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Difference between revisions of "2022 AIME I Problems/Problem 13"

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Then we need to find the number of positive integers less than <math>10000</math> that can meet the requirement. Suppose the number is <math>x</math>.
 
Then we need to find the number of positive integers less than <math>10000</math> that can meet the requirement. Suppose the number is <math>x</math>.
  
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Case <math>1</math>: <math>(9999, x)=1</math>. Clearly <math>x</math> satisfies. <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>
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Case <math>1</math>: <math>\gcd (9999, x)=1</math>. Clearly, <math>x</math> 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 <math>2</math>: <math>3|x</math> but <math>x</math> is not a multiple of <math>11</math> or <math>101.</math> Then the least value of <math>abcd</math> is <math>9x</math>, so that <math>x\le 1111</math>, <math>334</math> values from <math>3</math> to <math>1110.</math>
 
Case <math>2</math>: <math>3|x</math> but <math>x</math> is not a multiple of <math>11</math> or <math>101.</math> Then the least value of <math>abcd</math> is <math>9x</math>, so that <math>x\le 1111</math>, <math>334</math> values from <math>3</math> to <math>1110.</math>
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To sum up, the answer is <cmath>6000+334+55+3=\boxed{392}</cmath> mod <math>1000</math>.
 
To sum up, the answer is <cmath>6000+334+55+3=\boxed{392}</cmath> mod <math>1000</math>.
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==Video Solution==
 
==Video Solution==

Revision as of 16:20, 2 January 2023

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 $1$: $\gcd (9999, x)=1$. Clearly, $x$ satisfies the condition yielding \[\varphi \left( 9999 \right) =9999\times \left( 1-\frac{1}{3} \right) \times \left( 1-\frac{1}{11} \right) \times \left( 1-\frac{1}{101} \right)=6000\] values.

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 $1000$.

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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