sum of all three digit numbers which leave the remainder 3 when divided by 5?

• smallest 3 digit number which leave the remainder 3 when divided by 5 =103
  largest 3 digit number which leave the remainder 3 when divided by 5 =998
  3 digit number after 103 which leave the remainder 3 when divided by 5 =108
  an ap is formed: 103,108........998
  first term=103
  common difference=5
  let the total no. of three digit numbers which leave the remainder 3 when divided by 5=n
  998=103+(n-1)5
  180=n
  sum of ap=180/2 *(103+998)=90*1101
  =99090
• 13 years agoHelpfull: Yes(5) No(1)
• Numbers can be represented by 5x+3 103+108+......998 >>>>> 180 terms Sum = 103+108+......998 = 100+3+105+3+.....995+3 = 5*(20+21+...199)+3*180 = 5*19710 +3*180 = 99090
• 14 years agoHelpfull: Yes(4) No(1)
• 109090
• 14 years agoHelpfull: Yes(3) No(3)
• Numbers can be represented by 5x+3
  103+108+......998 >>>>> 180 terms
  
  
  Sum = 103+108+......998 = 100+3+105+3+.....995+3 = 5*(20+21+...199)+3*180
  = 5*18710 +3*180 = 94090
• 14 years agoHelpfull: Yes(3) No(4)
• 99090
• 14 years agoHelpfull: Yes(3) No(1)
• two ways we can solve
  first 103+108+113+........998=sum of this A.P series =99090
  
• 14 years agoHelpfull: Yes(3) No(2)
• 103,108,113,118......993,998
  Total numbers = 20X9=180
  sum=100X20X(1+2+3+...+9)+10X9X2X(0+1+2+..+9)+90X(3+8)=99090 ANS
• 14 years agoHelpfull: Yes(3) No(1)
• Just a correction to Dipin's sol
  
  Numbers can be represented by 5x+3 .........103+108+......998 >>>>> 180 terms
  ...
  Sum = 103+108+......998 = 100+3+105+3+.....995+3 = 5*(20+21+...199)+3*180 = 5*19710 +3*180 = 98540+ 540 =99090
• 14 years agoHelpfull: Yes(2) No(2)
• sum = 103+108+........+ 998 = 180*(103+998)/2 = 90*1101 = 99090
• 14 years agoHelpfull: Yes(2) No(1)
• 99090
• 13 years agoHelpfull: Yes(2) No(2)
• 6630 = (300*20)+3(20*21/2)
• 14 years agoHelpfull: Yes(1) No(6)