6+7+8+9)$ $ =1+97335+97335+92=194763$ Hence the result : $ \boxed{194763}$

Notice that the sum is essentially the same as 2(1+1+2+3+4+5+6+7+8+9)^3+2(1+1+2+3+4+5+6+7+8+9)-1=194763. The first one can be treated as a place holder for 0.