Question

The sum of a two digit numbers and the number obtained by reversing the digits is 66 . If the digits of the number differ by 2 find the number. How many such numbers are there?​

Answer 1

Step-by-step explanation:

this is the correct answer

Answer 2

$\huge\sf{\underline{\underline{Question:}}}$

The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?

_____________________

$\huge\mathfrak{\underline{Solution:}}$

• Let the ten’s digit in the first number be x.

• Unit's digit in the first number be y. 

Therefore, the first number can be written as 10x + y

When the digits are reversed, x and y becomes the unit's digit and ten's digit respectively.

Now, the number will be in the expanded notation is 10y + x.

$\bold{According\: to \:the \:given \:condition,}$

$\implies$(10x + y) + (10y + x) = 66

$\implies$11(x + y) = 66

$\implies$x + y = 66/11

$\implies$x + y = 6 ___ (1)

$\bold{Hence,}$

$\implies$ x – y = 2 ___(2)

$\implies$y – x = 2 ___(3)

If x – y = 2, then solving (1) and (2) by elimination, x = 4 and y = 2. In this case, the number is $\bold{42}$

If y – x = 2, then solving (1) and (3) by elimination, x = 2 and y = 4. In this case, the number is $\bold{24}$

Therefore,

the required such numbers are $\bold{42}$ and $\bold{24}$

$\huge\mathfrak{\underline{Verification:}}$

42 + 24 = $\bold{66}$