Difference between revisions of "1989 AIME Problems/Problem 3"
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== Solution == | == Solution == | ||
− | Repeating decimals represent [[rational number]]s. To figure out which rational number, we sum an [[infinite]] [[geometric series]], <math>0.d25d25d25\ldots = \sum_{ | + | Repeating decimals represent [[rational number]]s. To figure out which rational number, we sum an [[infinite]] [[geometric series]], <math>0.d25d25d25\ldots = \sum_{n = 1}^\infty \frac{d25}{1000^n} = \frac{100d + 25}{999}</math>. Thus <math>\frac{n}{810} = \frac{100d + 25}{999}</math> so <math>n = 30\frac{100d + 25}{37} =750\frac{4d + 1}{37}</math>. Since 750 and 37 are [[relatively prime]], <math>4d + 1</math> must be [[divisible]] by 37, and the only digit for which this is possible is <math>d = 9</math>. Thus <math>4d + 1 = 37</math> and <math>n = \boxed{750}</math>. |
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+ | (Note: Any repeating sequence of <math>n</math> digits that looks like <math>0.a_1a_2a_3...a_{n-1}a_na_1a_2...</math> can be written as <math>\frac{a_1a_2...a_n}{10^n-1}</math>, where <math>a_1a_2...a_n</math> represents an <math>n</math> digit number.) | ||
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+ | == Solution 2 == | ||
+ | To get rid of repeating decimals, we multiply the equation by 1000. We get <math>\frac{1000n}{810} = d25.d25d25...</math> We subtract the original equation from the second to get <math>\frac{999n}{810}=d25</math> We simplify to <math>\frac{37n}{30} = d25</math> Since <math>\frac{37n}{30}</math> is an integer, <math>n=(30)(5)(2k+1)</math> because <math>37</math> is relatively prime to <math>30</math>, and d25 is divisible by <math>5</math> but not <math>10</math>. The only odd number that yields a single digit <math>d</math> and 25 at the end of the three digit number is <math>k=2</math>, so the answer is <math>\boxed{750}</math>. | ||
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+ | ==Solution 3== | ||
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+ | Similar to Solution 2, we start off by writing that <math> \frac{1000n}{810} = d25.d25d25 \dots</math> .Then we subtract this from the original equation to get: | ||
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+ | <cmath> \frac{999n}{810} =d25 \Longrightarrow \frac{37n}{30} = d25 \Longrightarrow 37n = d25 \cdot 30</cmath> | ||
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+ | Since n is an integer, we have that <math>37 \mid d25 \cdot 30</math>. | ||
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+ | Since <math>37</math> is prime, we can apply Euclid's Lemma (which states that if <math>p</math> is a prime and if <math>a</math> and <math>b</math> are integers and if <math>p \mid ab</math>, then <math>p \mid a</math> or <math>p \mid b</math>) to realize that <math>37 \mid d25 </math>, since <math>37 \nmid 30</math>. Then we can expand <math>d25</math> as <math>25 \cdot (4d +1)</math>. Since <math>37 \nmid 25 </math>, by Euclid, we can arrive at <math>37 \mid 4d+1 \Longrightarrow d=9</math>. From this we know that <math>n= 25 \cdot 30 = \boxed{750}</math>. (This is true because <math>37n = 925 \cdot 30 \rightarrow n= 25 \cdot 30 = 750</math>) | ||
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+ | ~qwertysri987 | ||
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+ | ==Solution 4== | ||
+ | Write out these equations: | ||
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+ | <math>\frac{n}{180} = \frac{d25}{999}</math> | ||
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+ | <math>\frac{n}{30} = \frac{d25}{37}</math> | ||
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+ | <math>37n = 30(d25)</math> | ||
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+ | Thus <math>n</math> divides 25 and 30. The only solution for this under 1000 is <math>\boxed{750}</math>. | ||
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+ | -jackshi2006 | ||
== See also == | == See also == | ||
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 19:34, 25 April 2021
Problem
Suppose is a positive integer and is a single digit in base 10. Find if
Solution
Repeating decimals represent rational numbers. To figure out which rational number, we sum an infinite geometric series, . Thus so . Since 750 and 37 are relatively prime, must be divisible by 37, and the only digit for which this is possible is . Thus and .
(Note: Any repeating sequence of digits that looks like can be written as , where represents an digit number.)
Solution 2
To get rid of repeating decimals, we multiply the equation by 1000. We get We subtract the original equation from the second to get We simplify to Since is an integer, because is relatively prime to , and d25 is divisible by but not . The only odd number that yields a single digit and 25 at the end of the three digit number is , so the answer is .
Solution 3
Similar to Solution 2, we start off by writing that .Then we subtract this from the original equation to get:
Since n is an integer, we have that .
Since is prime, we can apply Euclid's Lemma (which states that if is a prime and if and are integers and if , then or ) to realize that , since . Then we can expand as . Since , by Euclid, we can arrive at . From this we know that . (This is true because )
~qwertysri987
Solution 4
Write out these equations:
Thus divides 25 and 30. The only solution for this under 1000 is .
-jackshi2006
See also
1989 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.