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

• remainder is 0.
  first of all calculate 1+2+3+4+5......41+42. It is A.P series.hence sum=n/2(a+l)=21(1+42)=21*43. so when this is divided by 43, the remainder is 0
• 9 years agoHelpfull: Yes(35) No(14)
• Ans. 4
  123/ 4 => rem=3,
  1234/5 => rem=4,
  12345/6 => rem=3,
  123456/7 => rem=4,
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  1234...141516....4142/43 => rem=4
• 9 years agoHelpfull: Yes(19) No(3)
• rem=0, formula=n/2[a+(n-1)d]
• 9 years agoHelpfull: Yes(4) No(6)
• shruti....ur sol is wrong as if we divide 1234 by 43 we get 30 as rem not
  1+2+3+4=10%43=10
• 9 years agoHelpfull: Yes(1) No(3)
• ans.18
  (42*43)/9
  rem=0
  (42*43)/5
  rem=3
  9k=5m+3
  k=(5m+3)/9
  if m=3,(5m+3)/9 is divisble.
  so 5m+3=5*3+3=18
  so ans=18
• 9 years agoHelpfull: Yes(0) No(30)