Difficulty: Medium
Correct Answer: 16
Explanation:
Introduction / Context:
This problem models a classic synchronization situation. Multiple bells ring at fixed time intervals. We are asked how many times they all ring together within a given time frame. Such questions are practical applications of the least common multiple (LCM) concept in time and periodic processes.
Given Data / Assumptions:
Concept / Approach:
All bells will toll together every time a moment occurs that is a common multiple of all their individual intervals. The smallest such time gap is the least common multiple of the intervals. Once that LCM is known, the number of simultaneous tolls is the count of multiples of this LCM that lie within the total time, including the starting moment at 0 seconds.
Step-by-Step Solution:
Find LCM of 2, 4, 6, 8, 10, 12.
Prime factorizations: 2 = 2, 4 = 2^2, 6 = 2 * 3, 8 = 2^3, 10 = 2 * 5, 12 = 2^2 * 3.
Take highest powers of each prime: 2^3, 3^1, 5^1.
LCM = 2^3 * 3 * 5 = 8 * 3 * 5 = 120 seconds.
Number of intervals inside 1800 seconds = 1800 / 120 = 15.
Include the initial toll at time 0 seconds, so total together tolls = 15 + 1 = 16.
Verification / Alternative check:
The bells toll together at times 0, 120, 240, 360, and so on up to 1800 seconds. Counting these gives 16 moments: this matches the calculation based on division and confirms the answer.
Why Other Options Are Wrong:
10, 15, 18, and 20 either ignore the initial together toll at time zero or assume a wrong LCM, or they miscount the number of multiples up to 1800 seconds.
Common Pitfalls:
Two common mistakes are forgetting to include the initial toll at time zero and confusing the sum of intervals with the LCM. Always remember that the bells are already together at the starting instant and that the LCM, not simple addition, determines when repeated alignment occurs.
Final Answer:
16
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