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Question

From a two digit number N, we subtract the number with the digits reversed and find that the result is a positive perfect cube. Then:


A
N cannot end in 5
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B
N can end in any digit other than 5
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C
N does not exist
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D
there are exactly 7 values for N
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E
there are exactly 10 values for N
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Solution

The correct option is D there are exactly 7 values for N
Let the two digit number be 10a+b
N = 10a+b
N' = reversed number = 10b+a
N-N' = 9(a-b)
N-N' is positive perfect cube.
$$\therefore$$ a>b
For 9(a-b) be perfect cube, a-b = 3
$$\therefore$$ b $$\epsilon$$ [0, 6] $$\longrightarrow$$ 7 values
$$\therefore$$ a $$\epsilon$$ [3, 9] $$\longrightarrow$$ 7 values
$$\therefore$$ Total 7 values are possible for N.

Mathematics

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