Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 4)
m (fixed solution numbering)
Line 3: Line 3:
 
<center><math>\frac{n}{810}=0.d25d25d25\ldots</math></center>
 
<center><math>\frac{n}{810}=0.d25d25d25\ldots</math></center>
  
−
== Solution ==
+
== Solution 1 ==
 
We can express <math>0.\overline{d25}</math> as <math>\frac{100d+25}{999}</math>. We set up the given equation and isolate <math>n:</math> \begin{align*} \frac{100d+25}{999} &= \frac{n}{810}, \\ \frac{100d+25}{111} &= \frac{n}{90}, \\ 9000d + 2250 &= 111n. \end{align*} We then set up the following modular congruence to solve for <math>d:</math> \begin{align*}9000d + 2250 &\equiv 0 \pmod {111}, \\ 9d + 30 &\equiv 0 \pmod {111}, \\ 3d + 10 &\equiv 0 \pmod {37}, \\ 36d + 120 &\equiv 0 \pmod {37}, \\ -d &\equiv -9 \pmod {37}, \\ d &\equiv 9 \pmod {37}.\end{align*} Since <math>d</math> is a digit, it must be equal to <math>9</math> based on our above constraint. When <math>d=9, n = \boxed{750}.</math>
 
We can express <math>0.\overline{d25}</math> as <math>\frac{100d+25}{999}</math>. We set up the given equation and isolate <math>n:</math> \begin{align*} \frac{100d+25}{999} &= \frac{n}{810}, \\ \frac{100d+25}{111} &= \frac{n}{90}, \\ 9000d + 2250 &= 111n. \end{align*} We then set up the following modular congruence to solve for <math>d:</math> \begin{align*}9000d + 2250 &\equiv 0 \pmod {111}, \\ 9d + 30 &\equiv 0 \pmod {111}, \\ 3d + 10 &\equiv 0 \pmod {37}, \\ 36d + 120 &\equiv 0 \pmod {37}, \\ -d &\equiv -9 \pmod {37}, \\ d &\equiv 9 \pmod {37}.\end{align*} Since <math>d</math> is a digit, it must be equal to <math>9</math> based on our above constraint. When <math>d=9, n = \boxed{750}.</math>
  
 
~vaisri
 
~vaisri
  
−
== Solution 1 ==
+
== Solution 2 ==
 
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>.
 
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>.
  
Line 14: Line 14:
 
(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.)
 
(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.)
  
−
== Solution 2 ==
+
== Solution 3 ==
 
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>.
  
−
==Solution 3==
+
==Solution 4==
  
 
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:
 
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:
Line 30: Line 30:
  
  
−
==Solution 4==
+
==Solution 5==
 
Write out these equations:
 
Write out these equations:
  

Revision as of 13:36, 11 August 2024

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 1

We can express $0.\overline{d25}$ as $\frac{100d+25}{999}$. We set up the given equation and isolate $n:$ \begin{align*} \frac{100d+25}{999} &= \frac{n}{810}, \\ \frac{100d+25}{111} &= \frac{n}{90}, \\ 9000d + 2250 &= 111n. \end{align*} We then set up the following modular congruence to solve for $d:$ \begin{align*}9000d + 2250 &\equiv 0 \pmod {111}, \\ 9d + 30 &\equiv 0 \pmod {111}, \\ 3d + 10 &\equiv 0 \pmod {37}, \\ 36d + 120 &\equiv 0 \pmod {37}, \\ -d &\equiv -9 \pmod {37}, \\ d &\equiv 9 \pmod {37}.\end{align*} Since $d$ is a digit, it must be equal to $9$ based on our above constraint. When $d=9, n = \boxed{750}.$

~vaisri

Solution 2

Repeating decimals represent rational numbers. To figure out which rational number, we sum an infinite geometric series, $0.d25d25d25\ldots = \sum_{n = 1}^\infty \frac{d25}{1000^n} = \frac{100d + 25}{999}$. 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 3

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 4

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


Solution 5

Write out these equations:


$\frac{n}{810} = \frac{d25}{999}$


$\frac{n}{30} = \frac{d25}{37}$


$37n = 30(d25)$


Thus $n$ divides 25 and 30. The only solution for this under 1000 is $\boxed{750}$.


-jackshi2006

Note: We know $n < 810$ since $\frac{n}{810} < 1,$ so it suffices to check for numbers under $810$ not $1000.$

Video Solution by OmegaLearn

https://youtu.be/1-iWPCWPsLw?t=600

~ pi_is_3.14

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_3&oldid=224932"