Anybody have any commentary on this? I'm confused because I don't know what those "places"/"developments" even are; have never seen anything quite like that

I can tell you it looks ugly. Is that what people dislike?

Life is good

Made Mr Poor's barbecue steak pizza

This question comes from the 27th Brazilian Mathematical Olympiad (2005). I, sadly, got this one wrong. Good luck.

We have four charged batteries, four uncharged batteries, and a radio which needs two charged batteries to work. We do not know which batteries are charged and which ones are uncharged, and we need to turn the radio on. What is the least number of attempts that suffices to guarantee that the radio will work even if we are maximally unlucky? (An attempt consists of putting two batteries in the radio and checking if the radio works or not.)

Congrats to the 3% of you that have gotten it right

Well, I got this one wrong, so explaining my thought process is a bit weird. But let's do it anyways, then I'll talk about what I missed.

I'll refer to tries where the radio doesn't work as "bad." When it does turn on, that's "good."

So, I wanted a starting point. The first thing to do for this type of problem is to find a method that works, even if it's not the best way (note: I didn't have access to the multiple choice answers, so setting myself a high ball helps even more). I noticed that you could fully test the first battery by pairing it with the other 7; if none of the combos work, that first battery must be uncharged (7 tries). We can use this same method to diagnose the batteries as uncharged one at a time. Next, test the second battery, except there's one less to test it with (6 tries). With two eliminated and 6 left, we test the third with the other 5 (5 tries) and then test one of the remaining batteries with the other 4 (4 tries). That would tell us which 4 are uncharged, by figuring it out one at a time, so we just need to put two of the remaining in the radio and it will work (1 try). So, 7 + 6 + 5 + 4 + 1 for a total of 23 tries.

23 is thereby a very basic answer, but I thought (correctly) that the best strategy would produce a lower result.

I recognized a pattern: if the first battery is bad, you actually don't need 7 tests to infer it. There are only 4 uncharged batteries. So, if my first battery is charged, it's possible that the first 4 tries could be bad results if the 4 after it were all uncharged. However, the 5th try being bad tells us that the first battery must be one of the ones without a charge, since a charged battery only has 4 uncharged batteries to be paired up with. Therefore, if we have n uncharged batteries, the amount of tries we need to show a certain battery is uncharged is n + 1. So, after we figure out the first is uncharged (5 tries), there are 3 uncharged batteries left, and so we need 4 tries to figure out the second battery is uncharged, then 3 tries, then 2 tries, and then we need that last try to put in 2 of the batteries we know are charged. 5 + 4 + 3 + 2 + 1 = 15. Progress!

But that's the best we can do with eliminating them one at a time. (At least I'm pretty sure, let me know if the one at a time method has a better way that I missed.)

"What would happen if I grouped them?"

If we split them into two groups, {1,2,3,4} and {5,6,7,8}, what would happen? Let's first try all the combinations in the first group. 1 and 2, 1 and 3, 1 and 4, 2 and 3, 2 and 4, and 3 and 4. That's 6 tries. What does it tell us? That the first group must have either 3 or 4 uncharged batteries. If it's 4, then any two from the second group work. If it's 3, then there's only one uncharged battery left, and, therefore, if in the other group we pair together 5 with 6 and 7 with 8 (2 tries), it's impossible that both pairs have an uncharged battery since there's only one left. 6 + 2 = 8 tries. More progress!

"What if we split them into pairs?"

{1,2}, {3,4}, {5,6}, {7,8}, for 4 total tries. What do we learn? Each pair must have one uncharged battery and one charged battery, because that's the only way all 4 tries would be bad. So, any two pairs will therefore have a working combination. If we take {1,2} and {3,4}, we must have 2 charged batteries in there. So, we make the pairs of 1 and 3, 1 and 4, 2 and 3, and 2 and 4 to guarantee a good try. That's 4 more for a total of 4 + 4 = 8 tries. The same as our previous attempt!
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Given this congruence and the lack of new ideas coming to me, I had settled on this as the best solution. However, I missed a better way to split them because I had not considered making uneven splits.

It turns out, the key is to split them into three groups; for example: {1,2,3}, {4,5,6}, {7,8}. This is the most efficient way. If 1 and 2, 1 and 3, and 2 and 3 are bad combos, this implies that 2 of them must be uncharged (3 tries). Ditto for the second group (3 tries).

With all 4 uncharged batteries accounted for, 7 and 8 (1 try) will turn the radio on and result in a good try. 3 + 3 + 1 = 7

So, the 3-3-2 split also works, and it seems to be the best answer. Very neat problem

@ArthurSchopenhauer1788 what have you done to me… I’m about to walk away from a career path because I realized I was chasing money and approval from others.

I’m glad you learned the hard lesson that the only person you need to seek approval from is Arthur Shopenjuper

Thales, so the story goes, because of his poverty was taunted with the uselessness of philosophy; but from his knowledge of astronomy he had observed while it was still winter that there was going to be a large crop of olives, so he raised a small sum of money and paid round deposits for the whole of the olive-presses in Miletus and Chios, which he hired at a low rent as nobody was running him up; and when the season arrived, there was a sudden demand for a number of presses at the same time, and by letting them out on what terms he liked he realized a large sum of money, so proving that it is easy for philosophers to be rich if they choose, but this is not what they care about

- Aristotle, Politics 1258