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? answer this question fast​

Answer 1

When the digits are reversed, x becomes the unit’s digit and y becomes the ten’s digit. This number, in the expanded notation is 10y + x.

According to the given condition.

(10x + y) + (10y + x) = 66

11(x + y) = 66

x + y = 6 ... (1)

You are also given that the digits differ by 2. Therefore,

either x – y = 2 ... (2)

or y – x = 2 ... (3)

If x – y = 2, then solving (1) and (2) by elimination, you get x = 4 and y = 2. In this case, the number is 42.

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

Thus, there are two such numbers 42 and 24...

Answer 2

Answer:

Original number = 24

Step-by-step explanation:

let the digits of the number be

x and y

so,

number = 10y + x

now,

given that,

The sum of the number and the number obtained by reversing the digits is 66.

so,

original number = 10y + x

reversed number = 10x + y

According to the question,

10y + x + 10x + y = 66

11x + 11y = 66

11(x + y) = 66

x + y = 66/11

x + y = 6. ....(1)

now,

Also given that,

the digits of the number differ by 2,

given digits = x and y

so,

x - y = 2 ...(2)

now,

we have,

x + y = 6. ....(1)

x - y = 2 ...(2)

adding both of equations,

x + y + x - y = 6 + 2

2x = 8

x = 8/2

x = 4

x + y = 6

4 + y = 6

y = 6 - 4

y = 2

so,

Original number = 10y + x

= 10(2) + 4

= 20 + 4

= 24

so,

Original number = 24