What is the remainder when the number 1234567.......424344 is divided by 45

• remainder will =9
  write number as (1234.............4142290+54)%45
  12345...............4142290 fully divisible by 45 as
  bcoz sum is ((44*45)/2-54)=936 which is divisible by 9
  and last digit is 0 hence also divisible by 5
  so just concern about 54
  hence 54%45=9 is answer
• 11 years agoHelpfull: Yes(21) No(1)
• for number to be divisible by 45 it should be divisible by both 5 and 9 to chk the divisiblity by 9 find the sum of the digits
  1-9=45+1(of 10)=46
  11-20=9+45+2=56
  21-30=18+45+3=66
  31-40=27+45+4=76
  41-44=16+10=26
  so sum of digits=270
  270 is divisible by 9
  since 5 and 9 are co-prime numbers...the number given number should also be divible by 5...on dividing the number by 5 we get a remainder of 4
  so ans is 4.
• 11 years agoHelpfull: Yes(9) No(7)
• sorry dres a mistake ans will be 9..!!!
• 11 years agoHelpfull: Yes(4) No(1)
• factors of 45 are 9 n 5.
  it is simple divisible by 9
  (clearly ie add all nos (1+44)*44/2)
  has reminder as 4 divided by 5..
  so continiously subtract 9 untill u can make it divisible by 5..
  which is 9*1(one time)..
  so reminder is 9.
• 11 years agoHelpfull: Yes(1) No(0)
• ans is 0 as sum of this series is 990 and i6t is completely divisible by 45
  
• 11 years agoHelpfull: Yes(1) No(4)
• the last digit of remainder will be 9
  
• 11 years agoHelpfull: Yes(0) No(0)
• please explain how..?
  
• 11 years agoHelpfull: Yes(0) No(0)
• pls explain..cant undrstand...
  
• 11 years agoHelpfull: Yes(0) No(1)
• plz @reema xplain how u add the no's divisble by 9.
  41-44=16+10??...
• 11 years agoHelpfull: Yes(0) No(0)