The total number of numbers that are divisible by 2 or 3 between 100 and 200(inclusive) are

• 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
• 9 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
• 9 years agoHelpfull: Yes(7) No(12)
• are bhayya between 100 and 200 nikalna h so 100 and 200 not include
  and get ans 65
• 7 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.
• 5 years agoHelpfull: Yes(0) No(0)