Difference between revisions of "1992 AIME Problems/Problem 5"
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<math>\Rightarrow x=\frac{abc}{999}</math> | <math>\Rightarrow x=\frac{abc}{999}</math> | ||
− | 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 <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 | + | 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 <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>. There aren't any numbers which are multiples of <math>372</math>, so we can't get numerators which are multiples of <math>37</math>. Therefore <math>648 + 12 = \boxed{660}</math>. |
{{AIME box|year=1992|num-b=4|num-a=6}} | {{AIME box|year=1992|num-b=4|num-a=6}} | ||
[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] |
Revision as of 13:48, 12 November 2007
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 and , so , which is the amount that are neither. The numbers that are multiples of reduce to multiples of . 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 | |
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