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2022 AIME I Problems/Problem 13

Revision as of 18:06, 13 January 2023 by Oxymoronic15 (talk | contribs) (→‎Solution)

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}$,$9999=9\times 11\times 101$.

Then we need to find the number of positive integers less than$ (Error compiling LaTeX. Unknown error_msg)10000$that can meet the requirement. Suppose the number is$x$.

Case$ (Error compiling LaTeX. Unknown error_msg)1$:$\gcd (9999, x)=1$. Clearly,$x$satisfies the condition yielding <cmath>\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</cmath> values.

Case$ (Error compiling LaTeX. Unknown error_msg)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$ (Error compiling LaTeX. Unknown error_msg)4$:$101|x$. None.

Case$ (Error compiling LaTeX. Unknown error_msg)5$:$3, 11|x$. Then the least value of$abcd$is$99x$,$3$values from$33$to$99.$To sum up, the answer is <cmath>6000+334+55+3=\boxed{392}</cmath> mod$1000$.

Video Solution

https://youtu.be/0FZyjuIOHnA

https://MathProblemSolvingSkills.com

See Also

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

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