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

1.	Replace A, B, C by suitable numbers:\(\begin{equation}\frac{\begin{array}[b]{r}A\,\,B\\\times\,\,\, B\,\,A\end{array}}{(B+1) C\,B}\end{equation}\)
Answer» Here, we can observe that B × A = B i.e. A = 1 Here, First digit = B+1 Thus, 1 will be carried from 1+B² and becomes (B+1) (B² -9) B. ∴ C = B² -1 Now, all B, B+1 and B² -9 are one digit number. This condition is satisfied for B=3 or B=4.For B< 3, B² -9 will be negative. For B>3, B² -9 will become a two digit number. For B=3, C = 32 - 9 = 9-9 = 0 For B = 4, C = 42 -9 = 16-9 = 7 Hence, A=1, B=3, C = 0 or A=1, B=4, C = 7

Replace A, B, C by suitable numbers:\(\begin{equation}\frac{\begin{array}[b]{r}A\,\,B\\\times\,\,\, B\,\,A\end{array}}{(B+1) C\,B}\end{equation}\)

Answer»

Here, we can observe that B × A = B

i.e. A = 1

Here,

First digit = B+1

Thus, 1 will be carried from 1+B² and becomes (B+1) (B² -9) B.

∴ C = B² -1

Now, all B, B+1 and B² -9 are one digit number.

This condition is satisfied for B=3 or B=4.For B< 3, B² -9 will be negative.

For B>3, B² -9 will become a two digit number.

For B=3, C = 32 - 9 = 9-9 = 0

For B = 4, C = 42 -9 = 16-9 = 7

Hence, A=1, B=3, C = 0 or A=1, B=4, C = 7