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Question
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.
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Solution
Let ab be any two-digit number.
Then, the digit formed by reversing it digits is ba.
Now, ab – ba = (10a + b) – (10b + a)
= (10a – a) + (b – 10b)
= 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
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