Question

$\frac{\begin{array}{cc}A& B\\ \times & B\end{array}}{CAB}$

Answer 1

$AB\times B=$ three digit number, that has unit digit $B$

Therefore $A$ can be $2-9$, As $10\times 9=90$ which is two digit

$B$ is a unit place digit such that its square has same unit place

So, possibilities of $B$ are $0,1,5,6$

But $0$ and $1$ are neglected because they cannot give three digit number when multiplied by $2$ digit number.

So, $B=5or6$

If $B=5$ then $2$ will be added in $A\times B$ to get $CA$ and If $B=6$ then $3$ will be added in $A\times B$ to get $CA$

So, for $B=5$, product $A\times B$ will produce unit place $0$ and $5$ and adding $2$ to it will give unit place $2$ and $7$ that means $A=2or7$

So, for $25\times 5=125$ which follows and for $75\times 5=375$ which also follows this condition.

Now, when $B=6$ product $A\times B$ will produce unit place $6,2,8,4$ and $0$ and adding $3$ to it will give unit place $9,5,1,7$ and $3$ that means $A=9,5,1,7or3$

So, for $96\times 6=576$ which does not follows

For $56\times 6=336$ which also does not follows this condition.

For $16\times 6=96$ which also does not follows this condition.

For $76\times 6=456$ which also does not follows this condition.

For $36\times 6=216$ which also does not follows this condition.

So, $A=2,B=5$ and $A=7,B=5$ follows given product.

hence, $A=25,B=5,C=1$ or $A=7,B=5,C=3$