Question

find the sum of numberswhich are divisible by 2 or 3 between 1 and 100(including 100)

Answer 1

Step-by-step explanation:

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Answer 2

Sum of the integers that are divisible by 2

= 2+4+6+......+100

2+4+6+......+100=2(1+2+3+....+50)

2+4+6+......+100=2(1+2+3+....+50)=2(50∗51)/2

2+4+6+......+100=2(1+2+3+....+50)=2(50∗51)/2=2550

2+4+6+......+100=2(1+2+3+....+50)=2(50∗51)/2=2550Sum of the integers that are divisible by 3

= 3+6+9+.........+99

3+6+9+.........+99=3(1+2+3+.....+33)

3+6+9+.........+99=3(1+2+3+.....+33)=3(33∗34)/2

3+6+9+.........+99=3(1+2+3+.....+33)=3(33∗34)/2=1683

Now if a numbers which are divisible by 2 or 3 is also divisible by 6

So the sum of integers that are divisible by both 2 and 3

= 6+12+18+.....+96

6+12+18+.....+96=6(1+2+3+...+16)

6+12+18+.....+96=6(1+2+3+...+16)=6∗(16∗17)/2

6+12+18+.....+96=6(1+2+3+...+16)=6∗(16∗17)/2=816

We know that

n(A∩B) = n(A)+n(B) - n(A∪B)

n(A∩B) = n(A)+n(B) - n(A∪B)= 2550 + 1683 - 816

n(A∩B) = n(A)+n(B) - n(A∪B)= 2550 + 1683 - 816= 3417

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