Where is that hen?

Note that if there is a lighter egg among $3$ eggs then we need only $1$ step to find out the lighter egg : we simply put any $2$ of those $3$ eggs on balance. If they balance perfectly then third egg is lighter, otherwise one which goes up is lighter. Now consider the case of $9$ or less number of eggs. In this case, lighter egg could be found in $2$ steps only. To do this, we keep $3 + 3$ eggs on balance and repeat the procedure given above. Given this, we can find lighter egg in $3$ steps if there are $27$ of them by putting $9+9$ on balance. Finally, for $80$ eggs, we put $27 + 27$ on balance. If they don't balance then pick those $27$ which contain lighter egg. Then locating that egg would take $3$ steps as explained above. If they do balance, then take remaining $26$ eggs and divide them as $9 + 9$ on balance and then its easy to see that this takes only $3$ steps. Hence in total we need $4$ steps to locate the lighter egg.

In general it takes $r$ steps if the number of eggs is less than or equal to $3^{r}$ . This I discovered in collaboration with @Abhijit Bendre .

Immediately rule out any answer choice that's 3 and below. I mean, yo, the table is like filled with eggs and stuff, 80 of them, and dang that's alot, and you be like, "You want ME to figure out which egg it is with THREE weightings? Aw hail nawww!" So yah, just imagine it for a sec, it AIN'T gonna happen.

So now we're down with 4 and 5. We either have to prove that A) 4 is not enough, or B) it can be done with 4.

But before we approach this problem, we need a strategy. Brute forcin is no man's land and so there's gotta be some nice logical concept of wonderfulness.

But yo, figuring out some awesome concept with a gazilllion eggs ain't gonna cut it. We gotta simplify things with lower numbers.

A princess, 17 rooms and only 30 days?? Simplify! 4 rooms and an infinite number of days and work your way up!

1000 bottles of wine? Can we just like, start with 3 bottles or somethin and see what we can do?

But sometimes, being simple doesn't cut it. 2 Levers and 50 prisoners? Not gonna do anythin! A chessboard, your friend and 64 randomly flipped coins? Eek.

But this balancin puzzle seems simple enough to use simplicity yo! Let's say there was like 1 egg, how would you-

Kay being too simply is bad. How about 2? Well duh, compare them once and dun...

How about 3? Well you have 3 eggs A, B, and C... Well, if you weigh like for example, AB against C, then that's completely useless! AB is of course greater! That lighter oddball egg must at least have SOME mass! If k > 0 then x+k > x!

...unless it has negative mass or something. Bleh.

So let's be a bit rational and weigh like, A with B. There are 3 things that can happen.

1. A > B, therefore B is odd.
2. A < B, therefore A is odd.
3. A = B, since A !< B and B !< A, therefore neither A or B is odd. Therefore C is odd.

So if there are 3 eggs you can win with 1 weighting.

You can totally just solve the puzzle by going up 1 egg each time like this. As soon as you reach 9 eggs and solve, you basically figure out the pattern. But,,, what if A, B, and C were groups of eggs? That means... every time you weigh, you eliminate like an ENTIRE TWO THIRDS of those eggs as possibilities!

So let's say you have like, 9 eggs. Separate into groups of 3 eggs A, B, and C. (3 * 3 = 9, fyi) Weigh A and B. Let's say like, B was less than A. The oddball egg can't be in A or C! It gotta be in B.

So group B has 3 eggs, D, E, and F. And well, we know this can be done in 1 weighing.

So you can kinda notice some powers of 3 stuff going on. 1 eggs --> 0 weightings, 3 eggs --> 1, 9 eggs --> 2, 27 eggs --> 3, 81 eggs --> 4...

Hey wait, weren't there only 80 eggs?

80 isn't a power of 3, so we can't cut it into thirds. However, we can PRETEND that there is another egg. We split into 27, 27, 26 and weigh the 27s. If they are equal, borrow a normal egg from the first 2 groups and stuff it into the group with the abnormal egg! Now instead of some oddball number of eggs like 26, you have 27! Therefore, 80 eggs can be done in 4 weightings.

Here's my solution. 1)First take any 40 eggs and weigh them 20-20. If Isn't balanced then take the lighter 20s. 2)now take any 5-5 eggs of them and weigh them. If isn't balanced then taje the lighter 5s. 3)now weigh the 2-2 of the lighter 5s. If balanced, then the answer is the remaining egg. If isn't balanced weigh the lighter 2s. 4) after weighing 2s the lighter one is the answer.

Since we are lacking a practical one,

step (i)

We take 27 eggs each on the scales and weigh them.

If it doesn't balance, we have the lighter egg on that side, so we take those 27 eggs to step (ii)

If both sides balance, we take the remaining 26 eggs to step (ii)

step(ii)

We take 9 eggs each on the balance.

If it doesn't balance, we have the lighter egg on that side, so we take those 9 eggs to step (iii)

If both sides balance, we take the remaining 8 (i.e. 26 - 18) or 9 (i.e. 27 - 18) eggs as the case may be to step (iii)

step (iii)

We take 3 eggs each on the balance.

If it doesn't balance, we have the lighter egg on that side, so we take those 3 eggs to step (iv)

If both sides balance, we take the remaining 3 or 2 or 1 eggs as the case may be to step (iv)

step (iv)

We take 1 eggs each on the balance.

We find the stupid egg. :-)

There is a flaw in the way the question is phrased, technically the minimum number of steps needed to identify the egg is one: Lucky pick of two eggs where one is the lighter one.

It should have been phrased as "given the optimal strategy what is the maximum number of steps needed to identify the egg."

Split into 40 and 40 then find side with lighter egg then split to 20 and 20 then find lighter side then split to 10 and 10 lighter side again. But this time split into 4 and 4 to avoid odd number if stable split the other 2 to 1 and 1 you will find the answer worst case senario it is step 4 is not stable then split into 2 and 2 find lighter egg then measure.

I simply categorized the 80 eggs into 2 groups then divide the one that was heavier and divided 3 more times to pin point which was heavier.

4 is possible as many of the solutions here explain how but in order to prove that this is the minimum we have to prove that 3 is not enough. Can anyone prove that?