What is the remainder when 6^17+17^6 is divided by 7?
a)1
b)6
c)0
d)3

• 6^17+17^6 = (7-1)^17 + (21-4)^6 = (7-1)^17 + (7*3 -4)^6....[eqn1]
  
  if the [eqn1] is expanded then every term of the expansion except [(-1)^17 + (-4)^6] will have 7 as one of its factors.
  Just think a little bit about the binomial expansion of both [(7-1)^17] and [(7*3 -4)^6] ,then u can readily point out that only the last term of both the expansions , i.e , [(-1)^17] & [(-4)^6] respectively,will don't have 7 as one of its factors.
  
  So, we have to calculate the remainder when [(-1)^17 + (-4)^6] is divided by 7.
  Now, clearly (-1)^17 = -1
  and, (-4)^6 = 4^6 = 2^12 = (2^3)^4 = (7+1)^4.
  Now, a same reasoning related to binomial expansion mentioned previously explains why, when (7+1)^4 is divided by 7 will leave a remainder 1.
  So, (7+1)^4 will be of the for (7*A + 1); where A is some +ve integer, to know whose value isn't important in this case.
  So, when
  [(-1)^17 + (-4)^6] will be divided by 7
  or, [-1 + (7+1)^4] will be divided by 7
  or, when [-1 + 7*A +1] will be divided by 7
  or, when [ 7*A ] will be divided by 7 , clearly therefore the remainder will be zero, i.e, 0.
  
  So, the answer is OPTION 3)0.
  
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• 11 years agoHelpfull: Yes(184) No(26)
• 6^17 mod 7=(7-1)^17 mod 7=(-1)^17 mod 7=-1
  17^6 mod 7=(7*2+3)^6 mod 7= (3)^6 mod 7=729 mod 7=1
  the remainder when 6^17+17^6 is divided by 7 is
  1-1=0(ans)
  
• 11 years agoHelpfull: Yes(132) No(16)
• 6^2 = 36
  6^3 = 216
  Thus 6 raise to anything will give you 6 in the decimal's place.
  
  Similarly 17^6 will give 9 in the decimal place.
  
  Adding the decimal place values of both i.e 6+9 = 15
  
  Remainder = 1. Ans (a)
• 11 years agoHelpfull: Yes(33) No(118)
• we can calculate this type of Questions with calculator.
• 11 years agoHelpfull: Yes(30) No(90)
• 6^17mod 7=((6^2 mod7)^6*6mod7)mod 7=((1^6)*6)mod 7=1*6mod 7=6 mod 7=6
  17^6 mod 7=((17^2mod 7)^3)mod 7=2^3mod 7=8 mod 7=1
  so,1+6=7 mod 7=1
  so,answer is (a)1
• 11 years agoHelpfull: Yes(8) No(43)
• remainder of 6^17 r1=6
  remaindr of 17^6 r2=(17^2)^3=(289)^3=(7*41+2)^3=2
  power should be oddexcept an case 1.
  total remainder=r1+r2=8
  now 8>7 again 8 is divide by 7
  
  then we get remainder=1ans
  
  short trick-
  
  6^17+17^6
  
  take6(6^n where n is odd)+17^2(17^n where n is even)=295 divide by 7
  
  remainder=1apply this method another qus..
  
• 11 years agoHelpfull: Yes(8) No(11)
• ANSWER C) 0 a congruent b (modm) means a-b is divisible by m
  
  6 CONGRUENT -1(MOD 7) THEREFORE 6¹⁷ CONGRUENT (-1)¹⁷(MOD 7) = -1(MOD 7)
  
  17 CONGRUENT 3(MOD 7) THEREFORE 17⁶ CONGRUENT 3⁶(MOD 7) = 1(MOD 7)
  
  HENCE 6¹⁷ + 17⁶ CONGRUENT -1+1(MOD 7)= 0(MOD 7)
  
  REMAINDER IS 0 .
• 11 years agoHelpfull: Yes(7) No(26)
• Remainder of (6^17) = 6............r1
  Remainder of (17^6) =(17^2)^3=(289)^3=(7*41+2)^3= 2^3 =8.......r2
  
  total remainder= r1 + r2 = 14
  now 14 is divide by 7 = 0 (Ans)
• 10 years agoHelpfull: Yes(6) No(2)
• 6 raised to power anything remainder will be 6
  again 17^6 is calculated in units place as (7^6) which will be (7^(6/4)) ie 7^2 its unit digit is 9
  as cyclicity of 7 as in units palce is
  7^1=7
  7^2=9
  7^3=3
  7^4=1
  so (6+9)%7=1 ans
  
• 8 years agoHelpfull: Yes(5) No(1)
• answer is 6..
  when 6,6^2,6^3....
  is divided by 7 gives the pattern of 6,1
  i.e,if power is odd it gives 6 and if it is even it is 1
  here 17 is odd that is 6+17^3 is divisible by 17 hence 0 is the remainder
  so in total 6+0=6mod17=6 is the answer
  hence option b is correct
• 11 years agoHelpfull: Yes(4) No(21)
• in this problem we use unit digit method
  6 power of any value we get the unit digit as 6 adding with 17^6 value is 9 total is 15 divided by 7 remaining is 1
  
• 10 years agoHelpfull: Yes(4) No(3)
• 6^17=...6(6 is the unit digit).....1
  17^6=...9(1 is the unit digit).....2
  
  (6^17)+(17^6)=6+9=15(which gives remainder as 1)
  So remainder is 0.........
  
  
  
• 9 years agoHelpfull: Yes(2) No(2)
• For Sachin Kumar... it works dear thanx alot...I had headache solving for remainder answers... U made it easy...IT WORKS... BEST ANSWER.
• 10 years agoHelpfull: Yes(1) No(0)
• 2.418097655x10^12
• 10 years agoHelpfull: Yes(1) No(0)
• 0. As the series is absolutely divisible by 7
• 10 years agoHelpfull: Yes(1) No(0)
• 6^anything the unit digit will be 6 so its remainder is 6 then for 2nd case by dividing the (17^2)^3 we get (7*41+2)^3 then find unit digit for 2^3 is 8
  its remainder now add two remainder and divide it by 7 so you vl get ans as 0
• 9 years agoHelpfull: Yes(1) No(0)
• (6^17)/7 by reminder method
  1*7=7
  6-7=-1
  (-1^17)=-1 (-1 power any odd number means = -1)
  
  (17^6)=7
  its formula like (n^(b-1))/b=1
  (17^(7-1))/7=1
  
  then -1+1=0
  answer is 0
  
• 8 years agoHelpfull: Yes(1) No(0)
• using -1 rule we get the value for (6)^17/7 as 6 because (a)^x/(a+1) gives remainder "a" when x is odd and gives remainder 1 when x is even..
  Similarly, using prime rule (a)^(p-1)/p gives remainder 1 where p is a prime number.. So value of (17)^6/7 gives remainder 1
  
  Now combining both both we get 6+1=7 which is completely divisible by 7
  Hence the remainder is 0.. :)
• 8 years agoHelpfull: Yes(1) No(0)
• 6^17 + 17^6
  1.For 6^17
  6^1 %7=6
  6^2 %7=1
  6^3 %7=6
  6^4 %7=1
  6^5 %7=6
  6^6 %7=1
  6,1 repeat for 2 times. So 17/2 = 1
  So 6^17 % 7 = 6
  2.For 17^6
  1st 17 % 7 = 3
  Therefore we go with 3 instead of 17
  3^6 = 729
  729 % 7 = 1
  So 17^6 % 7= 1
  Finally
  (6^17 + 17^6) % 7= (6+1) % 7= 7 % 7 =0
  Ans c)0
• 8 years agoHelpfull: Yes(1) No(0)
• 6^17+!7^6=6+9=15/7=1
• 4 years agoHelpfull: Yes(1) No(0)
• 
  ((-1)^17+(2)3)7=-1+1=0
• 8 years agoHelpfull: Yes(0) No(1)
• first find unit place digit so 6 to the power any thing always 6
  similarilly 17^6=9
  so 6+9=15/7=1 ans is a.
  
• 7 years agoHelpfull: Yes(0) No(0)