Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
Line 5: Line 5:
 
==Solution==
 
==Solution==
  
−
<math>0.abcd=\frac{\overline{abcd}}{9999}</math>, <math>9999=9\times 11\times 101</math>.
+
<math>0.</math>\overline{abcd}=\frac{abcd}{9999}<math>, </math>9999=9\times 11\times 101<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>.
+
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>.
  
−
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>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>
  
−
Case <math>3</math>: <math>11|x</math> but <math>x</math> is not a multiple of <math>3</math> or <math>101</math>. Then the least value of <math>abcd</math> is <math>11x</math>, so that <math>x\le 909</math>, <math>55</math> values from <math>11</math> to <math>902</math>.
+
Case </math>3<math>: </math>11|x<math> but </math>x<math> is not a multiple of </math>3<math> or </math>101<math>. Then the least value of </math>abcd<math> is </math>11x<math>, so that </math>x\le 909<math>, </math>55<math> values from </math>11<math> to </math>902<math>.
  
−
Case <math>4</math>: <math>101|x</math>. None.
+
Case </math>4<math>: </math>101|x<math>. None.
  
−
Case <math>5</math>: <math>3, 11|x</math>.  Then the least value of <math>abcd</math> is <math>99x</math>, <math>3</math> values from <math>33</math> to <math>99.</math>
+
Case </math>5<math>: </math>3, 11|x<math>.  Then the least value of </math>abcd<math> is </math>99x<math>, </math>3<math> values from </math>33<math> to </math>99.<math>
  
−
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$.
  
 
==Video Solution==
 
==Video Solution==

Revision as of 17:06, 13 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

$0.$\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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AIME_I_Problems/Problem_13&oldid=186538"