wht will be the remainder when expression
2^2+22^2+222^2+2222^2+....+22222...48times^2 divided by 9.

• WE can write it as 2^2(1+11^2+111^2+...),using remiander theorem 4*(1^2+2^2+3^2+..upto 48 terms)/9,using sum formula 1^2+2^2+3^2+....=n(n+1)(2n+1)/6 its comes to be=4*8*49*97/9=32*49*97/9=5*4*7/9=20*7/9=2*7/9=14/9=5 so ans is 5.
• 12 years agoHelpfull: Yes(34) No(22)
• First, let us consider a general case :
  
  (222222222222.....{2 is repeated n times})^2
  
  =[2(111111111111........{1 is repeated n times})]^2
  
  =4(111111111111........{1 is repeated n times})^2
  
  =4{(10)^(n-1) + (10)^(n-2) + (10)^(n-3) + (10)^(n-4) + .... +1}^2.......[(eqn1)]
  
  Now, we know 9 divides (10^k -1) , since (10^k -1)=(999999....k times);
  Where k is ofcourse a +ve integer. So,(10^k -1) is of the form (9A + 1);[A being a +ve integer].
  
  Now, looking at the [(eqn1)] ,
  
  we can find that here k =(n-1), (n-2), (n-3),....,1.
  
  Now,for 10^(n-1) = 9*(a1) + 1;
  
  10^(n-2) = 9*(a2) + 1;
  
  10^(n-3) = 9*(a3) + 1;
  
  10^(n-4) = 9*(a4) + 1;
  .
  .
  .
  .
  10^1 =9 +1;
  
  1 = 1
  
  So,now in [(eqn1)] , we can write
  
  (222222222222.....{2 is repeated n times})^2
  
  =4(9*A +{1 + 1 + 1 + 1 + .........[1 is added n times]})^2
  
  =4*(9*A + n)^2; here, A =(a1+a2+a3+a4+....+1),of course an integer.
  
  =4*81*(A^2) + 4*2*(9*A)*n + 4*(n^2);...............[(eqn 2)]
  
  Now this a general expression for (222222222222.....{2 is repeated n times})^2.
  
  For the given problem,we can find that n=1,2,3,4,5,6,....48.
  
  In the [(eqn 2)],since 1st two terms are already divisible by 9,so we only have
  
  to consider the last term,i.e, 4*(n^2);
  
  Summing up [4*(n^2)] for n=1,2,...48. we get
  
  4{(1^2)+(2^2)+(3^2)+(4^2)+(5^2)+...(48^2)}
  
  =4*[48*(48+1)*(2*48 +1)/6] ; using the formula[1^2 +2^2 +..+n^2=n(n+1)(2n+1)/6]
  
  =4*8*49*97
  
  =4(9- 1)(9*5 +4)(9*11 -2)
  
  =9*B + 4(-1)*(4)*(-2); B an integer, to know its value is not important
  
  =9*B + 32
  
  =9*B + 9*3 + 5;
  
  So, clearly the remainder will be 5.
  
  If any1 of u have any doubt/confusion or can find any loop-hole then never
  
  hesitate to mention it..
  
  Thanks to all..
• 12 years agoHelpfull: Yes(27) No(9)
• @ ALL:
  
  In my Solution,the notations used as a1 , a2 ,a3 ,a4 ...all are separate
  
  integers whose values are not important to know and moreover they
  
  DO NOT REPRESENT a^1, a^2 , a^3 , a^4 ....respectively, so plse dont get
  
  confused.Similarly B is also an integer knowledge about whose value is also not
  
  important,just look at the solution carefully and think a little bit then u will
  
  readily understand,still If any1 of u have any doubt/confusion or can find any
  
  loop-hole then never hesitate to mention it..
  
  Thanks to all..
• 12 years agoHelpfull: Yes(9) No(3)
• first of all we will divide 2^2+22^2 by 9 and check remainder.so 4+484=488
  and now 488/9.then remainder will be.2.now we will take first three number of series such as 2^2+22^2+222^2.and will check remainder ,when divide by 9.
  so 4+484+49284=49772.and now 49772/9.then remainder will again 2.so in my point of view remainder will be 2.
• 12 years agoHelpfull: Yes(3) No(26)
• its remainder will be 6....
  
• 11 years agoHelpfull: Yes(3) No(4)
• remainder will be 0.
  2^2/9 remainder will be 4
  22^2/9 remainder will be 7
  222^2/9 remainder will be 0
  2222^2/9 remainder will be 1
  and if we continue this further we will get same remainder 1,0,7,4
  means 12 times we get 1,0,7,4
  so total remainder is 12*(1+0+7+4)
  =144
  144/9
  remainder=0 ans
  
• 10 years agoHelpfull: Yes(3) No(3)
• 2^2/9 rem -5
  22^2/9 rem -2
  222^2/9 rem 0
  2222^2/9 rem 1
  (5 time 2)^2 rem 1
  (6 time 2)^2 rem 0
  (7 time 2)^2 rem -2
  (8 time 2)^2 rem -5
  (9 time 2)^2 rem 0
  (10 time 2)^2 rem -5
  
  from this we consider the cycle of the remainder is 9
  so the total remainder will be (-12*5 -6) =-66 = -3 =6
  hence 6 is the remainder
• 10 years agoHelpfull: Yes(3) No(0)