Q. What is the remainder when expression
2^2+22^2+222^2+2222^2+....+22222...48 times^2 divided by 9.

• [2^2+22^2+222^2+2222^2+....+22222...48 times^2]/9
  
  =[2^2(1^2+11^2+111^2+1111^2+....+11111...48 times^2 )]/9
  
  =4[1+(9*1+2)^2+(9*12+3)^2+(9*123+4)^2+(9*1234+5)^2+....]/9
  
  =Rem[4(1+2^2+3^2+4^2+...+48^2)]/9
  
  =Rem[(4*48*49*97/6)/9]
  
  =Rem[(32*49*97)/9]
  
  =Rem[(5*4*7)/9]
  
  =Rem(5)
  
  remainder = 5
  
  
• 10 years agoHelpfull: Yes(89) No(12)
• divide each term by 9
  then the remainder will be 4,7,0, 4,7,0 ....so on
  and these will be reapetd after 3 term
  total of 1st 3 term is 11.
  there have 48 term. so it will be reapeted 16 times.
  so total of all term is 16*11=176
  176/9 then qutient is 18 and remainder is 5.
• 10 years agoHelpfull: Yes(62) No(10)
• hai rakesh how become this Rem[(4*48*49*97/6)/9]
  explain me
• 10 years agoHelpfull: Yes(4) No(1)
• @manikandhan it is formula of 1^2+2^2+3^3.....n^2=n*(n+1)*(2n+1)/6
• 10 years agoHelpfull: Yes(3) No(0)
• sum of n^2 term is n(n+1)(2n+1)/6
• 10 years agoHelpfull: Yes(2) No(0)
• sorry! A little correction w.r.t previous answer. instead of "42 times" it's "21 times".
  
  rem[ (2^2+22^2+222^2+2222^2+....+22222...48 times^2) /9]
  rem[ (2^2+4^2+6^2+8^2+1^2 + 3^2+5^2+3^2+5^2+....(3^2+5^2)...21 times+ 3^2) /9]
  rem[ (4+16+36+64+1+ 21*(9+25) + 9 ) /9]
  rem[ (121+ 504 + 9) /9]
  rem[ (634) /9] = 4
  
  So Answer should be 4.
• 9 years agoHelpfull: Yes(2) No(1)
• ans is :- 6
• 9 years agoHelpfull: Yes(1) No(0)
• REM[4(1+2^2+3^2.........+48^2)-(9^2+18^2+27^2+36^2+45^2)]/9....
  4[38024-3455]/9
  0
  
• 10 years agoHelpfull: Yes(0) No(14)
• ans is :- 5
• 10 years agoHelpfull: Yes(0) No(0)
• ITS NOT 24 ITS INFINITY
  
• 10 years agoHelpfull: Yes(0) No(1)
• rem[ (2^2+22^2+222^2+2222^2+....+22222...48 times^2) /9]
  rem[ (2^2+4^2+6^2+8^2+1^2 + 3^2+5^2+3^2+5^2+....+3^2+5^2...42 times+ 3^2) /9]
  rem[ (4+16+36+64+1+ 42*(9+25) + 9 ) /9]
  rem[ (121+ 1428 + 9) /9]
  rem[ (1558) /9] = 1
  
  Ans: 1
• 9 years agoHelpfull: Yes(0) No(1)