help

ddm wrote:attachment

1 to 500 numbers.. the total number of numbers are

500-1+1 ( inclusive)

500

numbers divisible by 5 are

500/5
100

divisble by 11

500/11
45

divisbile by both

500/ 55 = 9

100 + 45 - 9 = 136

Not sure how u get 127....

Got doubt ! since 55 is (multiple of two integers rt ...ie . 5 * 11 ) so 500/55 is 9 , hence 9 * ( 2 integers) = 18 integers......
since 500 /5 = 100 ( becoz 5 is only 1 integer)
lly., 500/11 = 45 ( since 11 is only 1 integer ).

Hence 100+45 - 18 = 127 !!!!!

If iam wrong pls correct me !!!!!!!!!!

It has to be 136.

Let simplify the question how many integers from 1-20 inclusive are divisible by 2 or by 3 but not by both.

Total integers = 20-1+1 =20

divisible by 2 = 20/2=10

divisible by 3 = 20/3 = 6

divisible by both i.e. 6 = 20/6 = 3

10+6-3 = 13

if we go by "kshankker" method


divisible by 6 = 3 *(2) = 6 integers.

i dont think there are 6 integers divisible by 6 from 1-20 inclusive.

Answer has to 136.

hope this helps. Let me knowif anybody thinks otherwise.

parallel_chase wrote:It has to be 136.

Let simplify the question how many integers from 1-20 inclusive are divisible by 2 or by 3 but not by both.

Total integers = 20-1+1 =20

divisible by 2 = 20/2=10

divisible by 3 = 20/3 = 6

divisible by both i.e. 6 = 20/6 = 3

10+6-3 = 13

If you actually note down the numbers n check there are 10 and not 13 numbers that are divisible by 2 or 3 and not 6 (2,3,4,8,9,10,14,15,16,20)

we have to deduct 3 from both 10 and 6 separately hence kshankker's reasoning is correct IMO.

So in the actual question it should be (100-9)+(45-9) [which is the same as 100+45-18]
Hence the answer IMO should be 127.

As it was told:
500/5=100 numbers
500/11=45 numbers

We do not want those which are multiple of both 11 and 5. But these one number which are both multiple of 11 and of 5 are counted in the 100 and in the 45, so the are counted two times in 145.
For example,
5 10 15 20 25 30 35 40 45 50 55 60 65 ...
11 22 33 44 55 66 77 ...
Here 55 is counted two times but we don't want it, some we substract 2 times the number of divisible by 55.