What is the sum of all integers between 1 and 100 divisible

[Guest]

What is the sum of all integers between 1 and 100 which are divisible by 2 or 5.

[Guest]

Use the inclusion exclusion principle.
The sum of all integers which are divisble by 2 or 5
= The sum of all integers which are disible by 2
+ The sum of all integers which are disvisible by 5
- The sum of all integers that are divisible by 2 and 5 (i.e. are divisible by 10)

= (2 + 4 + 6 + ... + 100) + (5 + 10 + 15 + ... 100) - (10 + 20 + 30 + ... + 100)
= 2(1+2+3+...+50) + 5(1+2+3+...20) -10(1+2+3+...+10)
= 2(50x51/2) + 5(20x21/2) -10(10x11/2) (using the fact that 1+2+3+...+n = n(n+1)/2)
= 3050

[Guest]

Read the question again.....

"What is the sum of all integers between 1 and 100 which are divisible by 2 or 5. "

It is the sum of two arithmetic progressions

1..............from 2 to 100 with common difference of 2

2..............from 5 to 100 with common difference of 5

That means all integers that are divisible by 2 OR 5.......not 2 AND 5

and the answer is .......3600

................................................................................................

[HallsofIvy]

And YOU should read the response again! The sum of "all integers divisible by 2 or 5" is the sum of all integers divisible by 2 plus the sum of all integers divisible by 5 minus the sum of all integers divisible by 2 and 5. You, the third "guest", are wrong and the second "guest'' is right. Your adding "all numbers divisible by 2" and "all numbers that are divisible by 5" are counting number, such at 10, that are divisible by both 2 and 5, twice!

(This site should do something about all these "guests".)

[Baltuilhe]

Good afternoon!

Just to fan the flames

Between 1 and 100 ==> 2 till 99, 1 and 100 are out. So... 2950 is the correct answer

Have a nice weekend!