The sum of the digits of a two-digit number is 12. If the new number formed by reversing the digits is greater than the original number by 54, find the original number. Check your solution.
Let the units digit of the two-digit number be y and the tens digit be x.
Hence, the number will be 10x+y.
Given that the sum of digits is 12.
Hence, x+y=12.......(i)
It is also given that the new number formed by reversing the digits is greater than the original number by 54.
On reversing the digits, x becomes the units digit and y becomes the tens digit.
Hence, the new number will be 10y+x.
∴ 10y+x=10x+y+54
⇒9y−9x=54
⇒y−x=6
⇒−x+y=6.........(ii)
Adding (i) and (ii), we get:
(x+y)+(−x+y)=12+6
⇒2y=18
∴ y=182=9
Substituting y=9 in (i), we get:
x+9=12
∴ x=12−9=3
The units digit is 3 and the tens digit is 9.
Hence, the number is 39.
Checking the solution:
Original number =39
Hence, number obtained by reversing the digits =93
93=39+54
Hence, the number obtained by reversing the digits is greater than the original number by 54.
Hence, the answer is correct.