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

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== Solution 2 ==
 
== 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>.
 
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
  
 
== See also ==
 
== See also ==

Revision as of 13:16, 2 July 2019

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 $\boxed{750}$.

Solution 3

Similar to Solution 2, we start off by writing that $\frac{1000n}{810} = d25.d25d25 \dots$ .Then we subtract this from the original equation to get:

\[\frac{999n}{810} =d25 \Longrightarrow \frac{37n}{30} = d25 \Longrightarrow 37n = d25 \cdot 30\]

Since n is an integer, we have that $37 \mid d25 \cdot 30$.

Since $37$ is prime, we can apply Euclid's Lemma (which states that if $p$ is a prime and if $a$ and $b$ are integers and if $p \mid ab$, then $p \mid a$ or $p \mid b$) to realize that $37 \mid d25$, since $37 \nmid 30$. Then we can expand $d25$ as $25 \cdot (4d +1)$. Since $37 \nmid 25$, by Euclid, we can arrive at $37 \mid 4d+1 \Longrightarrow d=9$. From this we know that $n= 25 \cdot 30 = \boxed{750}$. (This is true because $37n = 925 \cdot 30 \rightarrow n= 25 \cdot 30 = 750$)

~qwertysri987

See also

1989 AIME (ProblemsAnswer KeyResources)
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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