Working out tallest formation possible In a stack of logs, each row has one less log than the row below.
With 2015 logs, what is the tallest possible stack?
I thought I'd start by doubling a 9-log stack and putting one upside up,one upside down, to make a parallelogram, but then I got stuck. How do I proceed?
 A: If I understand your trapezoid approach correctly, you are doubling a $9$-log stack so that you can algebraically calculate the number of logs as
$$
\frac{3 \textrm{ logs high } \times (4 \textrm{ logs at the bottom } + 2 \textrm{ logs at the top })}{2} = 9 \textrm{ logs } \\
$$
This is a similar approach used by the Pythagoreans to calculate triangular numbers because these can be arranged as an equilateral triangle just like your stack of logs. The only difference is that you seem to be chopping the triangle at the top, to $2$ logs in your $9$-log example.
Instead, it might be simpler to suspend our imagination for a moment and assume the logs are real numbers to calculate the $n$th partial sum for $2015$
$$
\begin{eqnarray}
\frac{n \times (n + 1)}{2} &=& 2015 \\
n^2 + n - 4030 &=& 0\\
\end{eqnarray}
$$
This is a quadratic equation so you can then use the quadratic formula to find $n$
$$
\begin{eqnarray}
n &=& \frac{-1 \pm \sqrt{1 - 4 \times (-4030)}}{2} \\
n &\approx& -63.98 \quad \textrm{ or } \quad n \approx 62.98 \\
\end{eqnarray}
$$
To resume our imagination, the number of rows cannot be negative and they cannot be fractional either so the tallest possible stack is $62$ logs high.
