Question

If from a two digits number, we subtract the number formed reversing its digits then the result so obtained is a perfect cube. How many such numbers are possible?
Write all of them.

Answer 1

let $ab$ is a two digit number.

Then reversing number is $ba$

$ab-ba=(10a+b)-(10b+a)$

$ab-ba=9(a-b)$

$ab-ba$ is perfect cube number and multiple of $9$

So, perfect cube can be $27$, therefore, $a-b=3$

For, bigger perfect cubes $a-b$ is two or three digit numbers which is not possible.

In above equation, $b=0$ to $6$, for $b>6$, $a$ becomes double digit number.

For $b=0,a=3$ and number is $30$

For $b=1,a=4$ and number is $41$

For $b=2,a=5$ and number is $52$

For $b=3,a=6$ and number is $63$

For $b=4,a=7$ and number is $74$

For $b=5,a=8$ and number is $85$

For $b=6,a=9$ and number is $96$

So, these are possible numbers.