If from a two digit number we subtract the number formed by reversing its digit then then the result so obtained is a perfect cube. how many such number are possible? write all of them.
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Let the tens digit be x and unit digit be y.
So the number is 10x + y
And the number formed by reversing its digits is 10y + x
Given, 10x +y - (10y + x) = a perfect cube
We assume that the perfect cube is a^3
10x + y - 10y - x = a^3
9(x -y) = a^3
3*3 * (x-y) =a^3
Hence x -y = 3
so the possible numbers are 96, 85, 74, 63, 52, 41 , 30.
So the number is 10x + y
And the number formed by reversing its digits is 10y + x
Given, 10x +y - (10y + x) = a perfect cube
We assume that the perfect cube is a^3
10x + y - 10y - x = a^3
9(x -y) = a^3
3*3 * (x-y) =a^3
Hence x -y = 3
so the possible numbers are 96, 85, 74, 63, 52, 41 , 30.
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0
Answer:
answer is 96,85,74,83,52,41,30
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