# Prime turns into Perfect Cube, if you reverse its digits and subtract…What is the number

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?

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

$$41$$. 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 :-)

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