All that remains

Since $67 \equiv -1 \pmod {68}$

Then $67^{67} \equiv (-1)^{67} \pmod {68}$

$(-1)^{67} \equiv -1 \pmod {68}$

Therefore, $67^{67} + 67 \equiv -1 + (-1) \pmod {68} \equiv -2 \pmod{68}$ .

Finally, $-2 \equiv 66 \pmod{68}$

67^67+67 can be wriiten as (68-1)^67+67 .. pllying bnomila exapnasion for the first term we have 67c068^0(-1)^67+67c168^1(-1)^66+67c268^2(-1)65...............................................67c6768^67 +67 -1+68(67c168^0(-1)^66+67c268^1(-1)65...........................67c6768^66) +67 66+68(67c168^0(-1)^66+67c268^1(-1)65...........................67c6768^66) which is of the form r+68q .... where r is remainder and q is qoutient .. hence r=66

67^2 when divided by 68 will leave a remainder 1. So 67^64 when divided by 68 will leave a remainder 1. And 67^3 when divided by 68 will leave a remainder 67. Therefore the remainder when 67^67 will be divided by 68 will be 67. Therefore when 67+67=134 when divided by 68 the remainder will be 66. That is the answer. :)