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Difference between revisions of "1989 AIME Problems/Problem 3"

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== Solution 2 ==
 
== Solution 2 ==
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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>750</math>.
+
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>.
  
 
== See also ==
 
== See also ==

Revision as of 23:09, 1 April 2018

Problem

Suppose $n$ is a positive integer and $d$ is a single digit in base 10. Find $n$ if

$\frac{n}{810}=0.d25d25d25\ldots$

Solution

Repeating decimals represent rational numbers. To figure out which rational number, we sum an infinite geometric series, $0.d25d25d25\ldots = \sum_{i = 1}^\infty \frac{d25}{1000^n} = \frac{100d + 25}{999}$. Thus $\frac{n}{810} = \frac{100d + 25}{999}$ so $n = 30\frac{100d + 25}{37} =750\frac{4d + 1}{37}$. Since 750 and 37 are relatively prime, $4d + 1$ must be divisible by 37, and the only digit for which this is possible is $d = 9$. Thus $4d + 1 = 37$ and $n = \boxed{750}$.


(Note: Any repeating sequence of $n$ digits that looks like $0.a_1a_2a_3...a_{n-1}a_na_1a_2...$ can be written as $\frac{a_1a_2...a_n}{10^n-1}$, where $a_1a_2...a_n$ represents an $n$ digit number.)

Solution 2

To get rid of repeating decimals, we multiply the equation by 1000. We get $\frac{1000n}{810} = d25.d25d25...$ We subtract the original equation from the second to get $\frac{999n}{810}=d25$ We simplify to $\frac{37n}{30} = d25$ Since $\frac{37n}{30}$ is an integer, $n=(30)(5)(2k+1)$ because $37$ is relatively prime to $30$, and d25 is divisible by $5$ but not $10$. The only odd number that yields a single digit $d$ and 25 at the end of the three digit number is $k=2$, so the answer is $\boxed{750}$.

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

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