For a positive integer $$n$$, let $$\langle n \rangle$$ denote the perfect square integer closest to $$n$$. For example, $$\langle 74 \rangle = 81$$, $$\langle 18 \rangle = 16$$. If $$N$$ is the smallest positive integer such that $$\langle 91 \rangle \cdot \langle 120 \rangle \cdot \langle 143 \rangle \cdot \langle 180 \rangle \cdot \langle N \rangle = 91 \cdot 120 \cdot 143 \cdot 180 \cdot N$$ find the sum of the squares of the digits of $$N$$.

Here $$\langle 91 \rangle = 100$$, $$\langle 120 \rangle = 121$$, $$\langle 143 \rangle = 144$$ and $$\langle 180 \rangle = 169$$, so the equation reduces to $$\frac{\langle N \rangle}{N} = \frac{91 \cdot 120 \cdot 143 \cdot 180}{100 \cdot 121 \cdot 144 \cdot 169} = \frac{21}{22}$$. Writing $$N = 22t$$ forces $$21t$$ to be a perfect square, so $$t = 21m^2$$ and $$N = 462m^2$$ with $$\langle N \rangle = (21m)^2$$, which is indeed the closest square since $$21m^2 < 22m^2$$. The smallest value is $$N = 462$$, and $$4^2 + 6^2 + 2^2 = 56$$.