Difference between revisions of "2022 AIME I Problems/Problem 13"
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Then we need to find the number of positive integers less than 10000 can meet the requirement. | Then we need to find the number of positive integers less than 10000 can meet the requirement. | ||
Suppose the number is x. | 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 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)=6000</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 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 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. | ||
Revision as of 21:36, 18 February 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 
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. 
Solution
![\[0.abcd=\frac{\overline{abcd}}{9999}\]](http://latex.artofproblemsolving.com/1/7/3/1731bdccd97b1502b9f13ead168a600b9bdc0e73.png)
, ![\[9999=9\times 11\times 101\]](http://latex.artofproblemsolving.com/8/7/f/87fe32a6f0ee36ddf028a538a8ba8522d1fc57af.png)
.
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. ![\[\varphi \left( 9999 \right) =9999\times \left( 1-\frac{1}{3} \right) \times \left( 1-\frac{1}{11} \right) \times \left( 1-\frac{1}{101} \right)=6000\]](http://latex.artofproblemsolving.com/8/a/e/8aeb9685536883c0ce85b966445df54c99fb90f9.png)
Case 2: 3|x but x is not a multiple of 11 or 101. Then the least value of abcd is 9x, so that ![\[x\le 1111\]](http://latex.artofproblemsolving.com/3/f/0/3f0cc6bc93c0ddd8c80efabaa8bc9dcafac2fc06.png)
, 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 ![\[x\le 909\]](http://latex.artofproblemsolving.com/5/c/1/5c1c0d7ce69ab41e81ff68fe47f1a28c47795d8f.png)
, 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 ![\[6000+334+55+3=6392\]](http://latex.artofproblemsolving.com/8/9/b/89b333a88cf2882a8495151f6f525a86cbdcda15.png)
.
