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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.


Solution

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.

Mathematics

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