what is the last two digits of
1!+2!+3!+4!+......50!

• @Rashmi It not 00.
  The answer is 13. Here is how.
  1!+2!+3!+4!+5!+6!+7!+8!+9! = 409113 (can check using calculator)
  last 2 digits of 10!,11!,12!...=00
  Therefore last 2 digits of first 9 or more facorials = 13
  
• 12 years agoHelpfull: Yes(64) No(0)
• the ans will be 00
  because after 4! factorial we will get 0 as the last digit and from 10! we will get two 0's as last 2 digits..
• 12 years agoHelpfull: Yes(8) No(12)
• take last two digit up to 9! and sum all
  from 10! to 50! last two digit only 00
  so ans is 13
  
  
• 12 years agoHelpfull: Yes(8) No(0)
• i guess the question is wrong
• 12 years agoHelpfull: Yes(1) No(9)
• 13.................
• 12 years agoHelpfull: Yes(1) No(1)
• ans=50! bcoz when we take 50! common then all value in the bracket are very small
  as compared to 50!.
  
• 12 years agoHelpfull: Yes(1) No(2)
• plz anyone mail me 81 patterns..tcs is cumg to our clg soon...
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• 12 years agoHelpfull: Yes(0) No(0)
• pls mail me also on
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• 12 years agoHelpfull: Yes(0) No(0)
• ans=25 as
  !1+....!50=50/2(!1+!50) ie n/2(a+l)
  =25(1+!50)
  =25+25!50
  so last two digit will be 25
• 12 years agoHelpfull: Yes(0) No(0)