The number of pennies, N, required to break the bridge length L is inversely proportional to the length. This is the basis for an equation. There may be a constant factor involved, so we call K this constant and we write N=p/L+K. Can we find a constant of proportionality to make this equation true for the values we have? Let's see.
24=p/4+K; 16=p/6+K; 13=p/8+K; 11=p/9+K; 9=p/10+K.
We can eliminate K by subtracting one equation from another. We need to remember that the equation may only be approximate, a guide, if you like. Let's take the first two equations: 24-16=p(1/4-1/6), so 8=p/12, and p=96. Now take the last two: 11-9=p(1/9-1/10), so 2=p/90 making p=180. Let's take the first and last equations: 24-9=p(1/4-1/10), so 15=3p/20 and p=100. Take the second and third equations: 16-13=p(1/6-1/8), so 3=p/24 and p=72.
So far p hasn't been particularly consistent having possible values between 72 and 180. Note also that if p=96, K=0; but if p=72, K=4. And K could be -1. Then there's something else: the breaking weight of pennies always involves a whole number of pennies, because they can't be split; so if N pennies is not quite enough to break the bridge, N+1 would certainly do it, so we can't expect N to be nicely rounded to a whole number. What we can say, though, is that if N is large it's more likely to be more accurate than if N is small. The first two equations show this. The third equation would be consistent with the first two if N=12 rather than 13. So we could go for N=96/L as the equation which fits the first two sets of figures and nearly fits the third.
Let's use N=96/L and see what we get for (N,L).
(24,4), (16,6), (12,8), (11,9), (10,10) and
(24,4), (16,6), (13,7), (11,9), (9,11),
depending on whether we start with N or L to find L or N. As a model, then, the formula gives reasonable results.