If a number ends in the digit 0, then for it to be of the form $2^m\times 5^n$ only, it must be divisible by:

**Theorem**: A rational number $\frac{p}{q}$ (in lowest terms) has a terminating decimal expansion if and only if $q$ has the form **2ᵐ × 5ⁿ** (m, n ≥ 0).

**Examples**:
- $\frac{3}{8}:q=8=2^3$ → terminates$.(\frac{3}{8}=\mathrm{0.375.})-\frac{7}{40}:q=40=2^3\times 5$ → terminates$.(\frac{7}{40}=\mathrm{0.175.})-\frac{1}{3}:q=3$ → does NOT terminate$.(\frac{1}{3}=0.333$… recurring.)
- $\frac{4}{15}:q=15=3\times 5$ → does NOT terminate (because of the 3).

**Why**: any decimal expansion is a fraction with denominator 10ⁿ = 2ⁿ × 5ⁿ. To convert $\frac{p}{q}$ to decimal form, you need $q$ to ONLY have 2s and 5s as primes — otherwise you can't multiply numerator/denominator into a power of 10.