To solve this question , first break the LCM i.e. find the remainder when 6^17 is divided by 7
and then find remainder when 117^6 is divided by 7 :
as follows:
See, if even power of 6 is divided by 7, remainder is 1 and
if odd power of 6 is divided by 7, remainder is 6
so here power of 6 is 17(odd) so its remainder is 6
Now for 117,
if even power of 117 is divided by 7, remainder is 4
if odd power of 117 is divided by 7, remainder is 5
so here power of 117 is 6(even) so remainder is 4
now add both the remainders obtained i.e. 6 + 4 = 10
and when 10 is divided by 7 remainder is 3,
so its answer is 3 (option d).

c) 0

(6^17+117^6)=(7-1)^17+(112+5)^6=(-1)^17+5^6=(-1)+15625=15624=0mod7

so (6^17+117^6)=0mod7

Thus the remainder when (6^17+117^6) is divided by 7 is 0

any time Parusharam

hai how are you medam