2022 AIME I Problems/Problem 13

Problem

Let $S$ be the set of all rational numbers that can be expressed as a repeating decimal in the form $0.\overline{abcd},$ where at least one of the digits $a,$ $b,$ $c,$ or $d$ is nonzero. Let $N$ be the number of distinct numerators obtained when numbers in $S$ are written as fractions in lowest terms. For example, both $4$ and $410$ are counted among the distinct numerators for numbers in $S$ because $0.\overline{3636} = \frac{4}{11}$ and $0.\overline{1230} = \frac{410}{3333}.$ Find the remainder when $N$ is divided by $1000.$

Solution

$0.abcd=\frac{\overline{abcd}}{9999}$ , $9999=9\times 11\times 101$ .

Then we need to find the number of positive integers less than $10000$ that can meet the requirement. Suppose the number is $x$ .

Case $1$ : $(9999, x)=1$ . Clearly $x$ satisfies. $\varphi \left( 9999 \right) =9999\times \left( 1-\frac{1}{3} \right) \times \left( 1-\frac{1}{11} \right) \times \left( 1-\frac{1}{101} \right)=6000$

Case $2$ : $3|x$ but $x$ is not a multiple of $11$ or $101.$ Then the least value of $abcd$ is $9x$ , so that $x\le 1111$ , $334$ values from $3$ to $1110.$

Case $3$ : $11|x$ but $x$ is not a multiple of $3$ or $101$ . Then the least value of $abcd$ is $11x$ , so that $x\le 909$ , $55$ values from $11$ to $902$ .

Case $4$ : $101|x$ . None.

Case $5$ : $3, 11|x$ . Then the least value of $abcd$ is $11x$ , $3$ values from $33$ to $99.$

To sum up, the answer is $6000+334+55+3=\boxed{392}$ mod $1000$ .

2022 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search

2022 AIME I Problems/Problem 13

Problem

Solution

See Also