Landing Strip Brazilian

Zadanie

Find the lowest four consecutive odd integers whose sum is divisible by 15.


solution to the problem:
four consecutive odd numbers is: 2k-3, 2k-1, 2k +1, 2k +3, where k> 1 to N
knalezy sum of these numbers is: 2k-3 +2 k-1 + 2k +1 +2 +3 = 8k k
smallest number of the form 8k 15 is divisible by 8 * 15 = 120, k = 15

Answer searched numbers satisfying conditions of the problem are: 27, 29, 31, 33