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(B) To find the sum of integers from $1$ to $100$ divisible by $2$ or $5$,we use the Principle of Inclusion-Exclusion: $S = S_2 + S_5 - S_{10}$,where

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$3050$

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(B) To find the sum of integers from $1$ to $100$ divisible by $2$ or $5$,we use the Principle of Inclusion-Exclusion: $S = S_2 + S_5 - S_{10}$,where $S_n$ is the sum of integers divisible by $n$.$1$. Sum of integers divisible by $2$ $(2, 4, ..., 100)$: This is an arithmetic progression with $n = 50$,$a = 2$,$l = 100$. Sum $= \frac{50}{2}(2 + 100) = 25 	imes 102 = 2550$.$2$. Sum of integers divisible by $5$ $(5, 10, ..., 100)$: This is an arithmetic progression with $n = 20$,$a = 5$,$l = 100$. Sum $= \frac{20}{2}(5 + 100) = 10 	imes 105 = 1050$.$3$. Sum of integers divisible by both $2$ and $5$ (i.e.,divisible by $10$) $(10, 20, ..., 100)$: This is an arithmetic progression with $n = 10$,$a = 10$,$l = 100$. Sum $= \frac{10}{2}(10 + 100) = 5 	imes 110 = 550$.Total sum $= 2550 + 1050 - 550 = 3050$.

The sum of integers from $1$ to $100$ that are divisible by $2$ or $5$ is

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