The total no. of numbers that are divisible by 2 or 3 between 100 and 200 (both inclusive) are

• 84.
  100+(n-1)2=200 // divisible by 2
  n=51
  102+(n-1)3=198 // divisible by 3
  n=33
  102+(n-1)6=198 // divisible by 6
  n=17
  so , numbers divisible by 2 or 3 is , by removing the numbers divisible by 2 & 3 ,,,
  33+51=84-17 = 67
  
• 9 years agoHelpfull: Yes(5) No(0)
• 51+33-17=67
• 9 years agoHelpfull: Yes(2) No(0)
• 67 is answer
• 9 years agoHelpfull: Yes(1) No(0)
• No of digits between 100&200 divisible by 2 is 50. Since 100 is included, it is 51. And the no of digits between 100 & 200 divisible by 3 is 33 (using arithmetic formula T(n)=a+(n-1)d. 198=102+(n-1)3.)
  The double counted digits must be divisible by both 2 and 3 ie it must be a multiple of 6(LCM of 2&3).
  No of digits divisible by 6 =17(using AP formula).
  Therefore Total =51+33-17 =67
  
• 9 years agoHelpfull: Yes(1) No(0)
• 84.
  100+(n-1)2=200
  n=51
  102+(n-1)3=198
  n=33
  33+51=84
  
• 11 years agoHelpfull: Yes(0) No(6)