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Replace A, B, C by suitable numbers:

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

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Best 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

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