Find the quotient when total no. Of divisors of 16! is divided by right most non-zero digit of 15!

• 16!= (2^4) * (3*5) * (2*7)* (13)*(2^2*3) * (11)*(2*5)*(3^2)*(2^3)*(7)*(2*3)*(5)*(2^2)*1
  =(2^15)*(3^6)*(5^3)*(7^2)*(11)*(13)
  so number of divisor= (15+1)*(6+1)*(3+1)*(2+1)*(1+1)*(1+1)
  =16*7*4*3*2*2
  =5376
  
  The generalized formula to find the non zero digit of N = [N/5]!×2^[N/5]×[N/5]R!
  (see http://www.campusgate.co.in/2013/10/finding-right-most-non-zero-digit-of.html )
  =[15/5]! * 2 ^[15/5] * 0!
  =48
  =8(unit digit)
  
  So Required Anwser=5376/8= 672
• 8 years agoHelpfull: Yes(14) No(0)
• first find the factors of 16!
  and then divide with non zero digit of 15!
  factors of 16!
  prime no below 16 are 2 3 5 7 11 13
  now highest power of 2 in 16! is 15
  3is 6,5 is 3,7 is 2,11,is 1,13 is 1
  so no of factors is 16*7*4*3*2*2=5376
  now last digit in 15!
  it can be written as 2^11*3^6*5^3*7^2*11^1*13*1
  5^3*2^3 gives 0 so remaining is 2^3 =8
  
  therefore 5376/8=672
• 8 years agoHelpfull: Yes(6) No(0)
• how did you get the number of divisors?
• 8 years agoHelpfull: Yes(0) No(0)