Step 1: Set up one combined formula before putting in any numbers.
Let $m_f$ be the mass of nuclear fuel, $c$ the speed of light, $M$ the mass of one coach and $v$ its speed. Only 10 percent of the fuel's energy is usable since 90 percent is wasted as friction, so the number of coaches $n$ that can be dragged satisfies
\[ n \times \frac{1}{2}Mv^2 = 0.1\, m_f c^2 \]
Solving for $n$ in one step:
\[ n = \frac{0.2\, m_f c^2}{M v^2} \]
Writing it this way means we substitute numbers only once, at the end, instead of computing two separate energies and then dividing.
Step 2: List the known values.
\[ m_f = 1 \ \text{g} = 10^{-3} \ \text{kg}, \quad c = 3\times10^{8} \ \text{m/s}, \quad M = 18000 \ \text{kg}, \quad v = 100 \ \text{m/s} \]
Step 3: Substitute into the single formula.
\[ n = \frac{0.2 \times 10^{-3} \times (3\times10^{8})^2}{18000 \times (100)^2} = \frac{0.2 \times 10^{-3} \times 9\times10^{16}}{18000 \times 10^{4}} \]
Step 4: Simplify numerator and denominator separately.
Numerator: $0.2 \times 9 = 1.8$, so the numerator is $1.8\times10^{13}$. Denominator: $18000 \times 10^4 = 1.8\times10^{8}$.
\[ n = \frac{1.8\times10^{13}}{1.8\times10^{8}} = 10^{5} \]
Step 5: Why this shortcut works as a check.
Because both the fuel-energy term and the coach-energy term share the same factor of $1.8$, that factor cancels cleanly, which is a useful check that the substitution was done correctly.
Final Answer:
\[ \boxed{100000 \ \text{coaches}} \]