Finding The Right Row

The triangle shown is in fact an infinite multiplication table of two positive integers $a$ and $b$ . The mulytiplands $a$ and $b$ are represented by the two sets of diagonals in opposite directions. Let $a$ be represented by the diagonals from the left and $b$ , from right. Then the first left diagonal represents $a=1$ ; the second left diagonal, $a=2$ ; the third left diagonal, $a=3$ and so on. Similarly for $b$ but with the right diagonal. The intersection of the $a$ -diagonal and $b$ -diagonal is the product $n=ab$ . For example, $1 \times 6 = 6$ is the intersection of the first $a$ -diagonal and the sixth $b$ -diagonal. While $2 \times 3 = 6$ is the intersection of the second $a$ -diagonal and the third $b$ -diagonal.

We note that when $a$ or $b$ increases by 1, the row number $r$ increases by 1. Now, let us assign the intersection be $(a,b)$ . To get to the intersection $n=ab$ , we go from $(1,1)$ $\rightarrow$ $(2,1)$ $\rightarrow$ $(3,1)$ $\rightarrow$ $...$ $\rightarrow$ $(a,1)$ and then from $(a,1)$ $\rightarrow$ $(a,2)$ $\rightarrow$ $(a,3)$ $\rightarrow$ $...$ $\rightarrow$ $(a,b)$ . At $(1,1)$ , $r=1$ ; at $(a,1)$ , $r=a$ ; at $(a,2)$ , $r=a+1$ $...$ at $(a,b)$ , $r=a+b-1$ .

Without loss of generality, let us assume $b>a$ and let $\delta=b-a$ be the difference between the two divisors $a$ and $b$ of $n$ . Then, we have:

$\begin{aligned} ab & = n \\ a(a+\delta) & = n \\ a^2 + \delta a -n & = 0 \\ \implies a & = \frac {-\delta+\sqrt{\delta^2+4n}}2 & \small {\color{#3D99F6}\text{As }a > 0 \text{, the negative root is rejected.}} \end{aligned}$

Now let us find the minimum $r$ . Since $r=a+b-1$ , let us find the minimum $a+b$ .

$\begin{aligned} a+b & = 2a + \delta \\ & = -\delta+\sqrt{\delta^2+4n} + \delta \\ & = \sqrt{\delta^2+4n} \end{aligned}$

So we can see that $a+b$ increases with $\delta$ , the difference between $a$ and $b$ . And for all positive real $a$ and $b$ , $a+b$ is minimum when $\delta = 0$ which is proven by AM-GM inequality : $a+b \ge 2\sqrt{ab} = 2\sqrt n$ . For integral $a$ and $b$ , minimum $a+b$ occurs when $\delta = b-a$ is minimum. For $n=2016$ , $a=42$ and $b=48$ and $r=42+48-1 = \boxed{89}$ .

A perhaps shorter but more arbitrary and impractical solution. We know that 2016 appears in every line $n = \frac{2016}{x} + x -1$ where is the line from which you start going inwards the triangle. With calculus we find the minimun value $n \approx 88.8$ , but of course n needs to be a natural numbers so we get a theoretical minimum of 89. Now we see if that can be achieved. Solve for x in the above equation which yields $x=42$ . Thus the theoretical minumun can be achieved hence $n=89$ .