1. **Problem statement:** Simplify the expression $$\frac{5^{m+3} + 125 \cdot 5^{m-1}}{5^{m+4} \div 25}$$. 2. **Recall important rules:** - Division by a number is multiplication by its reciprocal. - Powers of the same base add when multiplied: $$a^x \cdot a^y = a^{x+y}$$. - Powers of the same base subtract when divided: $$\frac{a^x}{a^y} = a^{x-y}$$. - Express constants as powers of the base 5 when possible: $$125 = 5^3$$ and $$25 = 5^2$$. 3. **Rewrite constants:** $$125 = 5^3$$ and $$25 = 5^2$$. 4. **Rewrite numerator:** $$5^{m+3} + 125 \cdot 5^{m-1} = 5^{m+3} + 5^3 \cdot 5^{m-1} = 5^{m+3} + 5^{3 + m - 1} = 5^{m+3} + 5^{m+2}$$. 5. **Rewrite denominator:** $$5^{m+4} \div 25 = 5^{m+4} \div 5^2 = 5^{(m+4) - 2} = 5^{m+2}$$. 6. **Rewrite entire expression:** $$\frac{5^{m+3} + 5^{m+2}}{5^{m+2}}$$. 7. **Factor numerator:** $$5^{m+2}(5^1 + 1) = 5^{m+2}(5 + 1) = 5^{m+2} \times 6$$. 8. **Simplify fraction:** $$\frac{5^{m+2} \times 6}{5^{m+2}} = 6$$. **Final answer:** $$6$$