Match Sticks

Let's try and solve the general case. Say there are $M$ matches in total, and we want to know the sum of the areas of the rectangles we can make with them with sides $a<b$ .

From the diagram, it's clear that $M=a(b+1)+b(a+1)=2ab+a+b$

Doubling and adding one allows us to factorise: $2M+1=4ab+2a+2b+1=(2a+1)(2b+1)$

Now, the quantity we're after is $S=\sum ab$ over all pairs $a<b$ satisfying $(2a+1)(2b+1)=2M+1$ .

Note that, from the first equation, $ab=\frac12 (M-a-b)$ for these pairs; so the sum is $S=\frac12 \sum (M-a-b)=\frac14 \sum \left(2M+2-(2a+1)-(2b+1)\right)$

So we're interested in the number of divisors of $2M+1$ , and their sum; the first gives twice the number of solution pairs $2a+1,2b+1$ , and the second their sum. These two functions are usually denoted $\sigma_0$ and $\sigma_1$ respectively.

If the prime factorisation of $x$ is $x=p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$

then we have $\sigma_0(x)=\left(1+k_1 \right)\left(1+k_2 \right)\cdots \left(1+k_r \right)$ and $\sigma_1(x)=\frac{p_1^{k_1+1}-1}{p_1-1} \cdot \frac{p_2^{k_2+1}-1}{p_2-1} \cdots \frac{p_1^{k_r+1}-1}{p_r-1}$

Returning to our formula for $S$ , $\begin{aligned} S&=\frac14 \sum \left(2M+2-(2a+1)-(2b+1) \right) \\ &=\frac14 \left[(M+1)\sigma_0(2M+1)-\sigma_1(2M+1) \right] \end{aligned}$

Note we would have to be careful if $2M+1$ happened to be a square number (why?), but in this case, it's not.

In the question, $M=337$ and $2M+1=675=3^3 \cdot 5^2$ . Using the above formulas, $\sigma_0(675)=12$ and $\sigma_1(675)=1240$ , so $S=\frac14 \left[338 \cdot 12 - 1240 \right]=\boxed{704}$

$a(b+1) + b(a+1) = 337 \\ \Rightarrow 2ab + a + b = 337 \\ \Rightarrow 2(2ab + a + b) + 1 = 2 \times 337 + 1 \\ \Rightarrow (2a + 1)(2b + 1) = 675 = 3^3 \times 5^2$

$\therefore Sum = 112 + 134 + 148 + 154 + 156 \\ = 704$