Question 129094: What is the sum of the integers between 1 and 300 that are divisible by 11 and/or 13?
Found 2 solutions by checkley71, solver91311:
Answer by checkley71(8403)   (Show Source):
You can put this solution on YOUR website! 11,22,33,44,55,66,77,88,99,110,121,132,143,154,165,176,187,198,209,220,231,242,
253,264,275,286,297 are all the integers divisable by 11.
13,26,39,52,65,78,91,104,117,130,143,156,165,182,195,208,221,234,247,260,273,286,299 are all the integers divisable by 13.
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website! First, divide 300 by 11 using integer division. You get 27 with a remaider of 3.
That means that there are 27 integers less than 300 that are divisible by 11. The first of these integers is quite obviously 11 itself, and the last is 11 times 27 = 297.
The sum of a series of numbers is given by  where a is the first number, l is the last number, and n is the number of numbers. For the 11s, we have a = 11, l = 297, and n = 27. So the sum of the numbers divisible by 11 is 
Performing the same process for 13, we get a = 13, l = 299, and n = 23. I'll let you figure out how I got those numbers. The sum of the numbers divisible by 13 is then 
Now that we have the sums of the numbers divisible by 11 and 13, we should just be able to add them together to get a grand total, right? Well, not quite. There is just one more detail to consider. There are two numbers divisible by BOTH 11 and 13 that are less than 300, namely 11 X 13 = 143 and 2 X 11 X 13 = 286. If we just add the two sums derived above, we will have counted these two numbers twice. So the sum desired is 4158 + 3611 - 143 - 286 = 7340.
Done.
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