Difference between revisions of "1992 AIME Problems/Problem 5"
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== Problem == | == Problem == | ||
| + | Let <math>S^{}_{}</math> be the set of all rational numbers <math>r^{}_{}</math>, <math>0^{}_{}<r<1</math>, that have a repeating decimal expansion in the form <math>0.abcabcabc\ldots=0.\overline{abc}</math>, where the digits <math>a^{}_{}</math>, <math>b^{}_{}</math>, and <math>c^{}_{}</math> are not necessarily distinct. To write the elements of <math>S^{}_{}</math> as fractions in lowest terms, how many different numerators are required? | ||
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== Solution == | == Solution == | ||
| + | We consider the method in which repeating decimals are normally converted to fractions with an example: | ||
| + | |||
| + | <math>x=0.\overline{176}</math> | ||
| + | |||
| + | <math>\Rightarrow 1000x=176.\overline{176}</math> | ||
| + | |||
| + | <math>\Rightarrow 999x=1000x-x=176</math> | ||
| + | |||
| + | <math>\Rightarrow x=\frac{176}{999}</math> | ||
| + | |||
| + | Thus, let <math>x=0.\overline{abc}</math> | ||
| + | |||
| + | <math>\Rightarrow 1000x=abc.\overline{abc}</math> | ||
| + | |||
| + | <math>\Rightarrow 999x=1000x-x=abc</math> | ||
| + | |||
| + | <math>\Rightarrow x=\frac{abc}{999}</math> | ||
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| + | If <math>abc</math> is not divisible by <math>3</math> or <math>37</math>, then this is in lowest terms. Let us consider the other multiples: <math>333</math> multiples of <math>3</math>, <math>27</math> of <math>37</math>, and <math>9</math> of both <math>3</math> and <math>37</math>, so <math>999-333-27+9 = 648</math>, which is the amount that are neither. The <math>12</math> numbers that are multiples of <math>81</math> reduce to multiples of <math>3</math>. We have to count these since it will reduce to a multiple of <math>3</math> which we have removed from <math>999</math>, but, this cannot be removed since the numerator cannot cancel the <math>3</math>.There aren't any numbers which are multiples of <math>37^2</math>, so we can't get numerators which are multiples of <math>37</math>. Therefore <math>648 + 12 = \boxed{660}</math>. | ||
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| + | {{AIME box|year=1992|num-b=4|num-a=6}} | ||
| − | + | [[Category:Intermediate Number Theory Problems]] | |
| − | + | {{MAA Notice}} | |
Latest revision as of 19:45, 4 March 2024
Problem
Let 
be the set of all rational numbers 
, 
, that have a repeating decimal expansion in the form 
, where the digits 
, 
, and 
are not necessarily distinct. To write the elements of as fractions in lowest terms, how many different numerators are required?
Solution
We consider the method in which repeating decimals are normally converted to fractions with an example:
Thus, let 
If 
is not divisible by 
or 
, then this is in lowest terms. Let us consider the other multiples: 
multiples of , 
of , and 
of both and , so 
, which is the amount that are neither. The 
numbers that are multiples of 
reduce to multiples of . We have to count these since it will reduce to a multiple of which we have removed from 
, but, this cannot be removed since the numerator cannot cancel the .There aren't any numbers which are multiples of 
, so we can't get numerators which are multiples of . Therefore 
.
| 1992 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
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