To solve this, we need to find the time at which all three traffic lights will change simultaneously, given that they have different intervals for changing. We do this by finding the least common multiple (LCM) of the intervals at which the lights change, which are 36, 42, and 72 seconds.
The least common multiple (LCM) of two numbers is the smallest number that is a multiple of both of them.
First, we find the LCM of 36 and 42.
The prime factors of 36 are \( 2 \times 2 \times 3 \times 3 = 2^2 \times 3^2 \).
The prime factors of 42 are \( 2 \times 3 \times 7 = 2^1 \times 3^1 \times 7^1 \).
To find their LCM, we take the highest powers of all prime numbers present: \( 2^2, 3^2, 7^1 \).
So LCM(36, 42) = \( 2^2 \times 3^2 \times 7 = 4 \times 9 \times 7 = 36 \times 7 = 252 \) seconds.
Next, we find the LCM of the result (252 seconds) and the third interval, 72 seconds.
The prime factors of 252 are \( 2 \times 2 \times 3 \times 3 \times 7 = 2^2 \times 3^2 \times 7^1 \).
The prime factors of 72 are \( 2 \times 2 \times 2 \times 3 \times 3 = 2^3 \times 3^2 \).
For their LCM, we again take the highest powers of all prime numbers present: \( 2^3, 3^2, 7^1 \).
So LCM(252, 72) = \( 2^3 \times 3^2 \times 7 = 8 \times 9 \times 7 = 504 \times 9 = 4536 \) seconds.
Thus, the three lights will change simultaneously every 4536 seconds after they are first switched on at 9.00 am.
Now we convert 4536 seconds into hours, minutes, and seconds to find out what time of day that will be. There are 3600 seconds in an hour, so:
Hours: \( 4536 \div 3600 = 1 \) hour with a remainder.
The remainder is \( 4536 - 3600 = 936 \) seconds.
We convert the remaining seconds into minutes:
Minutes: \( 936 \div 60 = 15 \) minutes with a remainder.
The remainder is \( 936 - 15 \times 60 = 936 - 900 = 36 \) seconds.
Finally, we add this to the original start time of 9.00 am:
9.00 am + 1 hour = 10.00 am
10.00 am + 15 minutes = 10.15 am
10.15 am + 36 seconds = 10.15 am and 36 seconds
So the traffic lights will all change simultaneously at 10:15:36 am.
