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.

By BYJU'S Exam Prep

Updated on: September 25th, 2023

The number is 18. It possesses the unique quality of having its digits summing up to 9. Additionally, when multiplied by nine, it results in a product that is precisely double the value obtained by reversing the order of its digits. One can easily find the number by applying logical reasoning and algebraic manipulation.

Let’s consider the two-digit number as “10a + b,” where a represents the tens digit and b represents the units digit. The sum of the digits being 9 can be represented as a + b = 9. Solve it further to find the values of a and b, and hence determine the number that satisfies the given conditions.

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1. 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.

Solution:

Let’s assume the two-digit number to be represented as 10a + b, where ‘a’ and ‘b’ denotes the tens and units digits, respectively.

We know that the sum of these digits is 9, so we can write the equation as

a + b = 9

Furthermore, we are given that nine times the original number is equal to twice the number obtained by reversing the digits. Mathematically, this can be expressed as:

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

Expanding both sides of the equation, we get:

90b + 9a = 20a + 2b

20a + 2b – 90b – 9a = 0

11a – 88b = 0

11(a – 8b) = 0

a – 8b = 0

a = 8b

Substituting this value, we find that a = 1.

Since, we know a = 1, the value of b will be

10a + b

10(1) + 8

Therefore, the number we seek is 18.

Summary: