the sum of a two digit number and the number obtained by reversing the digits is 66 if the digit of the number differ by 2 find the number how many such numbers are there?
the sum of a two digit number and the number obtained by reversing the digits is 66 if the digit of the number differ by 2 find the number how many such numbers are there?
Let the units digit be x and the tens digit be y.
The number is 10y+x
If we reverse the digits, the tens digit becomes x and the units digit becomes y
The number becomes 10x+y.
The sum of the 2 numbers 66. It means 10y+x+10x+y = 66
=> 11x+11y = 66 => x+y = 6. (1)
If the 2 digits differ by 2, it may mean x-y = 2 or y-x =2
Add x-y = 2 to equation (1) giving 2x = 8 so x = 4
x+y = 6
=> 4 + y = 6
=> y = 2 So one possibility is number = 42.
To get other number, add y-x=2 to (1) giving 2y = 8
=> y = 4
and x+y = 6=> x+4 = 6 => x =2 => number is 24
Thus, two numbers are 42 and 24
Let the units digit be x and the tens digit be y.
The number is 10y+x
If we reverse the digits, the tens digit becomes x and the units digit becomes y
The number becomes 10x+y.
The sum of the 2 numbers 66. It means 10y+x+10x+y = 66
=> 11x+11y = 66 => x+y = 6. (1)
If the 2 digits differ by 2, it may mean x-y = 2 or y-x =2
Add x-y = 2 to equation (1) giving 2x = 8 so x = 4
x+y = 6
=> 4 + y = 6
=> y = 2 So one possibility is number = 42.
To get other number, add y-x=2 to (1) giving 2y = 8
=> y = 4
and x+y = 6=> x+4 = 6 => x =2 => number is 24
Thus, two numbers are 42 and 24
