Difference between revisions of "1992 AIME Problems/Problem 5"
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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 <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>. | 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>. 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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+ | == Solution 2== | ||
+ | We can see that the denominator of the fraction will be 999b, where b is either 1, 1/3, 1/111, 1/9, or 1/333. This means that the numerator stays abc unless abc divides 999. Finding the numbers relatively prime to 999 gives us the numerators when the fraction cannot be simplified past 999. | ||
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+ | <math>\Rightarrow 999(1 - \frac{1}{3})(1- \frac{1}{111}) = 660</math> | ||
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+ | Now, look at a number that does divide 999: | ||
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+ | <math>\Rightarrow 555/999 = 5/9</math> | ||
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+ | We can see that if a number does divide 999, it will be simplified into a number which is relatively prime to 999 (since that is what simplifying is). There are 660 numbers less than or relatively prime to 999, so the answer is <math>\boxed{660}</math>. | ||
+ | - krishkhushi09 | ||
{{AIME box|year=1992|num-b=4|num-a=6}} | {{AIME box|year=1992|num-b=4|num-a=6}} |
Latest revision as of 18:27, 22 April 2020
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 . 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 .
Solution 2
We can see that the denominator of the fraction will be 999b, where b is either 1, 1/3, 1/111, 1/9, or 1/333. This means that the numerator stays abc unless abc divides 999. Finding the numbers relatively prime to 999 gives us the numerators when the fraction cannot be simplified past 999.
Now, look at a number that does divide 999:
We can see that if a number does divide 999, it will be simplified into a number which is relatively prime to 999 (since that is what simplifying is). There are 660 numbers less than or relatively prime to 999, so the answer is . - krishkhushi09
1992 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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