what is the remainder of (32^31^301) when it is divided by 9?

• ans:5
  solve by using remainder theorem.
  
• 9 years agoHelpfull: Yes(27) No(5)
• 32^31^301
  when 31 divided by 9 gives remainder 5
  5 5^2 5^3 all gives the same unit digit 5
  so 31^32 gives unit digit 5
  same rule applicable to 31^301
  when 31 divided by 9 gives remainder 4
  4 4^2 4^3 4^4 =4 6 4 6 unit place repeats for every 2 times i,e for even power its uint place is 6 and for odd its 4
  as 301 is odd its unit place is 4
  so 31^32^301=31^4=5^4=5 is the ans
  
• 9 years agoHelpfull: Yes(12) No(31)
• 32^31^301 /9=?
  STEP 1: 32/9= REMAINDER =5;
  
  STEP 2: 31^301/9; TAKE UNIT DIGIT'S CYCLE. THEREFORE NUM OF CYCLES =1, THUS 1/9;
  
  STEP 3: 5^1/9= 5/9;
  
  STEP 4: IF THE NUMERATOR IS LESS THAN DEMONINATOR THEN NUMERATOR IS THE ANSWER. ANS->5
• 7 years agoHelpfull: Yes(9) No(0)
• 32^31^301/9
  first step 32/9 the rem is 5
  second step 5^31^301/9
  take the unit digit : 31^301/9
  cyclicity of 1 is 1
  301/1 have rem =0 3^0 =1
  5^1/9=5/9
  ans 5
  
• 9 years agoHelpfull: Yes(4) No(3)
• SOLVING BY REMAINDER THEOREM
  32 wen divided by 9 gives 5 as remainder
  now 5^31^301
  now 31 divided by 9 gives 4 as remainder
  now 5^4^301
  4^301 shall give unit place as 4
  so now 5^4=20
  20/9=2 as remainder so 2 is rt
  
• 9 years agoHelpfull: Yes(3) No(29)
• 34/9 gives remainder 7
  (7^3)/9 gives remainder 1
  (((7^3)^3110)/9*7/9) = 1*7 = 7
• 9 years agoHelpfull: Yes(2) No(31)
• 301=4n+1
  31=4n+3
  so 32^3^1/9=5^3/9=8 as rem
• 9 years agoHelpfull: Yes(2) No(9)
• Use concept of negative remainder
  
  = 32^31^301/9
  = 32^9331/9
  = (-4)^9331/9 --------32/9 will give -4
  = ([(-4)^3]^3110 * (-4))/9
  = ([-64]^3110 * (-4))/9 --------(-64)/9 will give (-1)
  = ([-1]^3110 * (-4))/9
  = 1*(-4)/9
  = (-4)/9
  = 5
• 9 years agoHelpfull: Yes(2) No(3)
• ans - 5
  using a^(p-1)/p giver remainder 1
  so (32^8)/9 giver remainder 1
  unit digit of 31^301 is 1
  so , 32^31^301=32^(8k+1)= 32^8k.32
  (32^8k)/9 gives remainder 1
  32 divide by 9 gives remainder 5
  so 5.1=5
  hence ans=5
  
• 8 years agoHelpfull: Yes(2) No(0)
• rem.=2
  32^31^301=32^10k+1(since unit no. of 31^301=1)
  =>(2^6)^8k(32*4)/9
  =>128/9
  rem.=2
• 9 years agoHelpfull: Yes(1) No(13)
• (32^31^301) mod 9 = (27 + 5)(31^301) mod 9 = 5(31^301) mod 9
  
  Find out the last digit of 5n mod 9 till the sequence repeats.
  
  51 mod 9 = 5
  52 mod 9 = 7
  53 mod 9 = 8
  54 mod 9 = 4
  55 mod 9 = 2
  56 mod 9 = 1
  
  As the sequence repeats after every 6 terms so express exponential term in the form of : (6x + N) and the question will reduce from 5(6x + N) mod 9 to 5N mod 9.
  
  Therefore 31301 mod 6 = (30 + 1)301 = 1301 = 1
  
  Therefore twe need to find : 51 mod 9 = 5
  
  
• 8 years agoHelpfull: Yes(1) No(0)
• (32^31^301)/9
  (32^9331)/9
  32 can be written as 2^5
  so (2^5^9331)/9
  (2^46655)/9
  since the value of 2^3 is 8 which is closest to 9
  it can be written as ((8^15551)(2^2))/9
  so ((9-1)^15551)/9 can be written as ((-1)^15551)/9
  finally , (-1*4)/9
  so -4/9,then the remainder is 5
  Ans:5
• 7 years agoHelpfull: Yes(1) No(0)
• 32^31^301 = (32 raised to the power of 31) raised to the power 301.
  now 32 = 2^5. so 32^31 = 2^(5*31) = 2^155
  now we know that 2^3 = 8.
  write down 2 ^155 = (2^3)^51 * 2^2 = 8^51 * 4
  8/9 gives remainder 8 or -1. (-1)^51 = -1.
  so now 32^31^301 is simplified to (-4)^301
  now for odd power of 4 the unit digit is always 4.
  so (-4)^301 = -4. when -4 is divided by 9 remainder is -4 or 9-4=5.
  remainder can't be negative.
  So 5 is the answer
• 6 years agoHelpfull: Yes(1) No(0)
• (32^31)^301/9
  =2^5^31^301/9
  2^46655/9
  =(2^3)^15551 *4/9
  = rem 4
• 9 years agoHelpfull: Yes(0) No(2)
• Rem [34^31^301 / 9] = Rem [7^31^301 /9]
  
  Now, we need to observe the pattern
  7^1 when divided by 9, leaves a remainder of 7
  7^2 when divided by 9, leaves a remainder of 4
  7^3 when divided by 9, leaves a remainder of 1
  And then the same cycle of 7, 4, and 1 will continue.
  If a number is of the format of 7^(3k+1), it will leave a remainder of 7
  If a number is of the format of 7^(3k+2), it will leave a remainder of 4
  If a number is of the format of 7^(3k), it will leave a remainder of 1
  
  The number given to us is 7^31^301
  Let us find out Rem[Power / Cyclicity] t0 find out if it 7^(3k+1) or 7^(3k+2). We can just look at it and say that it is not 7^3k
  Rem [31^301/3] = Rem [1^301/3] = 1
  => The number is of the format 7^(3k + 1)
  => Rem [7^31^301 /9] = 7
• 6 years agoHelpfull: Yes(0) No(1)
• 9=3^2
  Euler number of 9=9*(1-1/3)=9*2/3=3*2=6
  301=6*50+1 => Remainder=1 => 32^31^301 is reduced to 32^31^1
  Similarly for 31, 31=6*5+1 => Remainder=1 => 32^31 is reduced to 32^1
  Now, 32=9*3+5
  Therefore remainder is 5
• 5 years agoHelpfull: Yes(0) No(0)