We know no eggs were left over in the groups of 7, so the total x is a multiple of 7. We know it's an odd multiple because it's not even. We know it ends in 1 or 6 because of the remainder of 1 with 5 but it can't end in the even 6.
So we make a list of odd multiples of 7 that end in 1.
21, 91, 161, 231, 301, ...
21 is a multiple of 3, nope
91 has a remainder of 3 when divided by 4
161 has a remainder of 2 when divided by 3
231 is a multiple of 3
301 has remainder 1 mod 2, 1 mod 3, 1 mod 4, 1 mod 5, 1 mod 6 and 0 mod 7. It's our answer; there are bigger answers as well.
Answer: 301
Multiple simultaneous congruences may be solved by the Chinese Remainder Theorem.
