Problem 37
Solution to Problem 37. Written in Python.
Problem: The number 3797 has an interesting property. Being prime itself, it is possible to continuously remove digits from left to right, and remain prime at each stage: 3797, 797, 97, and 7. Similarly we can work from right to left: 3797, 379, 37, and 3.
Find the sum of the only eleven primes that are both truncatable from left to right and right to left.
NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.
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| from problems.euler_lib import euler_lib as lib | |
| def trunked(number): | |
| trunked_left = [number] | |
| trunked_right = [number] | |
| left_digits = map(int, str(number)) | |
| right_digits = map(int, str(number)) | |
| length_digits = len(left_digits) | |
| for i in range(length_digits): | |
| left_digits.pop() | |
| right_digits.pop(0) | |
| if(len(left_digits) > 0): | |
| trunked_left.append(int(''.join([str(x) for x in left_digits]))) | |
| if(len(right_digits) > 0): | |
| trunked_right.append(int(''.join([str(x) for x in right_digits]))) | |
| return all(t in sieve for t in trunked_left) and all(t in sieve for t in trunked_right) | |
| ctr = 0 | |
| number = 10 | |
| sieve = lib.eratosthenes_sieve(1000000) | |
| answer = 0 | |
| while ctr != 11: | |
| if trunked(number): | |
| ctr += 1 | |
| answer += number | |
| number += 1 | |
| print answer |