1. **State the problem:** We have three lamps A, B, and C flashing at intervals of 20, 45, and 120 seconds respectively. They start flashing simultaneously at time zero. We want to find how many times in one hour (3600 seconds) all three lamps flash at the same time. 2. **Formula and concept:** The lamps flash together whenever the time elapsed is a common multiple of their flashing intervals. The first time is at 0 seconds. To find subsequent times, we find the Least Common Multiple (LCM) of the intervals 20, 45, and 120 seconds. 3. **Calculate the LCM:** - Prime factorization: - 20 = $2^2 \times 5$ - 45 = $3^2 \times 5$ - 120 = $2^3 \times 3 \times 5$ - LCM takes the highest powers of all primes: - For 2: max power is $2^3$ - For 3: max power is $3^2$ - For 5: max power is $5^1$ - So, $$\text{LCM} = 2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = 360$$ 4. **Interpretation:** The lamps flash together every 360 seconds. 5. **Find how many times in one hour:** - Total seconds in one hour = 3600 - Number of times they flash together = $\frac{3600}{360} + 1$ (including time zero) 6. **Simplify:** $$\frac{3600}{360} = 10$$ - So total flashes together = $10 + 1 = 11$ **Final answer:** The three lamps flash together 11 times in one hour.