There are 100 switches labeled from 1 to 100 corresponding to each of the bulbs. 10 persons are asked to report the number of defective bulbs they had found, such that the 1st person is allowed to check all the bulbs while the 2nd person is allowed to check bulbs corresponding to switches labeled 2,4,...,100 and so on, and the 10th person is allowed to check bulbs corresponding to switches labeled 10,20,...,100.
Each person reported the same number of defective bulbs. Also note that the number of defective bulbs can increase or remain the same between two consecutive testings and there is at least one defective bulb. Find the following just after the 10th person finished his test.
1) What can be the minimum number of defective bulbs? What are the defective bulbs in this case?
2) What can be the maximum number of defective bulbs? What can be the maximum number of bulbs that can be found to be defective?
If at least one bulb is defective, the most defective bulbs will be no more than 10, which is the most number of switches person 10 would check. Initially, I was tempted to consider person 1 finding a bulb at a switch having a prime number greater than 7, since no-one else would be checking those, but that just inflates the total...potentially a consideration for question 2.
For question 1, there are two solutions to find a minimum of only 2 defective bulbs, with each person finding just 1 bulb each:
1. Person 1 finds the first bulb at switch #72, which will also be tested in turn by persons 2, 3, 4, 6, 8 and 9 (all of which are factors of 72) consistent with the given testing sequence for each. After person 4's test, however, the second bulb at switch #70 goes on the fritz, which is then found by persons 5, 7 and 10 (factors again). No-one reports more than 1 bulb each!
2. Person 1 finds the first bulb at switch #90, followed by persons 2, 3, 5, 6, 9 and 10 (more factors). After person 3's test, the bulb at switch #56 goes out, which is then found in turn by persons 4, 7 and 8 (ditto). Again, no-one reports more than 1 bulb each.
Edited on December 12, 2008, 8:09 pm
Two solutions for Q 1.
