1994 AIME Problem 5

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

Write every number from

000

through

999

using three digits and let a zero digit contribute a factor of

1

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

000

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.

Problem 5 in other years