Latest TCS Aptitude Question SOLUTION: what is the remainder when 6^17+117^6 is divided by 7?

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what is the remainder when 6^17+117^6 is divided by 7?

Asked In TCS manjushree hossur (11 years ago)

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prashant  (11 years ago)

6/7 remainder = 6
6^2/7 = 36/7 remainder = 1
6^3/7 = 216/7 remainder = 6
similarly 6^17/7 remainder = 6
comes to 117^6 in number theory there is fermat's theorem
a^(p-1) = 1mod p here p = 7 a 117 so remainder will be 1
now above problem can be written as 6^17/7 + 117^6/7
or remainder 6/7+1/7
or 7/7
so 0 will be the remainder

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aditya gupta  (11 years ago)

as 6/7=remainder =-1
117/7=remainder=-2
so use remainder theorem
(-1)^17+(-2)^6/7
64-1/7
63/7
remainder=0

Is this solution Helpfull? Yes (6) | No (1)  

Ankit Kumar  (11 years ago)

6^17+117^6=(7-1)^17+(119-2)^6
after breaking it by binomial theorem we get the last term from
each one as:-17C17*(-1)^17 = -1 and another one 6C6*(-2)^6 = 64.
so sum is 63 and divided by 7 so reminder:-0.

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Sutripta Mukhopadhyay  (11 years ago)

remainder=0

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pendem  (11 years ago)

unit digit of in power for 6^17 is 7.
unit digit of in no for 117^6 is 7.
so 7/7 then remainder is zero.

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Dhanjeet kumar  (8 years ago)

(6*6^16+117^6)/6=(6(1^8)+1^6)/7=7/7,so remainder 0.

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Arun kamal Singh  (8 years ago)

(6^17/7)+(117^6)/7
(-1)17+(5^6)/7
-1+((7-2)^6)/7
-1+2^6/7
-1+((2^3)62)/7
-1+1
0
so remainder is 0

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prathna  (5 years ago)

6^17 mod 7=(7-1)^17 mod 7=(-1)^17 mod 7=-1
117^6mod 7=(7*16+5)^17mod7=(5)^17 mod7=1
the remainder when 6^17+117^6 is divided by 7 is
1-1=0(ans)

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