IIT JEE , AIEEE Forum - Mathematics , Algebra - the sum of integers from 1 to 100 are divisible by 2 or 5 is (chapter is ap.gp,hp.)

inderjeet.parihar

the sum of integers from 1 to 100 are divisible by 2 or 5 is (chapter is ap.gp,hp.)

ganesha1991

numbers divisible by 2
a=2, d=2, n=50,a_n =100
therefore
s_n = n/2[a+a_n]
     = 50/2 [2+100]
     = 25 *102
     = 2550

numbers divisible by 5 but not by 2
5,15,25...
a=5, d=10,a_n=95,n=10
s_n= 10/2 [ 5+95]
   = 5 * 100
   = 500

therefore sum of numbers divisible by 2 or 8 is 2550+500 = 3050

konichiwa2x

edited

shadowcat

Is the ans 3050

elastiboysai

edited

i got 3050 init. den changed it by mist.

heres wat i did

sum of int div by 2, +5 -10

so 50 *51/2 +5 *(10 *21) - 10*11 *5

sboosy

the favourable numbers are 2 4 6 8 10 ....96 98 100 ....and 5 15 25 35 ...95

sum of all these is

first set we get

50/2 [ 2 +100 ] = 25*102 = 2550

second part

19/2 [ 5+95] = 19 * 50 = 950

so answer is 3500 ...

puneet

hii

well from the sum of two n 5 .. u hav double counted few things . tat are multiples of 10 .. u must substract them ..

ok . ?

cheers

computer001

sandeepramesh

this is a simple applcn of the inclusion exclusion principle :)

computer001

@sboosy:
uve counted the nos divisible by 10 twice..

@sandeepramesh:
post solutions not comments..

sandeepramesh

no i will only lead solns but not post them :D let the people who posted try?

And i will say it again we shd let others try too mr. computer001 so i think i will give leading ideas only

computer001

the idea "this is a simple applcn of the inclusion exclusion principle :)" is not very helpful 1 feels

sandeepramesh

perhaps one may be more inclined to do research on the inclusion principle if one doesnt knw? Or shd i even give links fer that simple search also? Huh? Leading questions to the answer accord to me is the best method

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