AIME 1989 · 第 3 题

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 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