SOLUTION: Find the sum of all-natural numbers from 100 to 300:
i) Which are exactly divisible by 4.
ii) Excluding those which are divisible by 4.
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-> SOLUTION: Find the sum of all-natural numbers from 100 to 300:
i) Which are exactly divisible by 4.
ii) Excluding those which are divisible by 4.
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Question 1190794: Find the sum of all-natural numbers from 100 to 300:
i) Which are exactly divisible by 4.
ii) Excluding those which are divisible by 4. Found 2 solutions by math_helper, greenestamps:Answer by math_helper(2461) (Show Source):
Whenever you have a numbered set of things, the number of elements is the ending number minus the starting number, plus 1:
Examples:
If we number the elements of {ABCDE} starting from 0, we get:
A B C D E
0 1 2 3 4 --> number of elements = 4-0+1 = 5
and if we start at 1:
A B C D E
1 2 3 4 5 --> number of elements = 5-1+1 = 5
(you see here when you start counting with "1" the -1 and +1 always cancel and you can just pick the ending number for this common case)
i) 100 maps to 25 (100/4 = 25)
104 maps to 26
...
300 maps to 75 ---> number of elements of interest = 75-25+1 =
ii) Here, we are looking for the inverse. 300-100+1 = 201 total elements, and if we take out the 51 found in part (i), we are left with elements of interest.
For ANY set of numbers, the sum of the terms is the number of terms, multiplied by the average of the terms.
For an arithmetic sequence the average of the terms is the average of the first and last terms; the number of terms is (last minus first), divided by the common difference, plus 1.
(1) sum of ALL natural numbers from 100 to 300
number of terms:
average:
sum:
(2) sum of natural numbers from 100 to 300 that are divisible by 4
(note 100 and 300 are both divisible by 4)
number of terms:
average:
sum:
(3) sum of natural numbers from 100 to 300 that are NOT divisible by 4
