You have a flashlight and 8 batteries
7. If you break the batteries into 3 groups: Two groups of 3 and one group of 2. By doing this you guarantee that one of the groups has 2 working batteries. Both of the groups of 3 have 3 possible combinations of 2 batteries and the group of 2 only has 1 combination. So, 3 + 3 + 1 = 7 tries at most to find two working batteries.
Answer – 7
You are given a flash light which takes 2 good batteries to run, and 8 batteries, 4 good ones and 4 used up. What is the minimal number of trials needed to get the flashlight running?
Total batteries = 8.
Out of the 8 batteries, only 4 are in working condition.
We need 2 good batteries to run.
Lets try to divide the 8 batteries in a group of 3.
8 = 3 + 3 + 2.
Group 1 = 3 batteries,
Group 2 = 3 batteries,
Group 3 = 2 batteries.
Now you have made sure that one of the group has two working batteries.
How many combinations are possible for each group?
Group 1: B1 B2 B3.
Possible combinations → B1B2 , B2B3 or B1B3.
Total combinations = 3.
Group 2: B4 B5 B6.
Possible combinations → B4B5 , B5B6 or B4B6.
Total combinations = 3.
Group 3: B7 B8.
Possible combinations → B7B8.
Total combinations = 1.
Adding all the combinations = 3 (Group 1) + 3 (Group 2) + 1 (Group 3) =7.
Hence, minimum number of trails needed = 7.
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