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?


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Answer 1

Answer:

There are 10 such numbers .

Answer 2

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

•  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

According to the givencondition,

⟹ (10x + y) + (10y + x) = 66

⟹ 11(x + y) = 66

⟹ x + y = 66/11

⟹ x + y = 6 ___ (1)

Hence,

⟹ x – y = 2 ___(2)

⟹ 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 42

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

Therefore,

the required such numbers are 42 and 24