If $$p, q, r$$ are primes such that $$pq + qr + rp = pqr - 2025$$, find $$p+q+r$$.

If all three primes were odd then $$pqr - (pq+qr+rp)$$ would be even, but 2025 is odd, so one prime is 2. Putting $$p = 2$$ gives $$qr - 2q - 2r = 2025$$, that is $$(q-2)(r-2) = 2029$$. Since 2029 is prime the only possibility is $$q = 3$$ and $$r = 2031$$, so $$p+q+r = 2 + 3 + 2031 = 2036$$.