2018/19 AHSMC Part 1 Problem 8

Any number that is not divisible by $2$ and not divisible by $3$ has the form $6k + 1$ or $6k + 5$, where $0 \le k \le 20$. Therefore the sum is $$S = \sum_{k=0}^{20} (6k + 1) + \sum_{k=0}^{20} (6k + 5) = \sum_{k=0}^{20} (12k + 6) = 12(1+2+3+\cdots+20) + 6 \cdot 21 = 12 \cdot 21 \cdot 10 + 6 \cdot 21 = 2646.$$ The answer is (D).

Alternative Solution: The sum of all numbers smaller than $126$ is $$1+2+3+\cdots+125 = 125 \cdot 63.$$ The sum of all even numbers smaller than $126$ is $$2+4+6+\cdots+124 = 2(1+2+3+\cdots+62) = 62 \cdot 63.$$ The sum of all multiples of three smaller than $126$ is $$3+6+\cdots+123 = 3(1+2+\cdots+41) = 3 \cdot 41 \cdot 21.$$ The sum of all multiples of six smaller than $126$ is $$6+12+\cdots+120 = 6(1+2+\cdots+20) = 6 \cdot 21 \cdot 10.$$ Therefore, the sum is $$125 \cdot 63 - 62 \cdot 63 - 3 \cdot 41 \cdot 21 + 6 \cdot 21 \cdot 10 = 2646.$$