Let $x=(0.\overline{6})(0.\overline{60} )$. when $x$ is written out as a decimal, what is the sum of the first $15$ digits after the decimal point?

32

Step-by-step explanation:

We begin by writing both $0.\overline{6}$ and $0.\overline{60}$ as fractions.

Let $r=0.\overline{6}$. Then $10r=6.\overline{6}=6+r$, so $9r=6$ and $r=\frac{6}{9}=\frac{2}{3}$.

Let $s=0.\overline{60}$. Then $100s=60.\overline{60}=60+s$, so $99s=60$ and $s=\frac{60}{99}=\frac{20}{33}$.

Therefore, $x = r\cdot s = \frac{2}{3}\cdot \frac{20}{33} = \frac{40}{99}$.

We can convert $\frac{40}{99}$ to a decimal by long division:

$$

\begin{array}{*{8}{[email protected]{}} l}

&& 0 & . & 4 & 0 & 4 & 04 & \ldots \\

\cline{2-9}

99 & | & 40 & . & 0 & 0 & 0 & 00 & \ldots \\

&& 39 && 6 \\

\cline{3-5}

&&&& 4 & 0 & 0 \\

&&&& 3 & 9 & 6 \\

\cline{5-7}

&&&&&& 4 & 00 \\

&&&&&& 3 & 96 \\

\cline{7-8}

&&&&&&& 4 \\

&&&&&&& \vdots

\end{array}

$$

The repeating pattern of remainders explains why the decimal for $\frac{40}{99}$ will continue to repeat, giving us

$$\frac{40}{99} = 0.\overline{40} = 0.404040404040404\ldots~.$$The first $15$ digits after the decimal point are eight $4$s and seven $0$s, so their sum is $8\cdot 4 = \boxed{32}$.

what is the original number? you forgot to add it.