1 to 500 numbers.. the total number of numbers areddm wrote:attachment
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....
help
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sudhir3127
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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 !!!!!!!!!!
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 !!!!!!!!!!
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parallel_chase
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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.
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.
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mals24
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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)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
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.
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pepeprepa
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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.
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.
