Modular arithmetic task

Question

If $2^{100} \equiv 1$ (mod $125$) and $2^{100} \equiv 0$ (mod $8$) $\Rightarrow$ $2^{100} \equiv 376$ (mod $1000$).

1. $2^{100} = n_1\cdot 125 + r_1$ and $1 = 0\cdot 125 + 1$ $\Rightarrow$ $r = 1$
2. $\frac{2^{100} - 1}{125} = n_1 - 0$
3. $376 = 0 \cdot 1000 + 376$ and $2^{100} = m_1\cdot 1000 + r_2$ $\Rightarrow$ $r_2 = 376$
4. $\frac{2^{100} - 376}{1000} = m_1 - 0$ $\Rightarrow$ $\frac{2^{100} - (3\cdot 125 + 1)}{8 \cdot 125} = \frac{(2^{100} - 1) - 3\cdot 125 }{8 \cdot 125} = \frac{125\cdot n_1 - 0 - 3\cdot 125}{8 \cdot 125} = \frac{n_1 - 3}{8} = m_1$
5. $2^{100} = m_1 \cdot 1000 - 375$

Where is the mistake?

Answer 1

$\!\bmod 125\!:\ x := 2^{\large 100}\!\equiv 1\iff x = 1+125k.\ $ Substituting this into the congruence $\bmod 8$

$\!{\bmod 8\!: \ \ 0\equiv x = 1 + 125\,\color{#c00}k} \equiv 1-3k\iff 3k\equiv 1\equiv 9\iff k\equiv 3\iff \color{#c00}{k = 3+8n}$

So we deduce $\ x = 1 + 125(\color{#c00}{3+8n}) = \bbox[5px,border:1px solid orange]{376+1000n}\ $

Above is one of the standard versions of CRT = Chinese Remainder Theorem, e.g. see Easy CRT. You should study these methods to learn the correct way to solve such systems of congruences.

To debug your attempt you need to explain what you are attempting to do in your calculations (which is far from clear).