Problem 606

Solution:

The answer is $\mathbf{( C )}$. Since $3 m=A B C$ is obviously a multiple of 3, so is the sum of its digits $A+B+C$. Moreover, we know that $3 m=100 A+10 B+C$ and that $4 m=100 C+10 A+B$; by subtraction, therefore, $m=99 C-9 B-90 A=9(11 C-B-10 A)$. On the other hand $11 C-B-10 A=12 C-9 A-(A+B+C)$ turns out to be a multiple of 3, so $m$ must be divisible by 27. The only possibilities are therefore $m=27,54,81$ and a simple calculation reveals that they indeed verify the curious property that Veronica noticed.

Second solution.

We observe that $10(100 C+10 A+B)-(100 A+10 B+C)=999 C$, and by construction the expression $10(100 C+10 A+B)-(100 A+10 B+C)$ is equal to $10 \cdot 4 m-3 m=37 m$. One thus obtains $37 m=999 C$, that is $m=27 C$; from this it is immediate to deduce that the values sought are $27,54,81$.

Third solution.

Since $3 m<300$ we have $0 \leq A \leq 2$, and similarly $0 \leq C \leq 3$. The hypothesis tells us that $0=4(3 m)-3(4 m)=4(100 A+10 B+C)-3(100 C+10 A+B)=370 A+37 B-296 C$, and dividing by 37 one obtains $10 A+B=8 C$. This means that the number formed by the first 2 digits of $A B C$ is equal to 8 times the units digit; since, as already observed, $C$ does not exceed 3, the only possibilities are $000,081,162,243$, which correspond to $m=0$ (not acceptable), $m=27, m=54$ and $m=81$.