17^41+5^41 is divided by 11 then what is remainder?

• (x^n + y^n) is always divisible by (x+y) if n is odd number.
  
  17^41+5^41 is always divisible by 17+5 = 22 so also divisible by 11.
  
  17^41+5^41 / 11 => rem = 0
• 10 years agoHelpfull: Yes(98) No(1)
• 17^41+5^41=17^41+5^41/11
  =6^41/11+(-6)^41/11
  =(6/11)+(-6/11)
  =0
  
• 10 years agoHelpfull: Yes(11) No(1)
• 17^41+5^41/11
  
  here we can use the simple logic that 17+5=22 which is completely divisible by 11 and rem=0
  
  so the overall rem.is 0 Ans...
• 10 years agoHelpfull: Yes(10) No(6)
• The unit digit of 17 is repeating after 4 cycle.
  17^1= unit digit is 7
  17^2= unit digit is 9,,17^3= ---------- is 3, 17^4= 1, 17^5= again unit digit 7
  so 17^41= unit digit is 7.
  Similarly 5^41= Unit digit is 5
  so 5+7=12....12/11= remainder is 1
• 10 years agoHelpfull: Yes(5) No(13)
• yes 0 is ans. i have done calculation mistake
  17^41=(11+6)^41---->6^41
  5^41=(-6)^41
  0/11 remainder 0
• 10 years agoHelpfull: Yes(4) No(0)
• Unit digit approach is not applicable in remainder theorem.
  Exa. If unit digit is 7 and you want to divide by 3 then you get remainder 1 but its wrong because may be no. before 7 is 2 so 27/3 remainder 0 actually.
• 10 years agoHelpfull: Yes(4) No(0)
• 17 can be written as (11+6)^41/11 and 5 can be written as (11-6)^41/11
  =(11)^41/11+(6)^41/11+(11)^41/11-(6)^41/11
  =0+6+0-6
  =0
  
• 10 years agoHelpfull: Yes(2) No(3)
• Since the powers are same.
  We take powers common.
  (17+5)^41=22^41
  Dividing it by 11...
  22^41/11,.. Since 22 is divisible by 11, hence remainder is 0.
• 10 years agoHelpfull: Yes(2) No(2)
• ANSWER==> 0
  17^41+5^41/11=(17+5)(17^40+.......)/11= no remainder
  another way (x^n+y^n) is completely divisible by (x+y) if n= even
  so 17^41+5^41 is divisible by 17+5=22=2*11
  hence remainder=0
• 10 years agoHelpfull: Yes(1) No(0)
• 17^41,we can write it as 77.
  power=odd,keep 7 it as and multiply unit digits so 77
  5^41=55
  77+55=132/11=rem(0)
• 10 years agoHelpfull: Yes(1) No(0)
• 17^41+5^41/11 = (11+6)^41+5^41/11 = 6^41+5^41/11
  the unit digit of 6^41=6 and unit digit of 5^41=5
  (6+5)/11 = 11/11 =0
• 10 years agoHelpfull: Yes(1) No(0)
• the remanider is 1
• 10 years agoHelpfull: Yes(0) No(6)
• 17^41=(11+5)^41 11 is multiple so no remainder=5^41
  So 2*5^41=2*(11*2+3)^20 *5 =10*3^20 =10*(11-2)^10= 10*2^10=10*(33-1)^2=10
  So 10/11 remainder is -1 or 10
• 10 years agoHelpfull: Yes(0) No(4)
• remainder 0
  
  6+5 mod 11 = 0
• 10 years agoHelpfull: Yes(0) No(1)
• 17^41 means we can tke 17 as 7 n 7 has cyclic 4 so
  17^41=7 n 5^41=5 bcz 5 hve a cyclic 1.......
  so rem=7+5/11= 1 so rem=1
• 10 years agoHelpfull: Yes(0) No(4)
• the remainder is 1.
• 10 years agoHelpfull: Yes(0) No(6)
• 17^41+5^41/11
  first divide 17 by 11 the remainder is 6 and divide 5 by 11 the remainder is 5 therefore 6^41+5^41/5 the unit place of 6^41 is 6 and unit place of 5^41 is 5 therefore (6+5)/11 the remainder is so the remainder is 0
  
• 10 years agoHelpfull: Yes(0) No(3)
• 17^41+5^41
  
  =(Rem(17/11)^41+Rem(5/11)^41)
  =(6^41+5^41)
  =(11^41)
  11^41 is completely divisible by 11.Hence remainder is 0
• 10 years agoHelpfull: Yes(0) No(3)
• (17^41)/11= ((17-6)^41)= (-6^41)/11
  (5^41)/11= ((5+6)^41)= (6^41)/11
  (-6+6)^41/11 = 0.
  ans:0
• 10 years agoHelpfull: Yes(0) No(2)
• 17mod41 + 5 mod41
  6+9=15
  divided by 11 we get remainder is 4
  so ans is 4.
  
• 10 years agoHelpfull: Yes(0) No(4)
• Answer is 5
• 10 years agoHelpfull: Yes(0) No(5)
• i) 5^anything = 5
  ii) 17/11=> R 6 , 6^anytjing =6
  
  so ans is 5+6 = 11 and then 11/11=1 so R = 0
• 10 years agoHelpfull: Yes(0) No(1)
• 17=11+6;
  6^41=6*6^40=6(36^20)=6*3^20;
  5^41= 5*5^40=5(25)^20 = 5*3^20;
  [3^20(6+5)]/11 =[11*3^20]/11;
  rem=0
• 10 years agoHelpfull: Yes(0) No(0)
• ((17)^41 +5^41 )%11
  =
  (6^41+5^41)%11
  =
  ((-5)^41+5^41)%11
  =
  (-5^41+5^41)%11
  =0%11
  =0
  
• 10 years agoHelpfull: Yes(0) No(0)
• 0
  17^odd number +5 ^odd number giving number which is compeletly divisible by 11
• 10 years agoHelpfull: Yes(0) No(0)
• 11+6^41+11-6^41=6^41-6^41=0
• 10 years agoHelpfull: Yes(0) No(0)
• 17^41/11+5^41/11= 6^41+5^41/(6+5)
  
• 10 years agoHelpfull: Yes(0) No(0)
• 17^41
  7 power cycle 4
  41/4|rem=1
  unit digit is 7
  5^41
  5 power cycle 1
  unit digit is 5
  sum of unit digits=7+5=12
  12/11|rem=1
• 10 years agoHelpfull: Yes(0) No(0)
• (6)^41+(-6)^41=
  6-6=0
  0 ans
• 10 years agoHelpfull: Yes(0) No(0)
• simple logic 17+5=22 it is divisible by 11 hence rem =0
• 8 years agoHelpfull: Yes(0) No(0)
• Answer 0
  17^41+5^41/11
  So get remainder thorme
  17/11
  Remender 6
  (6+5)^41/11
  Answer 0
• 8 years agoHelpfull: Yes(0) No(0)
• Answer 0
  17^41+5^41/11
  So get remainder thorme
  17/11
  Remender 6
  (6+5)^41/11
  Answer 0
• 8 years agoHelpfull: Yes(0) No(0)