12341234..........400 digits when divided by 909 leaves reminder?

• 1234124341234.......upto 400 digits = 1234(10^0 + 10^4 + 10^8 + ... 10^396)
  1234(10^0 + 10^4 + 10^8 + ... 10^396) mod 909
  = 123400 mod 909 = 685 (As 10^4k mod 909 is always 1)
• 11 years agoHelpfull: Yes(9) No(2)
• Ans 325.
  last 4 digit is 1234 which when divide by 909 leave remainder 325......correct me if i wrong
• 11 years agoHelpfull: Yes(3) No(13)
• N=12341234…upto 400digits =1234[10^396+10^392+10^298+...+10^0] =1234[(10^4)^99+(10^4)^98+(10^4)^97+....+(10^4)^0] =A*B(let) So we can write N%909=(A*B)%909
  
  =1234%909*[1+1+...100 terms]%909[As10^4=1(mod909)]
  
  =32500%909=685[Here % implies Mod which determines the remainder]
• 11 years agoHelpfull: Yes(2) No(2)
• (1234)group of 4 digits whose sum of digits is 10, when repeated 100 times makes overall 400 digits like 12341234.....1234.
  check rem. for 9 which is = 1
  rem. for 101 =91
  ans.92
• 11 years agoHelpfull: Yes(2) No(2)
• 685 will the remainder !!!
• 11 years agoHelpfull: Yes(0) No(0)