Question

The sum of the digits of a two - digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Answer 1

Let the one’s digit be ‘a’ and ten’s digit be ‘b’.

therefore no is = 10b + a

Reversed no = 10a + b

Given,

Sum of the digits of a two - digit number is 9.

Also, nine times this number is twice the number obtained by reversing the order of the digits.

⇒ a + b = 9 ----- (1)

And 9(10b + a) = 2(10a + b)

⇒ 90b + 9a = 20a + 2b

⇒ 88b = 11a

⇒ a = 8b

Substituting value of a in eq1

⇒ 8b + b = 9

⇒ 9b = 9

⇒ b = 1

Thus,

a = 8(1) = 9

Hence, no is 10(1) + 8 = 18