Japan Mathematical Olympiad

$138$

$138$ is a multiple of $3$, but not of $9$, and since $138+1\times 3\times 8=162$ is a multiple of $9$, the number $138$ satisfies the conditions of the problem. We will show in the sequel that if $A<138$, then $A$ does not satisfy the conditions of the problem.

Now, if a number $A$ satisfies the conditions of the problem, then it is easy to see that the product of all the digits of $A$ must be a multiple of $3$ but not a multiple of $9$. Hence $A$ must satisfy the following condition, which we call the condition X:

One and only one of the digits of $A$ must be either $3$ or $6$ and none of the digits of $A$ can equal $0$ or $9$.

If $A$ has only one digit, then neither $3$ nor $6$ satisfies the conditions of the problem, so $A$ must have at least two digits. Now write $A=10a+b$ with $a$ a positive integer and $b$ an integer such that $0\le b\le 9$. From the condition that $A<138$, we see that $a\le 13$. Now consider the following cases:

(1) When $b=3$ or $6$: Since $10a=A-b$ must be a multiple of $3$ in this case, $a$ must also be a multiple of $3$. If $a=3$ or $6$ or $9$, then $A$ violates the condition X, so we must have $a=12$. But then since $126$ is a multiple of $9$, $A$ cannot be $126$. On the other hand, since $123+1\times 2\times 3=129$ is not a multiple of $9$, $A$ cannot be $123$ either. So, the case under consideration cannot occur.

(2) When $b$ is neither $3$ nor $6$: By the condition X, we see that one and only one of the digits of $a$ must be either $3$ or $6$. If $a=3$ or $a=6$, then $b=A-10a$ must be a multiple of $3$, but any choice of $b=0,3,6,9$ violates the condition X. Finally, if $a=13$, only number less than $138$ and is a multiple of $3$ and not of $9$ is $132$, but since $132+1\times 3\times 2=138$ is not a multiple of $9$, the case under consideration cannot occur either.

Thus we conclude that $138$ is the smallest possible positive integer satisfying the conditions of the problem.