Question

Given a positive integer $n,$ let $p(n)$ be the product of the nonzero digits of $n.$ (If $n$ has only one digit, then $p(n)$ is equal to that digit.) Let $\begin{aligned}S&=p(1)+p(2)+p(3)\\&\quad+\cdots+p(999).\end{aligned}$ What is the largest prime factor of $S?$

Answer

Answer: 103

Concepts:digits multiplication principle prime factorization

Difficulty rating: 2110

Small Hint:

Write every number from $000$ through $999$ using three digits and let a zero digit contribute a factor of $1$

Big Hint:

The sum factors by digit position; remember to remove the contribution of $000$

Solution:

For one digit position, the sum of its effective factors is $1+1+2+\cdots+9=46,$ where the first $1$ represents digit $0.$ Thus the sum over $000$ through $999$ is $46^3.$ Removing the artificial contribution $1$ from $000$ gives $\begin{aligned}S&=46^3-1\\&=45(46^2+46+1)\\&=45\cdot2163.\end{aligned}$ Since $2163=3\cdot721=3\cdot7\cdot103,$ the largest prime factor is $103.$

1994 AIME Problem 5

Answer: 103

Difficulty rating: 2110

Small Hint:

Big Hint:

Solution:

Problem 5 in Other Years