Question

Find the numerical value of the letters given below

Answer 1

Given, product of $DE$ and $3=FDE$
Since $DE$ is multiple of $3$ we need to find the number such that numbers $D,E$ exists in the product value. By trial and error method we get
$D=5$ and $E=0$
And the product of $50\times 3=150$
$\therefore F=1$ and $D,E$ numbers exists in the value $150$
Hence, $D=5,E=0$ and $F=1$