2022 AIME I Problems/Problem 13
Contents
[hide]Problem
Let be the set of all rational numbers that can be expressed as a repeating decimal in the form
where at least one of the digits
or
is nonzero. Let
be the number of distinct numerators obtained when numbers in
are written as fractions in lowest terms. For example, both
and
are counted among the distinct numerators for numbers in
because
and
Find the remainder when
is divided by
Solution
\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)10000x$.
Case$ (Error compiling LaTeX. Unknown error_msg)1\gcd (9999, x)=1
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)23|x
x
11
101.
abcd
9x
x\le 1111
334
3
1110.
3
11|x
x
3
101
abcd
11x
x\le 909
55
11
902$.
Case$ (Error compiling LaTeX. Unknown error_msg)4101|x$. None.
Case$ (Error compiling LaTeX. Unknown error_msg)53, 11|x
abcd
99x
3
33
99.
1000$.
Video Solution
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 | ||
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