part2 (885B)
1--- Part Two --- 2 3The final step in breaking the XMAS encryption relies on the invalid number you just found: you must 4[1m[37mfind a contiguous set of at least two numbers[0m in your list which sum to the invalid 5number from step 1. 6 7Again consider the above example: 8 935 20 [1m[37m15[0m [1m[37m25[0m [1m[37m47[0m [1m[37m40[0m 62 55 65 95 102 117 150 182 10127 219 299 277 309 576 11 12In this list, adding up all of the numbers from 15 through 40 produces the invalid number from step 131, 127. (Of course, the contiguous set of numbers in your actual list might be much longer.) 14 15To find the [1m[37mencryption weakness[0m, add together the [1m[37msmallest[0m and 16[1m[37mlargest[0m number in this contiguous range; in this example, these are 15 and 47, 17producing [1m[37m62[0m. 18 19[1m[37mWhat is the encryption weakness in your XMAS-encrypted list of numbers?[0m 20 21