Question

The sum of a two-digit number and the number obtained by reversing the digits is $88$. If the digits of the number differ by $4$, find the number. How many such numbers are there? 

OR 

The length of a rectangular field is $9$ m more than twice its width. If the area of the field is $810\ \text{m}^2$, find the length and width of the field.
 

Hint:

When a two-digit number and its reverse are involved, set the number as $10a+b$; reversing swaps $a$ and $b$. The equation $11(a+b)$ often appears from adding them.

Answer 1

Solution (Number with reverse):
 

Step 1: Let the number be $10a+b$
Here $a$ is the tens digit $(1\le a\le 9)$ and $b$ is the units digit $(0\le b\le 9)$. The reverse is $10b+a$.
 

Step 2: Use the given sum condition
Sum $=$ original $+$ reverse $\Rightarrow (10a+b)+(10b+a)=88 $$\Rightarrow$$ 11(a+b)=88 $$\Rightarrow$ $a+b=8$.
 

Step 3: Use the digit-difference condition
$\lvert a-b\rvert=4$. Solve the pair:
\(\hspace*{0.5cm}\)(i) $a-b=4$ with $a+b=8$ $\Rightarrow$ $2a=12$ $\Rightarrow$ $a=6,\ b=2$ $\Rightarrow$ number $=62$.
\(\hspace*{0.5cm}\)(ii) $b-a=4$ with $a+b=8$ $\Rightarrow$ $2a=4$ $\Rightarrow$ $a=2,\ b=6$ $\Rightarrow$ number $=26$.
\[ \boxed{\text{Numbers: }26 \text{ and }62;\ \ \text{there are }2\text{ such numbers.}} \]