Difference between revisions of "2022 AIME I Problems/Problem 13"
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Let <math>S</math> be the set of all rational numbers that can be expressed as a repeating decimal in the form <math>0.\overline{abcd},</math> where at least one of the digits <math>a,</math> <math>b,</math> <math>c,</math> or <math>d</math> is nonzero. Let <math>N</math> be the number of distinct numerators obtained when numbers in <math>S</math> are written as fractions in lowest terms. For example, both <math>4</math> and <math>410</math> are counted among the distinct numerators for numbers in <math>S</math> because <math>0.\overline{3636} = \frac{4}{11}</math> and <math>0.\overline{1230} = \frac{410}{3333}.</math> Find the remainder when <math>N</math> is divided by <math>1000.</math> | Let <math>S</math> be the set of all rational numbers that can be expressed as a repeating decimal in the form <math>0.\overline{abcd},</math> where at least one of the digits <math>a,</math> <math>b,</math> <math>c,</math> or <math>d</math> is nonzero. Let <math>N</math> be the number of distinct numerators obtained when numbers in <math>S</math> are written as fractions in lowest terms. For example, both <math>4</math> and <math>410</math> are counted among the distinct numerators for numbers in <math>S</math> because <math>0.\overline{3636} = \frac{4}{11}</math> and <math>0.\overline{1230} = \frac{410}{3333}.</math> Find the remainder when <math>N</math> is divided by <math>1000.</math> | ||
| − | ==Solution | + | ==Solution== |
<math>0.abcd=\frac{\overline{abcd}}{9999}</math>, <math>9999=9\times 11\times 101</math>. | <math>0.abcd=\frac{\overline{abcd}}{9999}</math>, <math>9999=9\times 11\times 101</math>. | ||
Revision as of 14:09, 20 November 2022
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
, 
.
Then we need to find the number of positive integers less than 
that can meet the requirement. Suppose the number is 
.
Case 
: 
. Clearly satisfies. ![\[\varphi \left( 9999 \right) =9999\times \left( 1-\frac{1}{3} \right) \times \left( 1-\frac{1}{11} \right) \times \left( 1-\frac{1}{101} \right)=6000\]](http://latex.artofproblemsolving.com/8/a/e/8aeb9685536883c0ce85b966445df54c99fb90f9.png)
Case 
: 
but is not a multiple of 
or 
Then the least value of 
is 
, so that 
, 
values from 
to 
Case : 
but is not a multiple of or 
. Then the least value of is 
, so that 
, 
values from to 
.
Case : 
. None.
Case 
: 
. Then the least value of is , values from 
to 
To sum up, the answer is ![\[6000+334+55+3=\boxed{392}\]](http://latex.artofproblemsolving.com/b/e/4/be41c40a4dc01d5dc7444e4320a4b9760c7742b6.png)
mod 
.
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. 
