Japan Mathematical Olympiad

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 $a\equiv b(\mathrm{mod}1000)$. But from the fact $a=mb$ it follows that $(m-1)b\equiv 0(\mathrm{mod}1000)$, 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 $a\equiv mb(\mathrm{mod}10)$ the fact that $b\equiv 6(\mathrm{mod}10)$ 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 $\mathrm{mod}100$, 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 $a\equiv mb(\mathrm{mod}10^n)$, 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 $a\equiv mb(\mathrm{mod}10)$ and $a\equiv b(\mathrm{mod}10)$ 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 $a\equiv mb(\mathrm{mod}100)$ and $a\equiv b(\mathrm{mod}100)$ 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.