What is the remainder when 100!/97^2.

• the ans is 8827
• 11 years agoHelpfull: Yes(6) No(6)
• 8827 IS CORRECT
• 11 years agoHelpfull: Yes(4) No(3)
• 100!/97^2=100*99*98*97*96!
  Now According to WILSON's Theorem
  (n-1)!/n gives remainder -1
  so,by applying it we have
  (96!*97*98*99*100)/97*97 = (96!*98*99*100)/97
  96!/97 gives remainder as -1
  And 98/97 gives rem. as 1
  sly,99 And 100 divided by 97 gives rem. as 2 and 3 respectvly,
  now total rem. is -1*1*2*3 = -6
  so rem. is 97-6 = 91
  remainder of the question is 91*97=8827
  ANS; 8827
  (we multiplied 91 by 97 because in previous steps we have divided by 97).
  
  
  
  
  
• 11 years agoHelpfull: Yes(4) No(0)
• 6 will be the remainder
• 11 years agoHelpfull: Yes(2) No(6)
• sharmi plz explain the answer
  
• 11 years agoHelpfull: Yes(2) No(1)
• i think it will solved using wilson's therom.
  
• 11 years agoHelpfull: Yes(1) No(0)
• please some1submit a proper solution.
• 11 years agoHelpfull: Yes(1) No(0)
• what is wilson's therom?
• 11 years agoHelpfull: Yes(0) No(0)