(72)^21/ 25

• using -ve remainder theorm
  (75-3)^21/25
  (-3)^21/25=(-2)^7/25=-3/25=22
  so ans is 22
• 10 years agoHelpfull: Yes(12) No(18)
• last 2 digits of 72^21 is 72.
  so xxx72/25=22
  so ans is 22.
  Explanation to find last 2 digits of 72^21 is
  72=(2^3)*(3^2) --> 72^21=(2^63)*(3^42) now find last 2 digits of 2^63 and 3^42
  
  2^63=(2^60)*(2^3)=((2^10)^6)*(2^3)
  last 2 digits of 2^10 is 24^6 6 is even so 76 is last 2 digits of 2^60
  76*8=08 is last 2 digits
  
  3^42=((3^4)^10)*9=(81^10)*9
  last 2 digits of 81^10 is 1
  1*9=9
  so now last 2 digits of 2^63*3^42 is 8*9=72
  
  so finally 72%25=22
• 10 years agoHelpfull: Yes(2) No(3)
• using -ve remainder theorm
  (75-3)^21/25
  (-3)^21/25=(-27)^7/25=(-25-2)^7/25=(-2)^7/25
  =-128/25=-3/25=22
  so ans is 22
• 10 years agoHelpfull: Yes(2) No(5)
• ans is 15.
  
• 10 years agoHelpfull: Yes(0) No(5)
• can u explain the -ve remainder theorem clearly khushboo
• 10 years agoHelpfull: Yes(0) No(2)
• 22.i think
• 10 years agoHelpfull: Yes(0) No(1)
• rem 22,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
  
• 10 years agoHelpfull: Yes(0) No(1)
• applying remainder theorem
  12^21/25
  12^3^7/25
  12*12*12)^7/25
  19*12)^7/25
  228^7/25
  3^7/25
  3^3*3^3*3/25
  2*2*3/25
  so remainder is 12
  
  
• 10 years agoHelpfull: Yes(0) No(0)
• (75-3)^21/25
  =remainder -3
  or we can say 22
• 10 years agoHelpfull: Yes(0) No(0)
• using -ve remainder theorm
  (75-3)^21/25
  (-3)^21/25=(-27)^7/25=(-25-2)^7/25=(-2)^7/25
  =-128/25=-3/25=22
  so ans is 22
• 10 years agoHelpfull: Yes(0) No(0)
• 22 is right answer
  Khushboo has done correct using -ve remainder theorem
• 9 years agoHelpfull: Yes(0) No(1)