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 1

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

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 $6000+334+55+3+0\equiv\boxed{392} \pmod{1000}.$

Clarification

In this context, when the solution says, "Then $abcd=9x$ to leave 3 uncancelled, and $x=3p$ ," it is a bit vague. The best way to clarify this is by this exact example - what is really meant is we need to divide by 9 first to achieve 1111, which has no multiple of 3; thus, given that the fraction x/y is the simplest form, x can be a multiple of 3.

Similar explanations can be said when the solution divides 9999 by 11, 101, and uses that divided result in the PIE calculation rather than 9999.

mathboy282

Solution 2 (similar(?) to solution 1 but reworded (not exactly for clarity))

$\text{To begin, we notice that all repeating decimals of the form }0.\overline{abcd}\text{ where }a,b,c,d\text{ are digits can be expressed of the form }\frac{\overline{abcd}}{9999}\text{.}$ $\text{However, when }\overline{abcd}\mid 9999\text{, the fraction is not in lowest terms.}$ $\text{Since }9999 = 3^2 \cdot 11 \cdot 101\text{, } x\mid 9999\iff x\mid 3\lor x\mid 11\lor x\mid 101\text{.}$ $\text{(For those of you who have no idea what that meant, it means every divisor of 9999 is a divisor of at least one of the following: )}$ $(3)$ $(11)$ $(101)$ $\text{(Also, I'm not going to give you explanations for the other logic equations.)}$ $\text{Let's say that the fraction in lowest terms is }\frac{x}{y}\text{.}$ $\text{If }x\mid 101\text{, then }99\mid y\text{ but that can't be, since }0\text{ is the only multiple of }101\text{ below }99\text{.}$ $\exists! f(f\in\mathbb{N}\land f\neq 1\land\exists g(g\nleq 0 \land x \mid f^g))\implies f=3\lor f=11 (1)$ $\text{If (1) is true, then we have two cases. If it isn't, we also have two cases.}$ $\textbf{\textit{Case 1: }}f=3$ $y=1111\land x=3z\implies 1\leq z\leq 370$ $370-33-3^{[1]}=334$ $\textbf{\textit{Case 2: }}f=11$ $y=909\land x=11z\implies 1\leq z\leq 82$ $82-27=55$ $\textbf{\textit{Case 3: }}\neg\exists! f(f\in\mathbb{N}\land f\neq 1\land\exists g(g\nleq 0 \land x \mid f^g))\land \exists f(f\in\mathbb{N}\land f\neq 1\land\exists g(g\nleq 0 \land x \mid f^g))=$ $\exists f_1(f_1\in\mathbb{N}\land f_1\neq 1\land\exists g_1(g_1\nleq 0 \land x \mid f_1^{g_1})\land\exists f_2(f_2\neq f_1\land f_2\in\mathbb{N}\land f_2\neq 1\land\exists g_2(g_2\nleq 0 \land x \mid f_2^{g_2}))\implies f_1=3\land f_2=11\lor f_1=11\land f_2=3$ $y=101\land x=33z\implies 1\leq z\leq 3$ $\textbf{\textit{Case 4: }}\neg\exists f(f\in\mathbb{N}\land f\neq 1\land\exists g(g\nleq 0 \land x \mid f^g))$ $\Phi (9999)=6000$ $\textbf{\textit{Grand Finale}}$ $\text{Adding the outcomes, }N=6000+334+55+3=6392\equiv\boxed{392}\text{ (mod 1000).}$

$\textit{[1] This is to make sure that 3 is the \textbf{only} factor of x}$

Note

$\text{When I tried to write LaTeX, AoPS kept putting the LaTeX on a new line so I gave up and put most of it in LaTeX instead.}$ $\text{Some of the text in this section is just normal.}$ $\text{Example:}$

Normal text $\text{This is some LaTeX.}$ More normal text

$\text{If any of you can fix this issue, please do so.}$

~ Afly (talk)

Video Solution

https://youtu.be/0FZyjuIOHnA

https://MathProblemSolvingSkills.com

2022 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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