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

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Since n is an integer, we have that <math>37 \mid d25 \cdot 30</math>.
 
Since n is an integer, we have that <math>37 \mid d25 \cdot 30</math>.
  
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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Since <math>37</math> is prime, we can apply [[Euclid's Lemma]] 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>)
  
 
~qwertysri987
 
~qwertysri987
 
  
 
==Solution 5==
 
==Solution 5==

Latest revision as of 13:38, 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 $n:$ 100d+25999=n810,100d+25111=n90,9000d+2250=111n. We then set up the following modular congruence to solve for $d:$ 9000d+22500(mod111),9d+300(mod111),3d+100(mod37),36d+1200(mod37),d9(mod37),d9(mod37). 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 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 (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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