Replace A, B, C by suitable numbers:\(\begin{equation}\frac{\begin{arr

1.	Replace A, B, C by suitable numbers:\(\begin{equation}\frac{\begin{array}[b]{r}A\,\,B\\\times\,\,\,\, 3\end{array}}{C\,A\,\,B}\end{equation}\)
Answer» Here, (B × 3) = B Here, B can be either 0 or 5, which satisfies above condition. If B is 5, then 1 will be carried, then, A×3+1 = A will not be possible for any number ∴ B = 0 Also, A×3=A is possible for either 0 or 5. If we take A=0, then all number will become 0, which is not possible ∴ A= 5 So, 1 will be carried. ∴ C = 1

Replace A, B, C by suitable numbers:\(\begin{equation}\frac{\begin{array}[b]{r}A\,\,B\\\times\,\,\,\, 3\end{array}}{C\,A\,\,B}\end{equation}\)

Answer»

Here, (B × 3) = B

Here, B can be either 0 or 5, which satisfies above condition.

If B is 5, then 1 will be carried,

then, A×3+1 = A will not be possible for any number

∴ B = 0

Also, A×3=A is possible for either 0 or 5.

If we take A=0, then all number will become 0, which is not possible

∴ A= 5

So, 1 will be carried.

∴ C = 1