1!+2!+3!+......+50!=?
a) 3.1035*10^64
b) 2.1021*10^65
c) 3.1035*10^63
d) 3.1035*10^62

• Note that 50! ≈ 3.04 x 10^64 already, so this eliminates (C) and (D) as they are less than 50!. Intuitively, you would expect that 1! + 2! + ... + 50!'s value would be significantly greater than 50!, but not by a factor of about 10 as (B) expects, so you would expect that 1! + 2! + ... + 50! < 10^65. Thus, the only answer that makes sense is (A).
  
  Therefore, the answer is (A).
  
  I hope this helps!
• 9 years agoHelpfull: Yes(25) No(15)
• Please can u explain me clearly
• 9 years agoHelpfull: Yes(8) No(2)
• 3.1035*10^64 should be the Ans.
• 9 years agoHelpfull: Yes(1) No(19)
• same que is in puzzle section..chk it from dere
• 9 years agoHelpfull: Yes(1) No(0)
• Sum of n factorials has
  first non-zero digit
  with least significance
  as 1
  when n>4
  
• 9 years agoHelpfull: Yes(1) No(0)
• Call the sum S, then 50! < S < 2*50! This is because 50! = 50*49! = 49! + 49! + 49! + ... +49!, which has 50 terms, each as large as the 49 other terms in 1!+2!+3!.......49!
  
  30 414 093 201 713 378 043 612 608 166 064 768 844 377 641 568 960 512 000 000 000 000 < S <
  60 828 186 403 426 756 087 225 216 332 129 537 688 755 283 137 921 024 000 000 000 000
  
  It factors as (2^E)(3^F)(5^G)*K, with E=0, F=1, G=0
  
  The least significant digits can be calculated, because they remain fixed! Starting at 1!+2!+3!+4! = 33, the units digit will remain 3 indefinitely. The sum at 9! will have the fixed last two digits (10! ends with two 0s).
  
  The most significant digits can be found in a similar way, pretty much done above, with 50! having the same leading digit as S. So
  S = 3. something...3 x 10^64
• 9 years agoHelpfull: Yes(1) No(3)
• calculator is allowed in tcs aptitude exam dont take tension
• 9 years agoHelpfull: Yes(1) No(3)
• can anyone please explain a good method to solve this.
• 9 years agoHelpfull: Yes(0) No(0)
• Call the sum S, then 50! < S < 2*50! This is because 50! = 50*49! = 49! + 49! + 49! + ... +49!, which has 50 terms, each as large as the 49 other terms in 1!+2!+3!.......49!
  
  30 414 093 201 713 378 043 612 608 166 064 768 844 377 641 568 960 512 000 000 000 000 < S <
  60 828 186 403 426 756 087 225 216 332 129 537 688 755 283 137 921 024 000 000 000 000
  
  It factors as (2^E)(3^F)(5^G)*K, with E=0, F=1, G=0
  
  The least significant digits can be calculated, because they remain fixed! Starting at 1!+2!+3!+4! = 33, the units digit will remain 3 indefinitely. The sum at 9! will have the fixed last two digits (10! ends with two 0s).
  
  The most significant digits can be found in a similar way, pretty much done above, with 50! having the same leading digit as S. So
  S = 3. something...3 x 10^64
• 9 years agoHelpfull: Yes(0) No(7)
• please explain it. i cant understand
• 9 years agoHelpfull: Yes(0) No(3)