let x is the number of the factors of 100

let y is the sum of the first 100 integers

let z is the number of zeroes in 100!

compute x+y+z

• 100=2^2*5^5
  so factors are (2+1)+(2+1)=9
  X=9
  
  Y=the sum of natural numbers = n*(n+1)/2
  Y=100*(100+1)/2
  Y=5050
  
  Z=(100/5)+(20/5)=24
  
  final answer
  X+Y+Z=9+5050+24=5083
• 10 years agoHelpfull: Yes(2) No(1)
• X = 9
  
  Y= the sum of natural numbers = n*(n+1)/2
  Y = 100*(100+1)/2
  Y = 5050
  
  Z = [100/5^1] + [100/5^2] = 24
  
  final ans
  X+Y+Z = 9+5050+24 = 5083
• 10 years agoHelpfull: Yes(1) No(1)
• 100=2^2*5^2
  so factors are (2+1)+(2+1)=9
  X=9
  
  Y=the sum of natural numbers = n*(n+1)/2
  Y=100*(100+1)/2
  Y=5050
  
  Z=(100/5)+(20/5)=24
  
  final answer
  X+Y+Z=9+5050+24=5083
• 9 years agoHelpfull: Yes(0) No(0)