If from a two-digit number, we subtract the number formed by reversing its digits then the result so obtained is a perfect cube. How many
such numbers are possible? Write all of them.
Let ab be any two-digit number. Then, the digit formed by reversing it digits is ba.
Now, ab-ba = (10a+b)-(10b +a)
= 9a – 9b = 9(a – b)
Further, since ab-ba is a perfect cube and is a multiple of 9.
.-.The possible value of a – b is 3.
i.e. a = b + 3
Here, b can take value from 0 to 6.
Hence, possible numbers are as follow.
For b = 0, a = 3, i.e. 30
For b = 1, a = 4, i.e. 41
For b =2, a = 5, i.e. 52
For b = 3, a = 6, i.e. 63
For b = 4, a = 7, i.e. 74
For b = 5,a = 8, i.e. 85
For b = 6, a = 9, i.e. 96