6. (10 points) A positive integer $x$ with $n$ digits, if appending it to the end of any two positive integers results in the product of the two new numbers still ending in $x$, then $x$ is called a “lucky number”. For example: 6 is a “lucky number”; but 16 is not, because $116 \times 216=25056$, and the end is no longer 16. The sum of all “lucky numbers” with no more than 3 digits is $\qquad$ .
6. (10 points) A positive integer $x$ with $n$ digits, if appending it to the end of any two positive integers, the product of the two new numbers still ends with $x$, then $x$ is called a "lucky number". For example: 6 is a "lucky number"; but 16 is not, because $116 \times 216=25056$, and the last digits are no longer 16. The sum of all "lucky numbers" with no more than 3 digits is $\qquad$.

【Solution】Solution: (1) The one-digit lucky numbers can only be $1, 5, 6$.
(2) Let the two-digit number that meets the condition be $\overline{\mathrm{ab}}$, satisfying $\overline{\mathrm{ab}}^{2}-\overline{\mathrm{ab}}$ is divisible by 100, $\overline{\mathrm{ab}}(\overline{\mathrm{ab}}-1)$ can be divided by 100. When the last digit $b=1$, there is no number that meets the condition.

When the last digit $b=5$, the number 25 meets the condition.
When the last digit $b=6$, the number 76 meets the condition.
(3) Let the three-digit number that meets the condition be $\overline{\mathrm{abc}}$, then $\overline{\mathrm{abc}}^{2}-\overline{\mathrm{abc}}$ must be divisible by 1000, i.e., $\overline{\mathrm{abc}}(\overline{\mathrm{abc}}-1)$ can be divided by 1000.
When the last two digits $\overline{\mathrm{bc}}=25$, $a=6$ meets the condition.
When the last two digits $\overline{\mathrm{bc}}=76$, $a=3$ meets the condition.
Therefore, the sum of the lucky numbers is: $1+5+6+25+76+625+376=1114$.
So the answer is: 1114.