Decimal cryptarithm

By converting recurring decimals to fractions we have:

$\frac { \overline { EVE } }{ \overline { DID } } =\frac { \overline { TALK } }{ 9999 } \Longrightarrow \overline { EVE } \times 9 \times 11 \times 101=\overline { DID } \times \overline { TALK }$

Since $\overline { EVE }$ and $\overline { DID }$ are coprime, and $\overline { TALK }$ is not divisible by $101$ ( $\overline { TALK } \neq \overline { abba }$ ), $\overline { DID }=101\times k$ with $k$ is a factor of $9$ , which then leads to $\overline { DID } =101, 303, 909$ ( $\overline { DID }$ cannot also be divisible by 11).

• If $\overline { DID } =101$ : since $E\neq D\neq I$ , $E\ge 2\Longrightarrow \overline { EVE } \times 99>10000>\overline { TALK }$ , a contradiction.

• If $\overline { DID } =909$ : the alphanumeric puzzle $\overline { EVE } \times 11=\overline { TALK }$ shows that $E=K$ , a contradiction.

• If $\overline { DID } =303$ : $\overline { EVE } \times 33=\overline { TALK } <10000,\quad E\neq D=3,\quad K\neq D=3$

$\Longrightarrow E<3,E\neq 1$ so we fix $E=2$ and $K=6$ .

Next, we try each value of $V$ in condition that $V\neq T\neq A\neq L\neq 0,2,3,6$ which then gives a unique result $V=4,T=7,A=9,L=8$

Hence, $\overline { EVE } +\overline { DID } +\overline { TALK } =242+303+7986=\boxed { 8531 }$

Since the repeating decimal is 4 places long, DID must divide 9999. 999 = 9 * 11 * 101. So DID is a multiple of 101.

Since DID is only 3 digits long, it cannot also be a multiple of 11. So EVE must be a multiple of 11. Since EVE is palindromic, it must be 121, 242, 363, or 484.

DID and EVE cannot have any digits in common. DID > EVE. It is now simple to try the possibilities. 242/303 is the only one that works.