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Difference between revisions of "2022 AIME I Problems/Problem 13"
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Case 5: 3, 11|x.  Then the least value of abcd is 11x, 3 values from 33 to 99.
 
Case 5: 3, 11|x.  Then the least value of abcd is 11x, 3 values from 33 to 99.
  
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To sum up, the answer is <cmath>6000+334+55+3=\boxed{6392}</cmath>.
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To sum up, the answer is <cmath>6000+334+55+3=\boxed{392}</cmath>.
  
 
==See Also==
 
==See Also==

Revision as of 19:02, 24 June 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 $1000.$

Solution 1

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

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. \[\varphi \left( 9999 \right) =9999\times \left( 1-\frac{1}{3} \right) \times \left( 1-\frac{1}{11} \right) \times \left( 1-\frac{1}{101} \right)=6000\]

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, the answer is \[6000+334+55+3=\boxed{392}\].

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

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