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

• lets take 2*2 away from question
  
  2*2 (1*1 + 11* 11 + 111 * 111 + 1111 * 1111 + 11111 * 11111 + ..................+ 11111..........48times * 1111.............48times)
  
  now solving step by step ,
  
  1) 1 %3 =1 and 1 % 3 = 1
  2) 11% 3 = 2 and 11 % 3 =2
  3) 111% 3 = 0 and 111 % 3 = 0
  4) 1111%3 = 1 and 1111%3 = 1
  5) 11111 % 3 = 2 and 11111% 3 = 2
  6) 111111%3 = 0 and 111111% 3 = 0
  
  as we know that any number whose sum of digit is divisible by 3 is actually divisible by 3.
  e.g. 1 + 1 +1 = 3%3=0 and 111 % 3 = 0
  
  so now as total digit are 48 then,
  
  (2*2) * 16 * (1 * 1 + 2 * 2 + 0 * 0) % 9 (here 16 is multiplied as total cycles are 16, 48/3 = 16 )
  
  4 * 16 * 5 % 9 = 320 % 9 = 5
  
  Hence remainder is 5
  
  
• 8 years agoHelpfull: Yes(8) No(4)
• 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.
• 8 years agoHelpfull: Yes(5) No(0)
• remainder will be 5
  
• 8 years agoHelpfull: Yes(3) No(6)
• Remainder 3 i calculated it using Excel formula so i got 3 as a remainder
• 8 years agoHelpfull: Yes(2) No(3)
• reminder will be 3
  
  2^2(1+11^2+111^2+.........)
  here,reminder of 111^2 = 0,111111^2 = 0...so on
  hence 1's which are multiple of 3 =0
  rem of 1^2=0,11^2=2,1111^2=4,....
  =2^2{(1+2+3+......+48)-(3+6+9+....48)}
  =2^2(1176-408)
  =2^2*768
  =rem of 3072/9
  =3
• 8 years agoHelpfull: Yes(2) No(3)
• 2^2 + 22^2 + 222^2 + 222.... 48 times ^ 2.
  
  Use Remainder theorem. This is a long problem as Two types of series will emerge.
  
  1st Series:
  2^2 /9 . rem = 2^ 2
  22^2 /9 . rem = 4 ^2
  222^3 / 9 rem = 6 ^2
  2222^2 /9 rem = 8 ^2
  22222^2 / 9 rem = 1 ^ 2
  
  2nd series:
  222222^2 9 rem= 3 ^2
  2222222^29 rem= 5^2
  22222222^2 9 rem= 7^2
  222222222^29 rem= 0
  
  After this, the remainders will start to repeat as per the above series. So after every 9 numbers, the remainders will repeat. So as there are 2222....48 means, 48 terms in the series,
  hence, 48/9 = 5 + rem: 3 .
  
  It means that the first 45 terms will be divisible by 9.
  Now we need to know rest 3 terms which will give, ( 2^2, 4^2, 6^2 ) are their remainders
  
  So ( 4 + 16 + 36 ) / 9 will give
  
  REMAINDER : 2
• 8 years agoHelpfull: Yes(1) No(9)
• Clearly answer is 5 ... its not 3.
  
  question = 2*2 + 22*22 + 222*222 + 2222*2222 + ..............................+ 22222......48 times * 2222...... 48times / 9
  
  take 2*2 from each number we get,
  
  2*2 (1 + 11*11 + 111*111 + 1111*1111 + ........................+ 1111........48 times * 11111.........48 times ) / 9
  
  now , 1%3 = 1 and 1%3 =1.
  11%3 = 2 and 11%3 = 2.
  111%3 = 0 and 111%3 = 0
  
  first cycle complete
  
  1111%3 = 1 and 1111%3 = 1
  
  11111%3=2 and 11111%3= 2
  
• 8 years agoHelpfull: Yes(1) No(0)
• 3
  4*48=192
  192/9 rem=3
  so rem=3
• 8 years agoHelpfull: Yes(1) No(1)
• remainder is 6
• 8 years agoHelpfull: Yes(0) No(4)
• how ??plz suggest the method..
  
• 8 years agoHelpfull: Yes(0) No(0)
• the remainder will be 4.
• 8 years agoHelpfull: Yes(0) No(1)
• 2^2(1^2+11^2+111^2+........1111..^48)=2^2*(2^48)=2^50
  then
  2^50= (1024*1024*1024*1024*1024)/9=(7*7*7*7*7)/9
  so we get remainder as 4
• 8 years agoHelpfull: Yes(0) No(3)
• In first term remainder is 4,in 2nd term 8, in 3rd term 12,so at 48th term remainder is 4*48=192.
  
• 8 years agoHelpfull: Yes(0) No(1)