We note that under the hypothesis the top digits of $a_{(10)}$ and $b_{(10)}$ must be different, since if they are the same, then we get $\frac{a}{b} < 2$ which violates the condition that $a$ is a multiple of $b$. Therefore, we must have the top digit of $a_{(10)}$ or $b_{(10)}$ to be 6, but if the top digit of $b_{(10)}$ is 6, then the fact that $a \ge 2b$ forces $a$ to have more digits than $b$, contradicting the assumption. Therefore, the top digit of $a_{(10)}$ must be 6. Furthermore, since the number of digits of $a$ and $b$ are the same, $\frac{a}{b}$ must be one of the numbers 2, 3, 4, 5, 6. Let us call this number $m$.

Next, we will show that if $c$ is the minimum number satisfying the conditions, then the 6 to be removed from $b_{(10)}$ to get $c_{(10)}$ must be one of the last three digits of $b_{(10)}$. First we note that the minimality of $c$ implies that $b$ cannot be a multiple of 10. For if the last digit of $b$ is 0, this forces the last digit of both $a_{(10)}$ and $c_{(10)}$ to be 0, and thus dividing every one of $a, b, c$ by 10, we get a triple satisfying all the conditions, thereby violating the minimality of $c$. Now if the 6 to be removed from $b_{(10)}$ to get $c_{(10)}$ is not among the last three digits of $b$, then we must have $a \equiv b \pmod{1000}$. But from the fact $a = mb$ it follows that $(m-1)b \equiv 0 \pmod{1000}$, from which it follows that $b$ is a multiple of 10, as $m-1 \in \{1, 2, 3, 4, 5\}$, but this is impossible.

Next, we separate the cases depending on which digit the 6 to be removed from $b_{(10)}$ lies. We keep in mind that $b$ is not a multiple of 10 in the following discussion.

(1) Case when 6 to be removed is the last digit of $b$.
Since $a \equiv mb \pmod{10}$ the fact that $b \equiv 6 \pmod{10}$ determines the last digit of $a_{(10)}$, which would be equal to the next to the last digit of $b_{(10)}$. The consideration in terms of $\mod 100$, then determines the next to the last digit of $a_{(10)}$, which equals the third digit from the bottom of $b_{(10)}$. Continuing in this fashion, one can determine the digits of $a_{(10)}$ one by one from the bottom up. If the $n$-th digit from the bottom of $a_{(10)}$ equals 6, then we let $a, b$ be what are determined thus far. If this pair satisfies $a \equiv mb \pmod{10^n}$, then with $c_{(10)}$ determined from $b_{(10)}$ by removing the 6 in the last digit of $b_{(10)}$, we have the triple $(a, b, c)$ satisfying all the conditions. If the value of $m$ is given, then the $c$ obtained in this way will be the minimum one under the condition (1).
Checking for each of the cases $m = 2, 3, 4, 5, 6$ we see that with $m = 4$, we get $a = 615383$, $b = 153846$ and the corresponding $c = 15384$ gives the minimum.

(2) Case when 6 to be removed is next to the last digit of $b$.
From $a \equiv mb \pmod{10}$ and $a \equiv b \pmod{10}$ we can determine the last digit of $a, b$ to be 2, 4, 6 or 8 if $m = 6$, and be 5 if $m = 3$ or 5. If we check each of these possibilities as we have done in the case (1), we see that in each case the $c$ obtained has at least 6 digits.

(3) Case when 6 to be removed is the third digit from the bottom of $b$.
From $a \equiv mb \pmod{100}$ and $a \equiv b \pmod{100}$ we can get that the last two digits of $a, b$ are 25 and 75 respectively. If we check these cases further as we have done in the case (1), we see that in each case we get a $c$ having at least 6 digits.

Consequently, we can conclude that the smallest possible $c$ desired is 15348.