what will be the remainder when 720*720*720.....upto 1200 times is divided by 13.

• 720= (715+5)=(13*55 + 5);
  So, (720)^(1200)=(13*55 + 5)^(1200);Now if we expand [(13*55 + 5)^1200],then clearly every term of the expansion except (5^1200), is divisible by 13, because except (5^1200),every other term of the expansion ocontains (13*55) as one of its factors.
  So,if we can calculate the remainder when(5^1200) is divided by 13,that will be the required answer.
  Now, (5^1200)=(5^2)^(600)=25^(600)=(26-1)^(600)=(13*2-1)^600;
  Like same reasoning as above, every term except [(-1)^(600)] of the expansion[(13*2-1)^600] will be divisible by 13.
  Therefore ,when (5^1200)is divided by 13,the remainder will be =[(-1)^(600)]=1.
  So, when (720)^(1200) is divided by 13,the remainder will be = 1.
  I've tried my best to explain the whole matter in very lucid style,still if u have any confusion to make out the solution,then plse dont feel hesitated to ask me for any further clarification..
  Thanks to all..:-):-)
• 12 years agoHelpfull: Yes(19) No(0)
• 720^1200/(13)
  (715+5)^1200/13
  715 IS DIVISIBLE BY 13. HENCE 5^1200/13 IS REMAINDER. 5^1200=>25^600=>625^300. (624+1)^300/(13). 624 IS DIVISIBLE. HENCE 1^300/13. HENCE REMAINDER =1
• 12 years agoHelpfull: Yes(2) No(0)