Answer:
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
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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.
