21!=510909x21y1709440000

Without calculating 21!, what are the digits marked x and y? explain.

• Sum of the digits should be divisble by 9
  
  Sum of the digits is 52 so r is 2 so (x+y)mod9=2
  Possibilities is (1,1) (2,0)
  
  Also sum of the digits will be divisble 11
  Since sum of the digits is 52 and x and y are at even and odd places
  
  so (x-y)mod11--- R is minimum=3
  So that rules out (x+y)mod9=2 and then it should be 11
  
  Various possibilities of 11 (7,4),(9,2),(8,3)....
  But 7-4=3
  
  So digits are 7,4
• 11 years agoHelpfull: Yes(4) No(0)
• x=4 and y=7
  
  because it should be divisible by 9 and 11.
  check for divisibility of 9 and 11 and solve the two equations.
• 11 years agoHelpfull: Yes(1) No(0)
• ans x=4 and y=7.because by put that value no.is completely divisible by 21.
• 11 years agoHelpfull: Yes(0) No(2)