Question

The sum of the digits of a two-digit number is 7. When the digits are interchanged, the reversed number is 5 times the ten’s digit of the original number. Find the original number.

Answer 1

Let the ten’s digit be a and the unit’s digit be b. Then the number is 10a + b,

Given:

a + b = 7    ...(1)

Reversed number = 10b + a and it equals 5 × (ten’s digit) = 5a:

10b + a = 5a

10b = 5a − a

10b = 4a     ...(Dividing this expression by 2)

5b = 2a    ...(2)

∴ a = `(5b)/2`

Here, Substituting a = `(5b)/2` in equation (1):

`(5b)/2 + b = 7`

`(5b + 2b)/2 = 7`

7b = 14

b = `14/7`

∴ b = 2

Now, b = 2 in equation (2):

5(2) = 2a

10 = 2a

a = `10/2`

∴ a = 5

Also,

Original number = 10a + b

= 10 × 5 + 2

= 52

Hence, the required number is 52.