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The total number of numbers that are divisible by 2 or 3 between 100 and 200(inclusive) are
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- Answer is 67.
Solotion- for number to be divisible by 2 it will form an A.P
100,102,104......200
a=100, d=2, L=200, Using L=a+(n-1)d we get n=51
for number to be divisible by 3 it will form an A.P
102,105,108.......198
a=102,d=3,L=198 Using same formula n=33
but in this some number are repeated in both A.P so to remove number which comes twice
we find the number of numbers divisible by 6 and then subtract it from total number.
102,108,114......198
a=102,d=6,L=198 so now n=17
So total number which 51+33-17=67 - 10 years agoHelpfull: Yes(81) No(0)
- total nos are 200-100+1=101
no that is divisible by 2 is =((200-100)/2)+1=51
now no that is divisible by 3 not by 2 =(195-105)/3 +1 =17
total=68 - 10 years agoHelpfull: Yes(7) No(12)
- are bhayya between 100 and 200 nikalna h so 100 and 200 not include
and get ans 65 - 8 years agoHelpfull: Yes(2) No(0)
- There will be (200-100)/2+1 = 51 numbers divisible by 2.
There will be Floor[(200 - 100)/3] = 33 numbers divisible by 3.
Multiples of 6 from the 17th through the 33rd will be counted twice by the sum of these numbers (51+33). There are 17 of them.
That means the count of numbers you are interested in is 51+33-17 = 67. - 6 years agoHelpfull: Yes(0) No(0)
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