5
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Given:

1) I have more than one digit.

2) Reversed number is subtracted to give at least 2 digit number which should be a cube.

2) you don’t even need a calculator to figure me out.

Who am I?

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    $\begingroup$ 'Reverse digits and subtract' is a little vague. There are already 2 valid interpretations of it below. Can you clarify? $\endgroup$ – TwoBitOperation May 15 at 16:35
  • $\begingroup$ Simple, take 11, $11 -11 = 0 = 0 ^3$ $\endgroup$ – Ak19 May 15 at 16:38
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    $\begingroup$ From the prime number subtract the reverse number, result should result in cube of at least 2 digits. $\endgroup$ – Uvc May 15 at 16:38
  • $\begingroup$ Ok got it now, thanks!!(+1) $\endgroup$ – Ak19 May 15 at 16:38
  • $\begingroup$ spoiler FWIW A080178 has such numbers (palindromic primes giving the trivial cube of zero) $\endgroup$ – Jonathan Allan May 15 at 18:02
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The answer is

$41$. It is prime and

$41 - 14 = 27 = 3^3$

Reasoning

Assuming the number has 2 digits, it can be written as

$10a+b$

Then

$(10a + b) - (10b + a) = 9(a-b)$

should be a non-zero cube. The only way to make it a non-zero cube with single digits $a$ and $b$, is to have

$a = b + 3$

The only 2 digit prime number that fits is $41$.

As far as uniqueness of this solution goes, it is easy to show that no 3 digit number fits. I guess this is as far as you can go without a calculator :-)

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3
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The other answer is

47

Same argument as ppgdev's answer, but

$47 - 74 = -27 = (-3)^3$

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    $\begingroup$ Nice twist! (+1)! $\endgroup$ – ppgdev May 16 at 2:20
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    $\begingroup$ True..as I did not specify it to be positive number. $\endgroup$ – Uvc May 16 at 3:18

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