Q. What remainder when 51^203 divided by 7?

• (51^203/7)
  51=49+2;
  (49+2)^203/7
  or simply we hv to calculate 2^203/7 by using binomial exp.(terms containing 49 is divisible by 7)
  now 2^203/7=
  4*(2^201/7)=
  4*(7+1)^67/7
  so remainder is 4
• 10 years agoHelpfull: Yes(7) No(3)
• 51/7 gives remainder 2
  so expr becomes 2^203
  Now looking at 2's power cycle when dividing 2 by 7 i.e. 2^n/7
  PowerCycle Remainder
  2/7 power 1 = 2
  4/7 power 2 = 4
  8/7 power 3 = 1
  16/7 power 4 = 2
  32/7 power 5 = 4
  64/7 power 6 = 1
  
  So by this we can conclude that, for every multiple of 3 in power of 2 we get the remainder 1
  
  Now expre becomes
  2*2(2^201) since 201 is divisible by 3
  So 4*1=4
• 9 years agoHelpfull: Yes(4) No(0)
• last two digit of 51^203 is 51
  so remainder will be 2
• 10 years agoHelpfull: Yes(1) No(4)
• Answer is 4
  it is solved using remainder theorm
  (51*51*......................203)/7=(2*2*........203)/7=(1*1*...67times*2*2)/7=4
• 10 years agoHelpfull: Yes(1) No(2)
• 2^6/7==1(rem)..next 51/7=2.2^203/7==((2^6)^33)*(2^5).so 1*2^5/7==1*4==4 anss
• 10 years agoHelpfull: Yes(1) No(2)