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
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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