1. **Problem Statement:** Security light poles are erected along both sides of a street. On the left side, poles are spaced $a$ meters apart, and on the right side, poles are spaced $b$ meters apart. At one end, the poles are directly opposite each other. We want to find the number of poles erected by the first time the poles are directly opposite each other again at an end of the road. 2. **Understanding the problem:** The poles on each side form sequences of positions: left side at multiples of $a$, right side at multiples of $b$. The poles are opposite each other when their positions coincide, i.e., when $n a = m b$ for some integers $n, m$. 3. **Key formula:** The first time poles align again at an end corresponds to the least common multiple (LCM) of $a$ and $b$: $$\text{LCM}(a,b) = \frac{a \times b}{\text{GCD}(a,b)}$$ where GCD is the greatest common divisor. 4. **Number of poles on each side:** The number of poles up to the first alignment is the number of intervals plus one: - Left side poles: $\frac{\text{LCM}(a,b)}{a} + 1$ - Right side poles: $\frac{\text{LCM}(a,b)}{b} + 1$ 5. **Total poles erected:** Sum the poles on both sides: $$\left(\frac{\text{LCM}(a,b)}{a} + 1\right) + \left(\frac{\text{LCM}(a,b)}{b} + 1\right) = \frac{\text{LCM}(a,b)}{a} + \frac{\text{LCM}(a,b)}{b} + 2$$ 6. **Explanation:** The $+1$ accounts for the pole at the starting end. The poles align at the LCM distance, so counting poles up to that point on both sides and adding them gives the total number of poles erected. **Final answer:** The total number of poles erected by the first time they are directly opposite each other at an end is: $$\boxed{\frac{\text{LCM}(a,b)}{a} + \frac{\text{LCM}(a,b)}{b} + 2}$$