5. Veronica observes that $81 \cdot 3=243$ and $81 \cdot 4=324$ and wonders how many numbers $m$ with $10 \leq m \leq 99$ exist such that $3 m=A B C$ and $4 m=C A B$, where $A, B$, and $C$ are decimal digits (cases where one or more of the digits $A, B, C$ are zero are also considered valid).
(A) 1
(B) 2
(C) 3
(D) 4
(E) 6
5. 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 $4 m=100 C+10 A+B$; thus, by difference, $m=99 C-9 B-90 A=9(11 C-B-10 A)$. On the other hand, $11 C-B-10 A=$ $12 C-9 A-(A+B+C)$ is 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 satisfy the curious property noticed by Veronica.

Second solution. 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 $10 \cdot 4 m-3 m=37 m$. We thus obtain $37 m=999 C$, or $m=27 C$; from here it is immediate to deduce that the sought values 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 we get $10 A+B=8 C$. This means that the number formed by the first 2 digits of $A B C$ is equal to 8 times the unit 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 $m=81$.