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

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

Then we need to find the number of positive integers $x$ that (with one of more $y$ such that $y|9999$ ) can meet the requirement $1 \leq {x}\cdot\frac{9999}{y} \leq 9999$ .

Make cases by factors of $x$ . (A venn diagram of cases would be nice here.)

Case $A$ :

$3 \nmid x$ and $11 \nmid x$ and $101 \nmid x$ , aka $\gcd (9999, x)=1$ .

Euler's totient function counts these: $\varphi \left(3^2 \cdot 11 \cdot 101 \right) = ((3-1)\cdot 3)(11-1)(101-1)= \bf{6000}$ values (but it's enough to note that it's a multiple of 1000 and thus does not contribute to the final answer)

Note: You don't need to know this formula. The remaining cases essentially re-derive the same computation for other factors of $9999$ . This case isn't actually different.

The remaining cases have $3$ (or $9$ ), $11$ , and/or $101$ as factors of $abcd$ , which cancel out part of $9999$ . Note: Take care about when to use $3$ vs $9$ !

Case $B$ : $3|x$ , but $11 \nmid x$ and $101 \nmid x$ .

Then $abcd=9x$ to leave 3 uncancelled, and $x=3p$ , so $x \leq \frac{9999}{9} = 1111$ , giving:

$x \in 3 \cdot \{1, \dots \left\lfloor \frac{1111}{3}\right\rfloor\}$ ,

$x \notin (3\cdot 11) \cdot \{1 \dots \left\lfloor \frac{1111}{3\cdot 11}\right\rfloor\}$ ,

$x \notin (3 \cdot 101) \cdot \{1 \dots \left\lfloor \frac{1111}{3 \cdot 101}\right\rfloor\}$ ,

for a subtotal of $\left\lfloor \frac{1111}{3}\right\rfloor - (\left\lfloor\frac{1111}{3 \cdot 11}\right\rfloor + \left\lfloor\frac{1111}{3 \cdot 101}\right\rfloor ) = 370 - (33+3) = \bf{334}$ values.

Case $C$ : $11|x$ , but $3 \nmid x$ and $101 \nmid x$ .

Much like previous case, $abcd$ is $11x$ , so $x \leq \frac{9999}{11} = 909$ ,

giving $\left\lfloor \frac{909}{11}\right\rfloor - \left(\left\lfloor\frac{909}{11 \cdot 3}\right\rfloor + \left\lfloor\frac{909}{11 \cdot 101}\right\rfloor \right) = 82 - (27 + 0) = \bf{55}$ values.

Case $D$ : $3|x$ and $11|x$ (so $33|x$ ), but $101 \nmid x$ .

Here, $abcd$ is $99x$ , so $x \leq \frac{9999}{99} = 101$ ,

giving $\left\lfloor \frac{101}{33}\right\rfloor - \left\lfloor \frac{101}{33 \cdot 101}\right\rfloor = 3-0 = \bf{3}$ values.

Case $E$ : $101|x$ .

Here, $abcd$ is $101x$ , so $x \leq \frac{9999}{101} = 99$ ,

giving $\left\lfloor \frac{99}{101}\right\rfloor = \bf{0}$ values, so we don't need to account for multiples of $3$ and $11$ .

To sum up, the answer is $6000+334+55+3+0\equiv\boxed{392} \pmod{1000}.$

Video Solution

https://youtu.be/0FZyjuIOHnA

https://MathProblemSolvingSkills.com

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

Contents

Problem

Solution

Video Solution

See Also