Beth and the Cryptogram!

$\text{ Notice the solution is a multiple of 11 and } 10\equiv -1 \pmod{11}\\ \text{Solution can be written as } \color{#E81990}7\times 10^{4} + 7\times 10^{3} + T\times 10^{2} + 8\times 10^{1} + \color{#E81990}E\times 10^{0}$ $10\equiv -1 \pmod{11} \\ \Rightarrow 10^{k}\equiv -1^{k} \pmod{11} \\ \Rightarrow 10^{k}\equiv 1 \pmod{11} \text{ for even k, and } 10^{k}\equiv -1 \pmod{11} \text{ for odd k } \\ \Rightarrow \color{#E81990} -7 + 7 -\color{#E81990}T + 8 - \color{#E81990}E \equiv 0 \pmod{11} \text{ so } \color{#E81990}T = 8 - E$ $3\times \color{#E81990}E\equiv \color{#E81990}E \pmod {10} \text{ and } 5\times 3 = 15 \text{ is the only integer which has this property}$ $\color{#E81990}E = 5 \Rightarrow T = 3$ $\text{ Now we have }$ $\large{\begin{array}{cccccc} & & \color{#E81990}B & \color{#E81990}T & \color{#E81990}H & \color{#E81990}E \\ \times & & & & 3 & 3 \\ \hline & 7 & 7 & 3 & 8 & 5 \\ \end{array}}$ $\color{#E81990}{B T H E} = \frac{77385}{33}\\ \color{#E81990}{B T H E} = 2345$ $\text{ However her name is Beth so that is the combination, from the cryptogram... }$ $\color{#E81990}B = 2$ $\color{#E81990}E = 5$ $\color{#E81990}T = 3$ $\color{#E81990}H = 4$ $\text{ So the combination is } \boxed{2534}$