The number of elements in the set $$\{n \in \{1, 2, 3, \ldots, 100\} | (11)^n > (10)^n + (9)^n\}$$ is ___.

We need to find how many $$n \in \{1, 2, 3, \ldots, 100\}$$ satisfy $$11^n > 10^n + 9^n$$.

Dividing both sides by $$10^n$$, the inequality becomes $$\left(\frac{11}{10}\right)^n > 1 + \left(\frac{9}{10}\right)^n$$.

Let $$a = \frac{11}{10} = 1.1$$ and $$b = \frac{9}{10} = 0.9$$. We need $$(1.1)^n > 1 + (0.9)^n$$.

Let us define $$f(n) = (1.1)^n - (0.9)^n - 1$$ and check small values of $$n$$.

For $$n = 1$$: $$f(1) = 1.1 - 0.9 - 1 = -0.8 < 0$$. The inequality does not hold.

For $$n = 2$$: $$f(2) = 1.21 - 0.81 - 1 = -0.6 < 0$$. The inequality does not hold.

For $$n = 3$$: $$f(3) = 1.331 - 0.729 - 1 = -0.398 < 0$$. The inequality does not hold.

For $$n = 4$$: $$f(4) = 1.4641 - 0.6561 - 1 = -0.192 < 0$$. The inequality does not hold.

For $$n = 5$$: $$f(5) = 1.61051 - 0.59049 - 1 = 0.02002 > 0$$. The inequality holds.

Now we show that once $$f(n) > 0$$, it remains positive for all larger $$n$$. Consider: if $$11^n > 10^n + 9^n$$, then $$11^{n+1} = 11 \cdot 11^n > 11(10^n + 9^n) = 11 \cdot 10^n + 11 \cdot 9^n$$.

We need this to be at least $$10^{n+1} + 9^{n+1} = 10 \cdot 10^n + 9 \cdot 9^n$$. Since $$11 \cdot 10^n > 10 \cdot 10^n$$ and $$11 \cdot 9^n > 9 \cdot 9^n$$, we get $$11^{n+1} > 10^{n+1} + 9^{n+1}$$.

So the inequality holds for all $$n \ge 5$$ and fails for $$n = 1, 2, 3, 4$$.

The number of valid values is $$100 - 4 = 96$$.