Starting with the series:
\[S(x) = (1 + x) + 2(1 + x)^2 + 3(1 + x)^3 + \dots + 60(1 + x)^{60}\]
Multiplying both sides by \( (1 + x) \), we get:
\[(1 + x)S = (1 + x) + 2(1 + x)^2 + 3(1 + x)^3 + \dots + 60(1 + x)^{61}\]
Now, subtracting \( S \) from \( (1 + x)S \), we obtain:
\[-xS = \frac{(1 + x)(1 + x)^{60} - 1}{x} - 60(1 + x)^{61}\]
Now, put \( x = 60 \):
\[-60S = \frac{61((61)^{60} - 1)}{60} - 60 \cdot (61)^{61}\]
Solving this equation gives:
\[S = 3660\]
Let $\left\lfloor t \right\rfloor$ be the greatest integer less than or equal to $t$. Then the least value of $p \in \mathbb{N}$ for which
\[ \lim_{x \to 0^+} \left( x \left\lfloor \frac{1}{x} \right\rfloor + \left\lfloor \frac{2}{x} \right\rfloor + \dots + \left\lfloor \frac{p}{x} \right\rfloor \right) - x^2 \left( \left\lfloor \frac{1}{x^2} \right\rfloor + \left\lfloor \frac{2}{x^2} \right\rfloor + \dots + \left\lfloor \frac{9^2}{x^2} \right\rfloor \right) \geq 1 \]
is equal to __________.