2022 AIME I Problems/Problem 13


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


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.


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

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

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


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

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~ Afly (talk)

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See Also

2022 AIME I (ProblemsAnswer KeyResources)
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

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