The resistance of the heating element is:
\[R = \rho \frac{\ell}{A}.\]
The power is inversely proportional to the length:
\[P \propto \frac{1}{\ell}.\]
From the relation \(P_1 \times t_1 = P_2 \times t_2\):
\[\frac{P_1}{P_2} = \frac{t_2}{t_1} = \frac{15}{20}.\]
Substituting \(P \propto \frac{1}{\ell}\):
\[\frac{\ell_2}{\ell_1} = \frac{t_2}{t_1} = \frac{15}{20} = \frac{3}{4}.\]
Thus, the new length should be decreased to:
\[\ell_2 = \frac{3}{4} \ell_1.\]
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is:
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to: