What is the sum of all multiples of 3 or 5 below 2,200?

Step-by-step explanation:

Three

Sum of all numbers beginning at 0 up to 2200 that are divisible by 3

a = 3

d = 3

L = 2199

Now we need to find out how many there are.

L = a1 + (n - 1)*d

2199 = 3 + (n-1)*3             Subtract 3

2199 - 3 = (n - 1)*3            Divide by 3          

2196/3 = n - 1

732 = n - 1                        Add 1 to both sides

733 = n                    

Sum = (a + L)*n /2            Substitute the values

Sum = (3 + 2199)*733/2   Combine and Divide by 2

sum = (1101)*733               Multiply

Sum = 807033

Five

Go through exactly the same steps.

a = 5

L = 2195  ( below 2200. You can't include 2200)

d = 5

L = a + (n-1)*d

2195 = 5 + (n-1)*5

2190 = (n - 1)*5

438 = n - 1

439 = n

Sum = (a + L ) * n

Sum = (5 + 2195)*439/2

Sum = 1100 * 439

Sum = 482900

The problem is ambiguous, so I'm going to leave the two answers that are there.

Here's the problem.

What do you do with the numbers that are divisible by both? I counted them in both sums, but that may not be correct.

As it stands your answer should 965800 + 482900

Nor does it seem right to count them only in 1 sum. So the answer is given by adding the two numbers above.

hbfhv.hjgfskhgfdbmx n

step-by-step explanation:

9-15%