Eggs in a basket

This problem can be written as a system of linear congruences:

$\begin{cases}\begin{aligned} x &\equiv 1 \pmod{2} \\ x &\equiv 2 \pmod{3} \\ x &\equiv 3 \pmod{4} \\ x &\equiv 4 \pmod{5} \\ x &\equiv 5 \pmod{6} \\ x &\equiv 0 \pmod{7}. \end{aligned}\end{cases}$

This can be reduced to the equivalent system of congruences:

$\begin{cases}\begin{aligned} x &\equiv 2 \pmod{3} \\ x &\equiv 3 \pmod{4} \\ x &\equiv 4 \pmod{5} \\ x &\equiv 0 \pmod{7}. \end{aligned}\end{cases}$

Note that the first three congruences can be rewritten:

$\begin{cases}\begin{aligned} x &\equiv -1 \pmod{3} \\ x &\equiv -1 \pmod{4} \\ x &\equiv -1 \pmod{5}. \end{aligned}\end{cases}$

Since these constants are the same, a "shortcut" can be taken with the Chinese Remainder Theorem. The LCM of these moduli is 60, so

$\begin{aligned} x &\equiv -1 \pmod{60} \\ &\equiv 59 \pmod{60} \end{aligned}$

Write this congruence as an equation, and substitute the expression for $x$ into the final congruence:

$\begin{aligned} x &= 60j+59 \\ 60j+59 &\equiv 0 \pmod{7} \\ j &\equiv 1 \pmod{7} \end{aligned}$

Then write this congruence as an equation, and substitute it into the expression for (x:)

$\begin{aligned} j &= 7k+1 \\ x &= 60(7k+1)+59 \\ x &= 420k + 119 \\ x &\equiv 119 \pmod{420} \end{aligned}$

Thus, the least amount of eggs that could be in Brahmagupta's basket is $\boxed{119}.$