Find d max value of n such that 157! is perfectly divisible by 18^n...can sm 1 solve dis wid xplanation..plz

• 52+17+5+1=75
  so ans=3^75=3(3^74)
  =3(9^37)
  ans=37
• 12 years agoHelpfull: Yes(27) No(4)
• Value of n will depend on value of power of 9 in 157!.
  
  number of powers of 3 in 157! = 157/3 +157/9+ 157/27 + 157/81
  = 52+ 17 + 5 + 1= 79
  so 157! = k* 3^79= 3k*(3^78) = 3k*9^39
  
  so max value of n= 39
• 12 years agoHelpfull: Yes(17) No(16)
• 18 = 3*3*2
  We need to find the power of 3 in 157!
  = [157/3] + [157/9] + [157/27] + [157/81]
  where [x] represents highest integer less than or equal to x.
  = 52 + 17 + 5 + 1
  = 75
  
  So, we have 37 pairs of 3.
  
  The number of 2s will definitely more than 37 so we do not need to calculate.
  
  Hence, ans will be 37
• 11 years agoHelpfull: Yes(7) No(1)
• n=18
  18 can be written as 9*2
  157/9=17 & 81=9*9
  so total 9 used in 157! are 18
• 12 years agoHelpfull: Yes(4) No(9)
• there are 8, 18 in 157
  and some more can be found by
  9*2 ,6*3,
  so total number of 8's will be 8+2=10
  so n max=10
• 12 years agoHelpfull: Yes(2) No(6)
• akhil is right ans is 18
  
• 12 years agoHelpfull: Yes(1) No(5)
• how 79???????
• 12 years agoHelpfull: Yes(1) No(5)
• use formula
  [n/p]+[n/P^2]+[n/P^3]+[n/P^4]+[n/P^5]........ soo on
  n is given no with factorial take on n and [2.33]=2 greatest integer
  p is prime no
  so 18=9*2=3^2*2
• 12 years agoHelpfull: Yes(1) No(2)