Christmas Streak 17/88: No Calculator Allowed!

Note that the explanation is assuming $\log 2\approx 0.30103$ and $\log 3\approx 0.47712.$

Take a common logarithm on $5^{10000}.$ Then,

$\log 5^{10000} \\ =10000\log 5 \\ =10000\left(\log \dfrac{10}{2}\right) \\ =10000(\log 10-\log 2) \\ =10000(1-0.30103) \\ \approx {\color{#E81990}6989.7}$

((((( $6989<\log 5^{10000}<6990.$ Putting all sides on the exponential of 10, )))))

Then $10^{\color{#E81990}6989}<5^{10000}<10^{{\color{#E81990}6989}+1}.$

((((( Note that $10^{6989}$ has 6990 digits and $10^{6990}$ is the first number to have 6991 digits. )))))

So, $5^{10000}$ has $\color{#E81990}6990$ digits.

Let's say that $k\cdot 10^{\color{#E81990}6989}<5^{10000}<(k+1)\cdot 10^{\color{#E81990}6989}.$

((((( Taking a common logarithm on both sides, $\log k<0.7<\log k+1.$ )))))

$\log 5 \approx \dfrac{\color{#E81990}6989.7}{10000}$ and $\log 6=\log 2 + \log 3 \approx 0.30103+0.47712=0.77815=\dfrac{\color{#E81990}7781.5}{10000}.$

We can infer that $k=\color{#E81990}5.$

Therefore, the leading digit of a $\color{#E81990}6990$ -digit number $5^{10000}$ is 5.

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From above, $A=6989.7,~B=6989,~C=6990,~D=7781.5,~E=5.$

$\lfloor 100{\color{#E81990}A}+{\color{#E81990}B}+{\color{#E81990}C}+100{\color{#E81990}D}+{\color{#E81990}E}\rfloor \\ = \lfloor 698970+6989+6990+778150+5 \rfloor = \boxed{1491104}.$