  # 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.

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Updated on: September 25th, 2023 Consider x and y as the digits in units and tens place of the given number

10y + x is the number

Given that sum of two-digit numbers is 9

x + y = 9

By interchanging the digits, the number becomes 10x + y

Nine times this number is twice the number obtained by reversing the order of the digits

9 (10y + x) = 2 (10x + y)

By further calculation

90y + 9x = 20x + 2y

20x + 2y – 90y – 9x = 0

11x – 88y = 0

Taking 11 as common

11 (x – 8y) = 0

x – 8y = 0

The system of equations is

x + y = 9 ….. (1)

x – 8y = 0 ….. (2)

x = 8y

Let us solve the system of equations for x and y

Substituting x = 8y in equation (1)

8y + y = 9

9y = 9

y = 1

Substituting y = 1 in equation (2)

x – 8 (1) = 0

x = 8

So the number will be 10 x 1 + 8 = 18

Therefore, the number is 18.

Summary:

## 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.

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. The number is 18.

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