What is the remainder left after dividing (1! + 2!+ 3!+ ..... + 100!) by 7 ?

• After 7! all the nos will be divisible by 7. So, we have to check the nos. before 7! i.e.
  (1! + 2!+ 3!+4!+5!+6!)%7
  873%7
  which gives 5 as remainder
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• (1!+2!+3!+4!+5!+6!)=873%7
  
  Answer is 5
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• my bro please xpln me in a proper way
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• SIMPLE ONE:After 7! all the nos will be divisible by 7. So, we have to check the nos. before 7! i.e.
  (1!+2!+3!+4!+5!+6!)=873%7
  =5 REMAINDER ON SOLVING IT
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• the ans should be 4
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• The Ans will be 5.
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• what about 7 factorial friends
  
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• use wolfram alpha online calculatot to verify the result is 9
  
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• as the terms after 6! are divisible by 7 so we need to consider the first 6 terms:
  let us group them as (1!+3!)+(2!+4!)+(5!+6!)
  =(1+6)+2!(1+12)+5!(1+6)
  =(7)+2(13)+120(7)
  except 26 the other 2 terms are divisible by 7
  so the remainder of 26/7 i.e.; 5 is the answer
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