Question

When a two-digit number is divided by the sum of its digits, the quotient is 4. If the digits are interchanged, the reversed number is 6 less than twice the original number. Find the number.

Answer 1

Let the ten’s digit be a and the unit’s digit be b,

Then, the number = 10a + b,

`(10a + b)/(a + b)=4`

10a + b = 4(a + b)

10a + b = 4a + 4b

10a − 4a = 4b − b

6a = 3b

b = `6/3a`

∴ b = 2a    ...(1)

When the digits are interchanged, the new number = 10b + a,

According to the given condition:

10b + a = 2(10a + b) − 6

10b + a = 20a + 2b − 6

10b − 2b = 20a − a − 6

8b = 19a − 6    ...(2)

Here, Substituting equation (1) in (2):

8(2a) = 19a − 6

16a = 19a − 6

3a = 6

a = `6/3`

∴ a = 2

Also,

b = 2a

= 2(2)

∴ b = 4 

So, the number is:

10a + b

= 10(2) + 4 

= 20 + 4

= 24

Hence, the required number is 24.