Australian Mathematics Competition 2015 Q30

We note that the number of logs of a "perfect" stack, that is the top row with $1$ log; second, $2$ ; third $3$ , ... until $n$ logs at the bottom is a triangular number which is given by $T_n = \dfrac{n(n+1)}{2}$ . For example: $T_4 =\dfrac{4\times 5}{2} = 10$ . Therefore, for $9$ logs, the tallest stack is $n-1=3$ rows.

To find the number of rows for $2015$ , we need to find the smallest $T_n$ greater than $2015$ and check $T_n - 2015$ . Therefore, we have:

$\begin{aligned} T_n & \ge 2015 \\ \Rightarrow \frac{n(n+1)}{2} & \ge 2015 \\ n^2 + n - 4030 & \ge 0 \\ n & \ge 62.98 \\ \Rightarrow n & = 63 \\ \Rightarrow T_{63} & = \frac{63\times 64}{2} = 2016 \end{aligned}$

Since $2016-2015 = 1$ , the number of rows of the tallest stack for $2015$ logs is $63-1 = \boxed{62}$ .

This stack of logs can be defined as the expression $T_m - T_n$ where $T_x$ represents the $x$ th triangular number, $m$ is the number of logs on the bottom row (with $T_m$ being the number of logs if the stack was 'perfect', that is have one log at the to, two logs in the next row, and so forth) and $n$ being one less than the number of logs on top of the stack (so as to remove a portion of $T_m$ from the top to reduce it to the stack required. To answer the question, we must maximise the value of $m - n$ .

We have

$T_m - T_n = 2015$

$\dfrac {m(m+1)}{2} - \dfrac {n(n+1)}{2} = 2015$

$m^2 + m - n^2 - n = 4030$

$(m+n+1)(m-n)=4030$

Since $m$ and $n$ are positive integers, we have to determine pairs of positive integers that multiply to give 4030.

$4030 = 1 \times 4030, 2 \times 2015, 5 \times 806, 10 \times 403, 13 \times 310, 26 \times 155, 31 \times 130, 62 \times 65$

Since $m$ and $n$ are positive integers, we have $m+n+1 > m-n$ . In order to maximise the value of $m-n$ , we must minimise $m+n+1$ . We will start with the combination $62, 65$ .

$m+n+1=65$

$m-n=62$

$m+n=64$

$m=63, n=1$

Since $m$ and $n$ both satisfy being positive integers, this combination works. Therefore, the maximum value of $m-n$ is $63-1=62$ . There are 62 rows of logs in the stack.

We can obtain possible stacking arrangements from factorizations of 2015. Specifically, if $2015=ab$ , then we have $2015=T_n-T_m$ where $n=\frac{a+2b-1}{2}$ and $m=\frac{a-2b-1}{2}$ , provided that $a-2b>0$ . For example, with $2015=ab=65*31$ , we obtain $2015=T_{63}-T_1$ , so that we have $\boxed{62}$ rows, and we can't do any better than that since 2015 is not itself a triangular number.