Find the remainder when 97! is divided
by 101

• Use this theorem...
  Wilson's Theorem is about primes, namely, for integer n>1,
  (n-1)! = -1 mod n iff n is prime.
  
  Since 101 is prime, we can use it, along with the fact that
  100*99*98*97! = 100!
  
  Then taking mod 101 and using Wilson's Theorem,
  100*99*98*97! = -1 mod 101
  (-1)(-2)(-3)97! = -1 mod 101
  
  so that, multiplying both sides by -1,
  6*97! = 1 mod 101
  
  Now, we'd like to find the inverse mod 101 of 6. To do this, use the fact that
  6*17 = 102 = 1 mod 101, to multiply both sides by 17, yielding
  
  17*6*97! = 97!mod 101 = 17 mod 101
  
  So it's 17.
• 11 years agoHelpfull: Yes(10) No(4)
• 97! can be found using calculator given in open seesame.
  97!=9.619 e+151
  97!/101=9.52376 e+149
  to find remainder:
  subtract the whole number part i.e 9
  =(9-9.52376)=0.52376
  multiply this by 101
  0.52376*101=52.89976
  this can be round of to 53
  Hence remainder is 53.
• 11 years agoHelpfull: Yes(5) No(1)
• we know that 97! = 97*96*95.........1
  it can be write as 97+96.....+2+1 because when you find the remainder 91/101 =97,96/101=96...as it is and so on..
  add all by using n/2[2a+(n-1)d] . ..you will find the total than easily you can divide by 101 ans==whatever according to approach
  as
• 11 years agoHelpfull: Yes(3) No(32)
• total sum of series n/2[2a+(n-1)d]
  then sum is 9312*97
  then divide by 101
  first be divide 9312 by 101 then reminder is 19 and multiply 19 and 97
  then divide by 101
  and ans is 25
  
• 11 years agoHelpfull: Yes(3) No(4)