find the remainder when 51^203 is divided by 7

• 51%7=2
  (2)^203/7=4*(2^3)^67/7
  =4*(7+1)^67/7
  =4*(1)^67/7
  =4*1 =4 rem(ans)
• 10 years agoHelpfull: Yes(4) No(0)
• 51^203/7 = (7*7+2)^203/7 = 2^203/7
  = (2^3)^67 * 2^2 / 7
  = (8)^67 * 4 / 7
  = (7*1+1)^67 * 4 /7
  = (1)^67 * 4 / 7
  = 4/7
  = 4
• 10 years agoHelpfull: Yes(2) No(0)
• ans is 4
  remainder will be 4
  
  
• 10 years agoHelpfull: Yes(0) No(0)
• 51^203, the power 203 is completely divide by 7(prime)
  wkt, Fermat little theorem:
  If P is a prime number then (a^p−1/p)=1 or (a^p/p)=a;
  
  (51^7)^29/7=(51)^29/7=(51^7)^4*51/7=(51^5)/7=(2^5)/7=32/7=4;
  
• 10 years agoHelpfull: Yes(0) No(0)