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>
  
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==Solution 1==
  
 
<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 15:12, 23 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 $1000.$

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, it is \[6000+334+55+3=6392\].

See Also

2022 AIME I (ProblemsAnswer KeyResources)
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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