A number theory problem by Victor Herlys

With $N$ being the number of eggs in the basket, we have that

$N \equiv (k - 1)\pmod{k}$ for each of $k = 2, 3, 4, 5, 6$ , and so

$(N + 1) \equiv 0\pmod{k}$ for each of these values of $k$ .

Thus $N + 1$ must be divisible by each of $2, 3, 4, 5$ and $6$ . The least positive value that meets these conditions is LCM $(2,3,4,5,6) = 60$ , and thus the least possible value for $N$ is $59$ , and in general the possible values for $N$ are of the form $59 + 60m$ for any non-negative integer $m$ .

Now we must deal with the condition that $N$ is divisible by $7$ . We thus need to find the least value of $m$ such that $59 + 60m$ is divisible by $7$ . This is the case when

$59 + 60m \equiv 0\pmod{7} \Longrightarrow 60m \equiv -59\pmod{7} \Longrightarrow (56m + 4m) \equiv 4\pmod{7} \Longrightarrow 4m \equiv 4\pmod{7}$ ,

the least positive solution for which is $m = 1$ .

Thus $N = 59 + 60*1 = \boxed{119}$ .