Write a $2$-digit number $ab$ and the number obtained by reversing its digits i.e. $ba$. Let their sum be a $3$-digit number $dad$.

$ab+ba=dad$

$\Rightarrow (10a+b)+(10b+a)=dad$

$\Rightarrow 11(a+b)=dad$

The sum $(a+b)$ cannot exceed $18$. Why?

Write a $2$-digit number $ab$ and the number obtained by reversing its digits i.e. $ba$. Let their sum be a $3$-digit number $dad$.

$ab+ba=dad$

$\Rightarrow (10a+b)+(10b+a)=dad$

$\Rightarrow 11(a+b)=dad$

Write all the $3$-digit numbers which are multiples of $11$ up to $198$.

Write a $2$-digit number $ab$ and the number obtained by reversing its digits i.e. $ba$. Let their sum be a $3$-digit number $dad$.

$ab+ba=dad$

$\Rightarrow (10a+b)+(10b+a)=dad$

$\Rightarrow 11(a+b)=dad$

Find the values of $a$ and $d$.

Find the values of the letters and give reasons for the steps involved:

Find the values of the letters and give reasons for the steps involved.

$\begin{array}{cc}1& \mathrm{A}\\ \times & \mathrm{A}\\ 9& \mathrm{A}\end{array}$

$$\begin{gathered} \mathrm{A} \mathrm{B} \\ + 3 7 \\ \mathrm{\_\_\_\_\_\_\_\_} \\ 6 \mathrm{A} \end{gathered}$$

Find the values of the letters and give reasons for the steps involved

$$\begin{gathered} \mathrm{A} \mathrm{B} \\ \times 3 \\ \mathrm{\_\_\_\_\_\_\_\_} \\ \mathrm{C} \mathrm{A} \mathrm{B} \end{gathered}$$

$$\begin{gathered} \mathrm{A} \mathrm{B} \\ \times 5 \\ \mathrm{\_\_\_\_\_\_\_\_} \\ C A \mathrm{B} \end{gathered}$$