Call a positive integer , if it is prime and the cube of the number in decimal representation can be split into 3 parts such that their sum adds up to the original number.
Let be all of the distinct numbers on the interval . Determine the value of if:
Details and Test cases :
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There are three good numbers between 1 and 1 0 9 , namely 1 3 0 2 5 7 1 7 7 7 6 7 7 7 7 9 4 5 1 3 4 6 3 9 Observe that 1 3 0 2 5 7 1 3 1 3 0 2 5 7 1 7 7 7 6 7 7 7 7 3 7 7 7 6 7 7 7 7 9 4 5 1 3 4 6 3 9 3 9 4 5 1 3 4 6 3 9 = = = = = = 2 2 1 0 0 6 0 7 6 6 1 5 4 3 1 5 4 1 1 2 2 1 0 0 6 + ( 0 ) 7 6 6 1 5 4 + 3 1 5 4 1 1 4 7 0 3 2 6 0 7 2 3 2 0 8 7 3 7 0 7 5 2 6 4 3 3 4 7 0 3 2 6 0 7 + 2 3 2 0 8 7 3 7 + ( 0 ) 7 5 2 6 4 3 3 8 4 4 2 6 9 3 8 4 3 7 3 2 8 2 7 0 0 0 9 2 9 5 9 1 1 9 8 4 4 2 6 9 3 + 8 4 3 7 3 2 8 2 7 + ( 0 0 0 ) 9 2 9 5 9 1 1 9 There are no others. I listed the 5 0 8 4 7 5 3 4 primes between 1 and 1 0 9 (an export from Mathematica), and tested these numbers for goodness, using the Delphi code:
where BigCube is a routine that provides the decimal expansion of the cube of a number less than 1 0 1 0 as a string.
Since, modulo 1 0 1 0 , we have 1 3 0 2 5 7 1 1 3 0 2 5 7 1 7 7 7 6 7 7 7 7 7 7 7 6 7 7 7 7 9 4 5 1 3 4 6 3 9 9 4 5 1 3 4 6 3 9 ≡ ≡ ≡ 7 1 2 8 2 7 3 5 7 1 1 3 9 3 1 7 7 6 9 7 7 2 6 4 6 5 7 3 5 9 the answer is 5 7 8 6 1 0 9 0 2 7 .