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"

(→‎See Also)
Line 6: Line 6:
 
{{AIME box|year=2022|n=I|num-b=12|num-a=14}}
 
{{AIME box|year=2022|n=I|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
==Solution==
 +
 +
<cmath>0.abcd=\frac{\overline{abcd}}{9999}</cmath>, <cmath>9999=9\times 11\times 101</cmath>.
 +
Then we need to find the number of positive integers less than 10000 can meet the requirement.
 +
Suppose the number is x.
 +
Case 1: (9999, x)=1. Clearly x 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)</cmath>
 +
Case 2: 3|x but x is not a multiple of 11 or 101. Then the least value of abcd is 9x, so that <cmath>x\le 1111</cmath>, 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 <cmath>x\le 909</cmath>, 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, it is <cmath>6000+334+55+3=6392</cmath>.

Revision as of 22:33, 18 February 2022

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 $1000.$

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

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 can meet the requirement. Suppose the number is x. Case 1: (9999, x)=1. Clearly x satisfies. \[\varphi \left( 9999 \right) =9999\times \left( 1-\frac{1}{3} \right) \times \left( 1-\frac{1}{11} \right) \times \left( 1-\frac{1}{101} \right)\] 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, it is \[6000+334+55+3=6392\].

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AIME_I_Problems/Problem_13&oldid=171467"