what is the sum of all integers between 1 to 100 which r divisible by 2 or 3? remember 2 or 3 not and 3?

Question

just solve this prob of progression

Answer 1

The numbers between 1 and 100 which are divisible by 2 are
2,4,6,8.........96,98,100
First term=2 and the last term is 100 and the common difference is 2.

Let n be the number of terms
Therefore,2+(n-1)*2=100
or,2+2n-2=100
or,2n=100

or,n=50
Therefore there are 50 terms in the series
The sum of all terms
=(2+100)*50/2
=2550
Again,all the numbers divisible by 3 are
3.6.9.12,15.........93,96,99
Let there be n terms in the series
therefore,3+(n-1)*3=99
or,3+3n-3=99
or,3n=99
or,n=33
Sum of the series
=(3+99)*33/2
=1683
in the above calculations,the nos divisible by both 2 and 3 ie.6,12,18.......90,96 have been included two times
in the series 6,12,18......96 ,let there be n terms
Therefore,6+(n-1)*6=96
or n=96 rn=16
The sum of the series
(6+96)*16/2
=102*8
=816
Therefore sum of all the numbers divisible by 2 or 3 is
2550+1683-816
=3417

Answer 2

You are summing

sum1 of 2, 4, 6, 8 , . .. , 100
and
sum2 of 3, 6, 9, 12, 15, ... , 99

Sum of integers from 1 to N
= N (N+1)/2

Sum1 = 2 x (sum of 1 to 50)
= 2 x 50 x (1 + 50)/2
= 50 x 51
= 2550

Sum2 = 3 x (sum of 1 to 33)
= 3 x 33 x (1 + 33)/2
= 3 x 33 x 34/2
= 99 x 17
= 1683

Sum1 + sum2 = 2550 + 1683 = 4233

Answer 3

i. Numbers divisible by 2:
2,4,...,100 (50 such numbers) their sum is:
(2 + 100) + (4 + 98) + .. + (50 + 52) = 25 * 102 =
= 2550

ii. Numbers divisible by 3:
3,6,...,99 (33 such numbers)
Their sum is:
(3 + 99) + (6 + 96) + ... + (48 + 54) + 51 =
= 102 * 16 + 51 = 1683

iii. Numbers divisible by 6 (i.e. 2 and 3):
6,12,...,96 (16 such numbers)
Their sum is (6 + 96) + (12 + 90) + ... + (48 + 54) =
= 102 * 8 = 816

Now, the sum of numbers divisible by 2 but not 6 is
(i) - (iii) = 2550 - 816 = 1734

The sum of numbers divisible by 3 but not by 6 is
(ii) - (iii) = 1683 - 816 = 867

The sum of the 2 last sums is 1734 + 867 = 2601
(If you meant that integers divisible by 6 are not included - exclusive OR)

Otherwise, subtract (iii) only once:
2601 + 816 = 3417.

Answer 4

6

Answer 5

of course, there is two hundred integers divisible with the help of three, and 3 hundred integers divisible with the help of two. yet a number of those numbers are divisible with the help of two and 3, which potential they are divisible with the help of 6. you will ought to choose what proportion are divisible with the help of 6 and subtract that huge form from the two hundred+3 hundred to get the respond.

Answer 6

3417 counting 100

Answer 7

3,417