3 ms·
7*n^3 is even when n is even and odd otherwise. odd + odd is even, as is even + even.
by BurningFrog 2y ago
7*n^3 is even when n is even and odd otherwise.
odd + odd is even, as is even + even.
- Spivak 2y agoThe easy way of seeing the first part is to do the prime factorization. The 7 doesn't matter since it's prime. If n has a 2 in its factorization it now has 2^3. But if it doesn't have a 2 it won't suddenly acquire one. All the symbol soup proofs aren't wrong but I don't think they satisfyingly explain the why.
- thaumasiotes 2y agoAll of these "always divisible by n" proofs are asking you to solve them case by case in modular arithmetic. For divisibility by two, there are only two cases. So if n is 1, then n³ is 1, and if n is 0, n³ is 0. 0+0 = 0; 1+1 = 0; and this completes the proof. I am not actually sure that doing a prime factorization on 7n³ for unknown n is easier than knowing that 1³ = 1.
- crabbone 2y agoI vote this the best proof. All you need to know to understand it is to know how multiplication, addition and exponentiation works. You could probably show this to a child in a sixth grade or so, and have them understand it. This is really good!