19. What is the remainder if the no. 123456789101112....143144 is divided by 35?

• 4, because by multiplying 5 with 1-9 numbers we can only get 0 and 5 in the very first position,so in this question 4 is placed in the very first position so by dividing it by 35 can give 0,so 4 becomes the remainder.
• 10 years agoHelpfull: Yes(8) No(13)
• n=144
  sum of natural no=(n(n+1))/2
  sum=10440
  when divided by 35 it gives remainder=10
  
• 10 years agoHelpfull: Yes(6) No(13)
• divide 35 means if the number unit digit is 0, or 5 so the remainder is 4
• 10 years agoHelpfull: Yes(3) No(6)
• n(n+1)/2
  n=44
  so 44(45)/2=990
  so remainder=990%35=10
  answer:-10
• 10 years agoHelpfull: Yes(1) No(13)
• remainder 2?
• 10 years agoHelpfull: Yes(0) No(1)
• this no after divided by 35 we get reamainder is 24.
• 10 years agoHelpfull: Yes(0) No(1)
• Let N = 1234567891011121314151617181920......424344
  Remainder when N is divided by 5 is 4. So N = 5K + 4 .....(1)
  Remainder when N is divided by 9 is Sum of the digits of N divided by 9. We know that 1+2+3+...44 = 990 Which
  gives digit sum as 9. So remainder when N is divided by 9 is 0.
  So N = 9L .....(2)
  Equation (1) and (2) we 9L = 5K + 4
  For K = 1 this equation gets satisfied. So least possible number satisfies the condition is 9
  So The general format of N = w(LCM of (9, 5)) + Least number satisfies the condition.
  So N = w.45 + 9
  When N is divided by 45, we get 9 as remainder.
• 8 years agoHelpfull: Yes(0) No(2)