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Does 4^9 + 6^10 + 3^20 happen to be a prime number?
I cheated and calculated the sum and checked it against known primes so I have the answer. But I’m sure there’s a clever answer.
I have the clever solution. They could have made this problem with larger numbers so the brute force method was not possible.
:::
(a-b)^2 = a^2-2ab+b^2.
It's a square number. End of story.
Yep, and to my point that they could have made it not solvable through brute force, it works for any 3^n and 2^(n-1).
So ask about 4^999 + 6^1000 + 3^2000
2 weeks in a row they gave easy problems. Kinda disappointing.
But, to be honest, it was easier to derive the "clever" solution from the "brute force" solution than arriving directly at the "clever" soluton.
