Let P be the product of any three consecutive positive odd integers each of which is less than 146. Then the largest integer dividing all such P is

1
6
3
7
15

• Anurag..what about 17*19*21,its not divisible by 15..
  (17*19*21)%3 ==0 and divisible by 7..
  (23*25*27)%3==0 but not divisible by 7
  and so on...
  so ans is 3...
• 11 years agoHelpfull: Yes(38) No(5)
• 15 is the ans...
  let the three nos. be a,a+2,a+4
  1st we take lower case i.e a=1,b=3,c=5
  So a*b*c=15 which is divisible by 15
  Now we take the upper case a=141,b=143,c=145
  So again a*b*c=2923635 which is divisible by 15
  
• 11 years agoHelpfull: Yes(19) No(23)
• anurag what about 7,9,11=693,i;e not divisible by 15
  thus correct answer is 3......
• 11 years agoHelpfull: Yes(5) No(0)
• plss write the correct ans
• 11 years agoHelpfull: Yes(3) No(0)
• Ans will be 3, because if we consider any possible consecutive odd numbers then one of them will be divisible by 3 thus making theri product divisble by 3.
  eg {1,3,5} or {17,19,21} or {53,55,57} or {101,103,105} or {141,143,145}
  3, 21, 57, 105, 141 all are divisble by 3 thus making their product divisble by 3
• 9 years agoHelpfull: Yes(2) No(0)