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"

m (→‎Solution)
m (→‎Solution: added some detail and clarification)
Line 5: Line 5:
 
==Solution==
 
==Solution==
  
−
<math>0.\overline{abcd}=\frac{abcd}{9999}</math>, <math>9999=9\times 11\times 101</math>.
+
<math>0.\overline{abcd}=\frac{abcd}{9999} = \frac{x}{y}</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 <math>x</math> that (with one of more <math>y</math> such that <math>y|9999</math>) can meet the requirement <math>1 \leq {x}\cdot\frac{9999}{y} \leq 9999</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.
+
Make cases by factors of <math>x</math>. (A venn diagram of cases would be nice here.)
  
−
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>A</math>:
  
−
Case <math>4</math>: <math>101|x</math>. None.
+
<math>3 \nmid x</math> and <math>11 \nmid x</math> and <math>101 \nmid x</math>, aka <math>\gcd (9999, x)=1</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>
+
Euler's totient function counts these:  
 +
<cmath>\varphi \left(3^2 \cdot 11 \cdot 101 \right) = ((3-1)\cdot 3)(11-1)(101-1)= \bf{6000}</cmath> values (but it's enough to note that it's a multiple of 1000 and thus does not contribute to the final answer).
  
−
To sum up, the answer is <cmath>6000+334+55+3\equiv\boxed{392} \pmod{1000}.</cmath>
+
The remaining cases have <math>3</math> (or <math>9</math>), <math>11</math>, and/or <math>101</math> as factors of <math>abcd</math>, which cancel out part of <math>9999</math>.
 +
Note: Take care about when to use <math>3</math> vs <math>9</math>!
 +
 
 +
 
 +
Case <math>B</math>: <math>3|x</math>, but <math>11 \nmid x</math> and <math>101 \nmid x</math>.
 +
 
 +
Then <math>abcd=9x</math> to leave 3 uncancelled, and <math>x=3p</math>,
 +
so  <math>x \leq \frac{9999}{9} = 1111</math>, giving:
 +
 
 +
<math>x \in 3 \cdot \{1, \dots \left\lfloor \frac{1111}{3}\right\rfloor\}</math>,
 +
 
 +
<math>x \notin (3\cdot 11) \cdot \{1 \dots \left\lfloor \frac{1111}{3\cdot 11}\right\rfloor\}</math>,
 +
 
 +
<math>x \notin (3 \cdot 101) \cdot \{1 \dots \left\lfloor \frac{1111}{3 \cdot 101}\right\rfloor\}</math>,
 +
 
 +
for a subtotal of <math>\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}</math> values.
 +
 
 +
 
 +
Case <math>C</math>: <math>11|x</math>, but <math>3 \nmid x</math> and <math>101 \nmid x</math>.
 +
 
 +
Much like previous case, <math>abcd</math> is <math>11x</math>, so  <math>x \leq \frac{9999}{11} = 909  </math>, 
 +
 
 +
giving <math>\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}</math> values.
 +
 
 +
 
 +
Case <math>D</math>: <math>3|x</math> and <math>11|x</math> (so <math>33|x</math>), but <math>101 \nmid x</math>. 
 +
 
 +
Here, <math>abcd</math> is <math>99x</math>, so  <math>x \leq \frac{9999}{99} = 101  </math>,
 +
 
 +
giving <math>\left\lfloor \frac{101}{33}\right\rfloor - \left\lfloor \frac{101}{33 \cdot 101}\right\rfloor = 3-0 = \bf{3}</math> values.
 +
 
 +
 
 +
Case <math>E</math>: <math>101|x</math>.
 +
 
 +
Here, <math>abcd</math> is <math>101x</math>, so  <math>x \leq \frac{9999}{101} = 99  </math>,
 +
 
 +
giving <math>\left\lfloor \frac{99}{101}\right\rfloor = \bf{0}</math> values, so we don't need to account for multiples of <math>3</math> and <math>11</math>.
 +
 
 +
To sum up, the answer is <cmath>6000+334+55+3+0\equiv\boxed{392} \pmod{1000}.</cmath>
  
 
==Video Solution==
 
==Video Solution==

Revision as of 16:16, 20 December 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

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

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).

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

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=208005"