Let the digits of the required number be x and y. 10x + y = 4(x + y) So,6x − 3y = 0
\[\Rightarrow\]2x − y = 0
\[x = \frac{y}{2}\] .....(1) Also, \[10\left( \frac{y}{2} \right) + y = 3\left( \frac{y}{2} \right)y\]\[ \Rightarrow 5y + y = \frac{3}{2} y^2 \] \[ \Rightarrow 6y = \frac{3}{2} y^2 \] \[\Rightarrow y^2 - 4y = 0\]\[ \Rightarrow y(y - 4) = 0\] \[ \Rightarrow y = 0, 4\] So, x = 0 for y = 0 and x = 2 for y = 4. Hence, the required number is 24. |