find the sum of numberswhich are divisible by 2 or 3 between 1 and 100(including100)
Answers
From 1 to 100,all numbers are either divisible by 2 or by 3 except
(a) All
Prime numbers (excluding 2)
(b) Number 1 (not coming under calculations )
So in order to have the sum of all numbers divisible either by 2 or 3 First we have to get the sum of all Prime numbers & sum of all numbers from 2 to 100,then we will subtract the former from the latter to arrive at the ANSWER.
SUM OF ALL PRIME NUMBERS = 1058
SUM OF NUMBERS 1 TO 100
{(100)*(100+1)}/2=5050
5050–1= 5049
ALL NUMBERS FROM 1 TO 100 DIVISIBLE BY 2 OR 3=5049 -1058=3991 ANSWER
List of 25 Prime numbers from 1 to 100
2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89&97
I suppose there is no formula to add all prime numbers within a certain range of number.
ANSWER ::
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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