AIME 1994 · 第 5 题

Solution

Solution 1

Suppose we write each number in the form of a three-digit number (so $5 \equiv 005$), and since our $p(n)$ ignores all of the zero-digits, replace all of the $0$s with $1$s. Now note that in the expansion of

$(1+ 1 +2+3+4+5+6+7+8+9) (1+ 1 +2+3+\cdots+9) (1+ 1 +2+3+\cdots+9)$

we cover every permutation of every product of $3$ digits, including the case where that first $1$ represents the replaced $0$s. However, since our list does not include $000$, we have to subtract $1$. Thus, our answer $\boxed{103}$.

Solution 2

Note that $p(1)=p(11), p(2)=p(12), p(3)=p(13), \cdots p(19)=p(9)$, and $p(37)=3p(7)$. So $p(10)+p(11)+p(12)+\cdots +p(19)=46$, $p(10)+p(11)+\cdots +p(99)=46*45=2070$. We add $p(1)+p(2)+p(3)+\cdots +p(10)=45$ to get 2115. When we add a digit we multiply the sum by that digit. Thus $2115\cdot (1+1+2+3+4+5+6+7+8+9)=2115\cdot 46=47\cdot 45\cdot 46$. But we didn't count 100, 200, 300, ..., 900. Thus, we have to add another 45 to get $45\cdot 2163$. The largest prime factor of that is $\boxed{103}$.