The sum of k consecutive natural numbers is k(2a + k – 1) / 2 = 100, meaning k(2a + k – 1) = 200. Testing odd factors gives only k = 5, yielding 18 + 19 + 20 + 21 + 22 = 100.
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The sum of k consecutive natural numbers is k(2a + k – 1) / 2 = 100, meaning k(2a + k – 1) = 200. Testing odd factors gives only k = 5, yielding 18 + 19 + 20 + 21 + 22 = 100.