2006 AIME II Problems/Problem 14

Problem

Let $S_n$ be the sum of the reciprocals of the non-zero digits of the integers from $1$ to $10^n$ inclusive. Find the smallest positive integer $n$ for which $S_n$ is an integer.

Solution

Let $K = \sum_{i=1}^{9}{\frac{1}{i}}$ . Examining the terms in $S_1$ , we see that $S_1 = K + 1$ since each digit $n$ appears once and 1 appears an extra time. Now consider writing out $S_2$ . Each term of $K$ will appear 10 times in the units place and 10 times in the tens place (plus one extra 1 will appear), so $S_2 = 20K + 1$ .

In general, we will have that

$S_n = (n10^{n-1})K + 1$

because each digit will appear $10^{n - 1}$ times in each place in the numbers $1, 2, \ldots, 10^{n} - 1$ , and there are $n$ total places.

The denominator of $K$ is $D = 2^3\cdot 3^2\cdot 5\cdot 7$ . For $S_n$ to be an integer, $n10^{n-1}$ must be divisible by $D$ . Since $10^{n-1}$ only contains the factors $2$ and $5$ (but will contain enough of them when $n \geq 3$ ), we must choose $n$ to be divisible by $3^2\cdot 7$ . Since we're looking for the smallest such $n$ , the answer is $\boxed{063}$ .

2006 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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2006 AIME II Problems/Problem 14

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