For a positive integer $n$, let $⟨n⟩$ denote the perfect square integer closest to $n$. If $N$ is the smallest positive integer such that $$⟨91\times 120\times 143\times 180\times N⟩=91\times 120\times 143\times 180\times N$$findthesumofthesquaresofthedigitsof$N$.

The closest perfect square $⟨x⟩$ equals $x$ if and only if $x$ itself is a perfect square. Thus, the quantity $91\times 120\times 143\times 180\times N$ must be a perfect square.

We find the prime factorization of $91\times 120\times 143\times 180$:

• $91=7\times 13$

• $120=2^3\times 3\times 5$

• $143=11\times 13$

• $180=2^2\times 3^2\times 5$

Multiply them together:

For $91\times 120\times 143\times 180\times N$ to be a perfect square, $N$ must supply a prime factor for each prime base that currently has an odd exponent in the factorization.

• Primes with odd exponents: $2$ (exponent 5), $3$ (exponent 3), $7$ (exponent 1), and $11$ (exponent 1).

Thus, the smallest positive integer $N$ is:

The digits of $N=462$ are $4,6,$ and $2$.
The sum of the squares of the digits is: