remainder of 2^47/47

• ans 2
  a^p-a is divisible by p when p is prime
• 11 years agoHelpfull: Yes(9) No(0)
• answers is= 2
• 11 years agoHelpfull: Yes(6) No(3)
• ANS WILL BE=18
• 11 years agoHelpfull: Yes(5) No(13)
• ans is 2
  it is because 2^23 leaves remainder 1 when divided by 47
  so we can express 2^23 as 47x+1
  so 2^46= (47x+1)^2 = (47x)^2 + 2*47x + 1
  and 2*47 = 2*2^46=2*(47x)^2 + 2*2*47x + 2
  we see that first two terms are divisible by 47 but last term i.e. 2 is not ..
  so the remainder is 2
  
• 11 years agoHelpfull: Yes(5) No(2)
• mistake earlier
  2^47/47 =(2^6)^7*(2^5)/47=((47+17)^7*(2^5))/47
  =(17)^7*32/47=17*((17^2)^3)*32/47=17*((282+7)^3)*32/47 as 282 is divisible by 47 so
  17*(7)^3*32/47=186592/47=2 hence 2
• 11 years agoHelpfull: Yes(4) No(3)
• lets start
  2^47/47 =(2^6)*(2^5)/47=((47+17)^7*(2^5))/47
  =(17)^7*32/47=17*((17^2)^3)*32/47=17*((282+7)^3)*32/47 as 282 is divisible by 47 so
  17*(7)^3*32/47=186592/47=2 hence 2 is and uph had to do a lot :) thanks bfr
• 11 years agoHelpfull: Yes(3) No(0)
• answers is= 24
• 11 years agoHelpfull: Yes(2) No(2)
• 2mod47=2
  2^2mod47=4
  2^4mod47=(2^2)*(2^2)mod47=4*4mod47=16
  2^8mod47=(2^4)*(2^4)mod47=16*16mod47=21
  2^16mod47=(2^8)*(2^8)mod47=21*21mod47=18
  2^32mod47=(2^16)*(2^16)mod47=18*18mod47=42
  2^47mod47=(2^32)*(2^8)*(2^4)*(2^2)*2mod47=42*21*16*4*2mod47=2
  ANS=2
• 9 years agoHelpfull: Yes(2) No(0)
• 2^47/47=3*10^12
• 11 years agoHelpfull: Yes(0) No(3)
• Pls explain it clearly and crctly
• 9 years agoHelpfull: Yes(0) No(0)
• 2^47/47 =2
  
  
• 9 years agoHelpfull: Yes(0) No(0)
• Apply fermat's little theorem
  a^(p-1) is congruent to 1 (mod p)
  here a = 2, p = 47
  Thus, 2^46 yields remainder 1,
  So by property of congruence we can say
  2^46 * 2 yields remainder 1 * 2. Answer is 2
• 6 years agoHelpfull: Yes(0) No(0)