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(D) Let $S$ be the sum of all integers from $1$ to $100$.$S = \frac{100}{2}(1 + 100) = 50 	imes 101 = 5050$.Let $S_1$ be the sum of integers divisibl

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$2632$

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(D) Let $S$ be the sum of all integers from $1$ to $100$.$S = \frac{100}{2}(1 + 100) = 50 	imes 101 = 5050$.Let $S_1$ be the sum of integers divisible by $3$ up to $100$: $3, 6, 9, \dots, 99$.$S_1 = 3(1 + 2 + 3 + \dots + 33) = 3 	imes \frac{33 	imes 34}{2} = 3 	imes 33 	imes 17 = 1683$.Let $S_2$ be the sum of integers divisible by $5$ up to $100$: $5, 10, 15, \dots, 100$.$S_2 = 5(1 + 2 + 3 + \dots + 20) = 5 	imes \frac{20 	imes 21}{2} = 5 	imes 10 	imes 21 = 1050$.Let $S_3$ be the sum of integers divisible by both $3$ and $5$ (i.e.,divisible by $15$) up to $100$: $15, 30, 45, 60, 75, 90$.$S_3 = 15(1 + 2 + 3 + 4 + 5 + 6) = 15 	imes \frac{6 	imes 7}{2} = 15 	imes 21 = 315$.Using the Principle of Inclusion-Exclusion,the sum of integers divisible by $3$ or $5$ is $S_1 + S_2 - S_3 = 1683 + 1050 - 315 = 2418$.The sum of integers not divisible by $3$ or $5$ is $S - (S_1 + S_2 - S_3) = 5050 - 2418 = 2632$.

The sum of the integers from $1$ to $100$ which are not divisible by $3$ or $5$ is

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