Find the maximum value of n such that
570*60*30*90*100*500*700*343*720*81 is perfectly divisible by 30^n.

• 30=2*3*5
  similarly
  570=2*3*5*19
  60=2*2*3*5
  30=2*3*5
  90=2*3*3*5
  100=2*2*5*5
  500=2*2*5*5*5
  700=2*2*5*5*7
  343=7*7*7
  720=2*2*2*2*3*3*5
  81=3*3*3*3
  no of 5`s is 12 but no of 3's is limited to 11 thus the answer is 11
• 8 years agoHelpfull: Yes(0) No(0)