What will be the remainder of the following division:
(1234567....141516...20212223....404142)/43?

• 0
  first add the digits
  1+2+3+.........+42
  =>903
  then 903 divided by 43
  because it is the rule of division by 43( a prime number)
  so 903%43=0
  % is modulo operator it returns the remainder part
• 10 years agoHelpfull: Yes(17) No(8)
• numerator forms an a.p so
  here a=1 L=42 n=42
  sum=n(a+L)/2
  =21*43
  remainder=num/denominator
  =(21*43)/43=0 remainder
  
• 10 years agoHelpfull: Yes(7) No(2)
• 123/ 4 => rem=3,
  1234/5 => rem=4,
  12345/6 => rem=3,
  123456/7 => rem=4,
  .
  .
  .
  .
  1234...141516....4142/43 => rem=4
  
• 10 years agoHelpfull: Yes(4) No(1)
• 0 as sum of the digits is 903 and it is divisible by 43
• 10 years agoHelpfull: Yes(1) No(2)
• Ans :0
  1+2+3.........42
  s=903
  r=903%43=0
• 10 years agoHelpfull: Yes(0) No(14)
• ANS MAY BE 37
• 9 years agoHelpfull: Yes(0) No(0)