Question

A two digit number is equal to 7 times the sum of its digits

Answer 1

Let the digits in ones place be x

and ten’s place be y.

The original number = 10y + x Number formed by interchanging the digits = 10x + y (Given)

(10y + x) = 7(x + y)

10y + x = 7x + 7y

7x + 7y – x – 10y = 0

6x – 3y = 0

Dividing by 3 on both the side

2x – y = 0 → (1)

It is also given that on reversing the digits the number obtained is 18 less than the original number

Hence , 10x + y = (10y + x) – 18

10x + y – 10y – x = – 18

9x – 9y = – 18

Dividing by 9 on both the sides

x – y = – 2 → (2)

Subtract (2) from (1)

2x – y = 0

x – y = – 2

x = 2

Put x = 2 in 2x – y = 0

2(2) – y = 0

∴ y = 4

Therefore the original number is 42.

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