Oh baby a triple

Number Theory Level 2

If $n$ is a positive integer, which of these 3D vectors has a component divisble by 3?

$\begin{array}{lc} \text{A:} & \begin{pmatrix} n \\ n+1 \\ 2n+1 \end{pmatrix} \\ \text{B:} & \begin{pmatrix} n \\ 2n+1 \\ 2^{2015}n+1 \end{pmatrix} \\ \text{C:} & \begin{pmatrix} n \\ 2^{2015}n+1 \\ 2^{2015^{2}}n+1 \end{pmatrix} \end{array}$

A C B

1 solution

Jake Lai
May 8, 2015

Mortal's solution : By considering cases, prove that: $\text{B}$ and $\text{C}$ fail when $n+1$ is a multiple of 3, keeping in mind that $2^{\text{odd number}} \equiv 2 \pmod{3}$ ; and that $\text{A}$ covers $n, n+1, n+2$ so that, in fact, exactly one is a multiple of 3.

Jake's solution :

$\sum_{k=0}^{n} k^{2} = \frac{n(n+1)(2n+1)}{6}$

Oh baby a triple

1 solution

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