P is a no. which has exactly 3 factors which divide 25! Then whats is the greatest power of 3 in (p-25)!

• suppose p=125 has three factor 125==5*5*5;
  (125-25)=100;
  power of 3=100/3=33.33/3+11.11/3+3.3/3=47+11+3+1=48
• 7 years agoHelpfull: Yes(6) No(10)
• Only perfect squares of prime numbers can have three factors, all other numbers will have even number of factors .
  
  Hence factors of P are 1,p and p^2, where p is a prime number.
  
  Since P divides 24!, the greatest value p can have is 11.( Since 11^2 is 121, which can be obtained from the terms 22 and 11 from 24!. If you try to take 13, a single 13 can be obtained but for a second 13, we need 26, which is clearly not available.)
  
  Hence the number P is 121.
  
  Therefore (P-25) is 96.
  
  The highest power of 3 in (p-25)! is obtained by as follows:
  
  [96/3]+[96/3^2]+[96/3^3]+[96/3^4]= 32+10+3+1=46
  
  Hence the answer is 46.
• 7 years agoHelpfull: Yes(3) No(0)
• p=125 has three factor 125==5*5*5;
  (125-25)=100;
  
  power of 3 = 100/3 + 33.33 /3 + 11.11 /3 + 3.3 /3
  = 33 + 11 + 3 + 1 = 48
• 7 years agoHelpfull: Yes(1) No(7)
• [100/3]+[100/3^2]+[100/3^3]+[100/3^4]= 33+11+3+1=48
• 7 years agoHelpfull: Yes(1) No(1)