What will remainder when 128^1000 is divided by 153..?

• 128^1000/153
  =128^1000/(9*17)
  now, 128^1000/9 = 2^7000/9=(2^3)^2333*2/9=(9-1)^2333*2/9=(-1)^2333*2/9=-2/9=7/9 rem= 7 so rem will be of form (9x+7)
  
  and 128^1000/17=(2^7)^1000=(2^4)^1750/17=(17-1)^1750/17 gives remainder 1 so rem will be of form (17y+1)
  
  solve, 9x+7=17y+1, for least +ve integral value of x & y
  or 9(x-y)=8y-6
  or y=3,x=5
  hence remainder is 9x+7=9*5+7=52
  or 17y+1=17*3+1=52
  
  ANS 52
• 10 years agoHelpfull: Yes(49) No(7)
• We have to find out Rem [128^1000/153]
  
  153 = 9*17
  128^1000 = 2^7000
  
  Let us find out Rem[2^7000/9] and Rem[2^7000/17]
  We will combine them later.
  
  Rem[2^7000/9]
  = Rem [ 2^6999 x 2 / 9]
  = Rem [ 8^2333 x 2 / 9]
  = Rem [ (-1)^2333 x 2 / 9]
  = Rem [ (-1) x 2 / 9]
  = - 2 from 9
  = 7
  
  Rem[2^7000/17]
  = Rem [16^1750 / 17]
  = Rem [ (-1)^1750 / 17]
  = 1
  
  So, our answer is a number which leaves a remainder of 7 when divided by 9 and it should leave a remainder of 1 when divided by 17.
  
  Let us start considering all numbers that leave a remainder of 1 when divided by 17
  => 18 (leaves a remainder of 0 from 9. Invalid)
  => 35 (leaves a remainder of 8 from 9. Invalid)
  => 52 (leaves a remainder of 7 from 9. Valid) .............ans
• 8 years agoHelpfull: Yes(8) No(0)
• (128^998)16384/153
  153*107=16371 WHICH HAS DIFFERENCE OF 13 AS REMAINDER.
• 10 years agoHelpfull: Yes(1) No(1)
• Ist method:
  1281000 =(2^7)^1000= 27000 = 2(96*72 + 88) = 288mod153.
  [As φ(153) = 96 and 296 = 1mod153]
  As also 27 = -25mod153
  => 288 = 2(7*12 + 4) = 24*2512mod153 = 24*136 = 24*163 = 216 = 2(7*2 + 2) = 4*252 = 4*13 = 52mod153.
  [As 252 = 13mod153 and 132 = 16mod153]
  
  
  IInd method:
  153 = 32*17 = 9*17
  and 1281000 = 21000mod9 = 2(3*333 + 1)mod9 = -1*2mod9 = 7mod9.
  also 1281000 = (-8)1000mod17 = 23000mod17 = 24*750mod17 = 1mod17.
  Also 9*2 - 17*1 = 1
  Using Chinese Remainder Theorem
  1281000 = (9*2*1 - 17*1*7)mod153 = -101mod153 = 52mod153.
  
• 10 years agoHelpfull: Yes(1) No(2)
• NOT 5, ITS 24
  
• 9 years agoHelpfull: Yes(1) No(1)
• i think answer will be 1 bcs on dividing this number by 3 and 17 we get the common remainder 1
• 9 years agoHelpfull: Yes(0) No(1)
• 
  
  The main observation is that the powers an of an integer a^n must repeat modulo m because there are only m possible remainders. Once it does repeat, it repeats periodically. The period may be less than m.
  
  Since 153=9×17, consider the powers of 2 modulo 9 and 17.
  
  Modulo 9, the powers of 2 repeat with period 6.
  
  Modulo 17, the powers of 2 repeat with period 8.
  
  Therefore, modulo 153 the powers of 2 repeat with period lcm(6,8)=24.
  
  Finally, note that 1281000=27000 and 7000≡16mod24.
  
  Hence, 1281000 ≡ 216 ≡ 52mod153.
  
• 8 years agoHelpfull: Yes(0) No(0)
• Ans = 52.
  
  
• 8 years agoHelpfull: Yes(0) No(0)
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• 8 years agoHelpfull: Yes(0) No(2)