WHAT WILL BE THE REMANDER IF (34^31)^301 IS DIVIDED BY 9?

• units digit of 31^301 is 1.
  hence when 34^1/9 remainder is 7.
• 10 years agoHelpfull: Yes(39) No(17)
• 34^31^301/9
  (34*34*34)^8729/9
  39304)^8729/9
  =(9*4367+1)/9
  1^8729/9
  ans 1
• 10 years agoHelpfull: Yes(18) No(13)
• first of all we have 2 see the cycle of 34....
  34*34=6
  34*34*34=4
  34*34*34*34=6
  therefore cycle of 34 is 2..
  301/2 remainder is 1.....
  31/2 remainder is 1....
  hence 34^1/9 has remainder 7
• 10 years agoHelpfull: Yes(8) No(3)
• 34^6k divide by 9 gives 1
  31^301 is 6k+1 type no.
  so 34^6k+1= 34^6k * 34^1
  hence 34^1 by 9 gives remainder 7
  ans= 7
• 10 years agoHelpfull: Yes(2) No(2)
• 31^301 =1 bcz unit digit is 1
  therefore 34^1is 34
  34%9=7
• 8 years agoHelpfull: Yes(2) No(0)
• First we need to get the remainder of 34/9 which is 7. Now if we divide 7's power series by 9 we get these remainders:-
  7^1/9=7,7^2/9=4,7^3/9=1,7^4/9=7 and so on repeats this pattern of remainder of 7,4,1. so using this logic we get 7^31/9=7. so now we get 34^31/9=7 and similarly 7^301/9=7. So, we get the result as 7.
• 8 years agoHelpfull: Yes(2) No(0)
• 34/9 gives remainder of 7 or -2
  thus 34^2 gives remainder of 4(-2*-2)
  34^3 gives remainder of (-2*-2*-2)=-8 or 1
  and, 34^31=34^(3k+1), so now we have to find remainder of (34)^301/9
  since 34^301=34^(3k+1), so we r left with 34/9, as 34^3k gives 1 as remainder.
  
  so finaly remainder is 34 mod 9= 7
  
• 10 years agoHelpfull: Yes(1) No(0)
• 34 mod 9 = 7
• 9 years agoHelpfull: Yes(1) No(0)
• no of triangle=(87-1)/2=43
  no of rectangle=43/2=21
  
• 10 years agoHelpfull: Yes(0) No(22)
• remainder will be 7. :)
• 10 years agoHelpfull: Yes(0) No(2)
• (32^1)/9=7 asw
• 9 years agoHelpfull: Yes(0) No(1)
• (17*2)^31^301
  =(17*2)^9331/9
  =17^9331/9*2^9331/9
  =-1*(2^3)^3110 *2/9
  = -1*1*2/9
  = rem -2 so 9-2 =7 rem
• 8 years agoHelpfull: Yes(0) No(1)
• first note that 34=3x9+7 so the remainder will the same as for 7^31^301.
  When we look at the 7 table (modulo 9) we see that we have
  7^1->7
  7^2->4
  7^3->1
  and so on.
  So the question is now related to 31^301 and how much 3 is in it.
  And following the same idea, 31=10x3+1, we see that the remainder of the division of 31^301 by 3 is 1. So we are left with 7^1.
  the answer is 7.
• 8 years agoHelpfull: Yes(0) No(0)
• It will be 7
  
  --->---->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
• 8 years agoHelpfull: Yes(0) No(0)