SOLUTION: Find the sum of all even numbers from 4 to 100 inclusive, excluding those which are multiples of 3.

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Question 1139868: Find the sum of all even numbers from 4 to 100 inclusive, excluding those which are multiples of 3.
Found 3 solutions by MathLover1, ikleyn, MathTherapy:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
the sum of all even numbers from 4 to 100 inclusive, excluding those which are multiples of 3.

We have to find the number of terms that are divisible by 2 but not by 6+( as the question asks for the even numbers only which are not divisible by 3)
For 2,
4,6,8,.........,100
using AP formula, we can say 100+=+4+%2B+%28n-1%29+%2A2
100-4=%28n-1%29+%2A2
96%2F2=n-1
48%2B1=n
n=49+
For 6,
6,12,18,......96
using AP formula, we can say 96+=6%2B+%28n-1%29+%2A6
96-6=%28n-1%29%2A6
90%2F6=n-1
15%2B1=n
n=16

49-16=33+ is the number of terms that are divisible by 2 but not by 6
here are they: all divisible by 2, highlighted divisible by 6
4, highlight%286%29, 8, 10, highlight%2812%29, 14, 16,highlight%28+18%29, 20, 22, highlight%2824%29, 26, 28, highlight%2830%29, 32, 34, highlight%2836%29, 38, 40, highlight%2842%29, 44, 46, highlight%2848%29,+50, 52, highlight%2854%29, 56, 58,highlight%28+60%29, 62,+64,highlight%28+66%29,+68, 70,highlight%28+72%29, 74, 76, highlight%2878%29, 80,+82, highlight%2884%29,+86, 88, highlight%2890%29,+92,+94, highlight%2896%29,98,100

now add all 33+ numbers that are not highlighted





Answer by ikleyn(52776) About Me  (Show Source):
You can put this solution on YOUR website!
.

            The  RIGHT  WAY  to solve the problem is  THIS : 

            Calculate the sum of all even integers from 4 to 100 inclusively, and then subtract from it 
            the sum of all multiples of 6 from 6 to 96.

            Below is how I implemented it. 


(a)  the sum of all even integers from 4 to 100 inclusively is


         4 + 6 + 8 + . . . + 100 = %28%284+%2B+100%29%2F2%29%2A49 = 2548.    


              (49 is the number of terms in this arithmetic progression).

 
(b)  the sum of all multiples of  6  from  6 to  96  inclusively is


         6 + 12 + 18 + . . . + 96 = %28%286+%2B+96%29%2F2%29%2A16 = 816.    


              (16 is the number of terms in this arithmetic progression).



     In both cases (a) and (b), I applied the formula for the sum of the first "n" terms of an arithmetic progression.



(c)  The difference  2548 - 816 = 1732  is your  ANSWER.

Solved.

Surely,  counting  "by hands"  (as the other tutor did)  is not the way to solve this problem.

-----------------

There is a bunch of lessons on arithmetic progressions in this site:
    - Arithmetic progressions
    - The proofs of the formulas for arithmetic progressions
    - Problems on arithmetic progressions
    - Word problems on arithmetic progressions
    - One characteristic property of arithmetic progressions
    - Solved problems on arithmetic progressions
    - Mathematical induction and arithmetic progressions

Also,  you have this free of charge online textbook in ALGEBRA-II in this site
    - ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic "Arithmetic progressions".


Save the link to this textbook together with its description

Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson

into your archive and use when it is needed.



Answer by MathTherapy(10551) About Me  (Show Source):
You can put this solution on YOUR website!
Find the sum of all even numbers from 4 to 100 inclusive, excluding those which are multiples of 3. 
Number of EVEN numbers, from 4 to 100, inclusive: matrix%281%2C3%2C+%28100+-+4+%2B+2%29%2F2%2C+%22=%22%2C+49%29

Sum of the 49 EVEN numbers, from 4 to 100, inclusive: 



Number of EVEN numbers, from 4 to 100, inclusive, that're MULTIPLES of 6 (2 * 3): matrix%281%2C3%2C+%2896+-+6+%2B+6%29%2F6%2C+%22=%22%2C+16%29

Sum of the 16 EVEN numbers, from 4 to 100, inclusive, that're MULTIPLES of 6: 



Sum of ALL EVEN numbers, from 4 to 100, inclusive, that're NOT MULTIPLES of 6: