The Denominator Is Ridiculously Large!

$\frac{n}{810} = 0.d25d25 \ldots$ ( $0 \leq d \leq 9$ )

So, $\frac{n}{810}\times 1000 = d25.d25 \ldots = d25 +\frac{n}{810}$ .

Therefore, $\frac{n}{810} \times 999 =d25$

or, $n = (d25) \times \frac{30}{37}$ .

Since n is an integer, $37 | (d25)$ .

So, $d = 9$ and $n = 750$ . $\square$

n/810 = (100d+25)/999

n = 810(100d + 25)/999 = 30(100d+25)/37

So, to become n be an integer, 100d+25 must divide 37

So, d must be: 9

so, therefore,

n = 30(925)/37

n = 750

We know that : n/810 = 0.d25d25...

So 1000 * (n/810) = d25.d25...

then 999 * (n/810) = d25

and since 25 | d25 then 25 | n

because 810=5 * 9^2 and 999=111 * 9

so n=750

0.d25... = d25/999 & 810:999 is 30:37 as ratio.

So ratio n:d25 is 30:37.

d25 is divisible by 25 and lowest common factor is 25x37 which = 925.

Therefore ratio 30:37 makes 750:925. n = 750 and d25 is 925

First I realized since all three digits are repeating could convert $\frac{d25}{999}$ into a fraction. Then I started to prime factor 999 and came up with 9 x 111 = 9 x 3 x 37. Next, I prime factored 810 which equals 3 x 3 x 3 x 3 x 5 x 2. Next I tried to simplify a little, so I got rid of 9 on both sides, which left me with $\frac{d25}{111}$ = $\frac{n}{90}$ and then I remembered that if the digits of a number add up to a multiple of 3, it can be divided again. So I divided by 3 again: $\frac{n}{30}$ = $\frac{d25}{37}$ lol... then I solved for d because I had a feeling that either 125, 225, 325, 425, 525, 625, 725, 825, or 925 is divisible by 37. It look a long time, in fact, it took 9 tries lol, but I solved for d = 9. Then I plugged d into the original equation and i came up with $\frac{n}{810}$ = $\frac{925}{999}$ . Then I simplified and it turned out 750 was the answer!

Solution without trial and error

• Let $d\in\{0;\:\ldots;9\}$ be the missing digit. Turn the periodic decimal into a fraction: $\begin{aligned} \frac{n}{810}&=0.\overline{d25}=\frac{100d+25}{999}&\left|\cdot999\right.&&\left|\cdot 30\right.&&\Rightarrow&&37n-3000d=750&&(*) \end{aligned}$

• The integers $37,\:-3000$ are relatively prime and we can be sure the Diophantine Equation $(*)$ has (infinitely many) solutions. Let's use Euclid's Algorithm to solve it: $\begin{aligned} \begin{array}{r|r|r|r} n&p_n&a_n&x_n\\ \hline-2&-3000&\%&\red{973}\\ -1&37&\%&\blue{12}\\ 0&-3&-81&1\\ 1&1&-12&0\\ 2&0&-3&1 \end{array}&&\Rightarrow&&\begin{aligned} n&=750\cdot \red{973}+3000N\\[.5em] d&=750\cdot \blue{12}+37N \end{aligned},&&N&\in\mathbb{Z} \end{aligned}$

• The restriction $d\in\{0;\:\ldots;\:9\}$ identifies the correct solution: $\begin{aligned} d&=750\cdot 12+37N\overset{!}{\in}\{0;\:\ldots;\:9\}&\Rightarrow&&N&=-243,&d&=9,&&&n&=750\cdot 973+3000N=\boxed{750} \end{aligned}$

Repeated decimals are expressed in fraction as m/9, m/99, m/999... And for 3 decimal repetition it is m/999. m/999= m/9 111= m/9 3 37 = m/27 37. If the divider is made of anything else that means it cannot have the 3 decimal pattern! So, let's look at the 810. 810 = 9 90= 27 30. The 30 shouldn't be there which means the denominator is also a product of 30 to cancel the divider's 30. n = #30. Looking at the pattern we see that m is a product of 25; m=d 100 +25 = d 4*25 +25. m = #25.
In order to display the 25 pattern with n/810 , n also needs to be a product of 25. n = #30 & n = #25, therefore n needs to be a product of 150. Which from the choice is 750.