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

Number divisible by 2 = even numbers between 1 and 100 (both inclusive)

Sum of even numbers = $50(51) = 2550$

Now, numbers divisible by 5 either end in 5 or 0.

But the numbers ending in 0 are also divisible by 2.

Sum of numbers ending in 5 = $5+15+25+...+95$

$= 5(1+3+5+...+19)$

$= 5\times10^2 = 5\times100 = 500$

$(2n -1 ) = 19 \Rightarrow n = 10$

Therefore, required sum = $2550 + 500 = 3050$

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