This problem requires students to fully understand what the problem is asking, to use logic, and to find the lowest common multiple of the numbers 1, 2, 3, 4, and 5.
A prisoner sits in his cell planning his escape. The prisoner is kept in by 5 laser beams, which operate along a corridor. Each laser is switched off at a specific time interval for just long enough to allow a person to walk through. The time between being switched off for each laser is shown below:
Laser One = every 3 minutes
Laser Two = every 2 minutes
Laser Three = every 5 minutes
Laser Four = every 4 minutes
Laser Five = every 1 minutes
The guard patrols and checks the prisoner each time all the laser beams are off simultaneously. Because each laser only switches off for a short time the prisoner knows he can only get past one laser at a time. He has to get past the five lasers from 1 to 5 in order. Laser One is at the entrance of the prisoner’s cell and laser Five is at the door to the outside. He also knows that if he spends longer than 4 minutes 12 seconds in the corridor an alarm will go off.
Can the prisoner escape without the alarm in the corridor going off?
If he can escape, how many minutes should he wait before passing Laser One?
How much time will he have after passing Laser Five before the guard raises the alarm?
- Introduce the problem and let the students read and think about it.
- In groups, have the students discuss the problem and write down what they think the main pieces of information are.
What information is given? What conditions apply? What strategy could we use to get started?
- Groups report key information back to the class, which the teacher records on the board.
- Let the class go back and tackle the problem.
- Help groups/class when needed:
Is there another way of writing the problem?
Is there a sequence of times when the laser beams are off that will help us?
Look for a sequence of common multiples.
How can we use the information?
What can we do to solve the problem?
- Groups should report back to the class at regular intervals.
Have you considered all the information? Is there another solution?
- Allow time for the students to write down their solutions.
Extension to the problem
Can the groups devise a similar problem using different time sequences? (Use times in either minutes or minutes and seconds.)
Can you give an algebraic solution to the above problem?
The prisoner can escape and will have 23 minutes before the guard will sound the alarm.
When the guard leaves then let that time equal 0 units’ time. Each laser turns off in a sequence of 1, 2, 3, 4, and 5 minutes. The guard will return every 60 minutes. This is the lowest common multiple of 1, 2, 3, 4 and 5.
The prisoner can only spend 4 minutes 12 seconds in the corridor. This means that the prisoner must wait for a sequence when all the laser beams go off one after the other. The only five successive times possible between 0 and 60 minutes are 33, 34, 35, 36 and 37.
This can be found by listing the laser times.
Laser 1 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33...
Laser 2 = 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, ...
Laser 3 = 5, 10, 15, 20, 25, 30, 35, ...
Laser 4 = 4, 8, 12, 16, 20, 24, 28, 32, 36, ...
Laser 5 = 1, 2, 3, ..., 35, 36, 37..
Or realising you are looking for a sequence of numbers where the first number is a multiple of 3, the second is an even number, the third is a multiple of 5, the fourth is a multiple of 4, and the fifth can be any number as it is a 1s number.
The prisoner must therefore wait until 33 minutes after the guard leaves before entering the corridor. He clears the final door 23 minutes before the guard returns.