Find the remainder when 1^2+11^2+111^2+1111^2+...+(2001 times 1)^2
is divided by 100.

• 1^2+11^2+111^2+1111^2+...+(2001 times 1)^2
  11^2=121
  111^2=12321
  1111^2=1234321
  every term have reminder 21 when divide by 100
  remainder =21*2000+1( 1 in first term)=42001
  rem=42001/100
  ans=1
  so final ans is 1
• 10 years agoHelpfull: Yes(9) No(2)
• ans:1
  
  (1+11^2+111^2+1111^2+....+(2001 times 1)^2)%100
  = (1+121+12321+1234321+.....+123...321)%100
  = (1+100+21+12300+21+1234300+21+.....123...4300+21)%100
  = (1+21*2000 + 100 + 12300+1234300+....+1234...4300)%100
  = (1 + 0 + 0 + 0+....+0)
  = 1
• 10 years agoHelpfull: Yes(7) No(3)
• answer:22
  (1^2+11^2+...........)%100
  =1+21+21+21+.........upto 2001 times...(bcz 11*11%100=21,111*111%100=21, so on )
  ={1+21*((2000)*2001)/2}%100
  ={1+21*1000*2001}%100
  =1+21*1(bcz 1000%100=0,2001%100=1)
  =22
  So answer is 22
• 10 years agoHelpfull: Yes(0) No(4)
• unit place 1 so 1
• 9 years agoHelpfull: Yes(0) No(0)