Finding the Largest Number Using 2,2,2,2

If you need to post a picture of your solution or idea:
#ChallengingMathProblems #Puzzles

PLAYLISTS 🎵 :

[angelmendez-rivera]

This problem is best solved by solving it for the general case. If the number of 2s available is 1, then the largest quantity is simply 2 itself. If the number is 2, then there are only two options: 2^2, and 22, and the largest is just 22. If the number is 3, there are multiple arrangements to consider, but since 22 is by far the largest arrangement when the number is 2, we only care about the arrangements that involve 22 in some form or other. Ergo, 2^22, or 22^2. Of these, 2^22 is naturally the biggest. So if the number is 4, we care about arrangements involving 2^22. The only possible arrangements are (2^22)^2 or 2^(2^22). 2^(2^22) is the largest. In the case that the number is k + 1, you take the previous largest quantity q in stage k, and there are only possibly two options: q^2, or 2^q, and 2^q > k^2 for all q > 4, and q > 4 by the time k > 2.

If we were to define a function f : N\{0} —> N\{0} such that f(k) is the largest possible quantity in stage k, then clearly, f(1) = 2, f(2) = 22, and f(k + 1) = 2^f(k) for all k > 2.

To an extent, you can even be generalize this logic if you are using other binary operators that are monotonically increasing in their inputs, instead of exponentiation. Suppose d is the repeating digit in some arbitrary base b. Here, dd stands for d·b + d = d·(b + 1), ddd stands for d·(b^2 + b + 1), and so on. Consider some arbitrary binary operator # that is monotonically increasing in all its inputs. Let f(k, d, b, #) denote the largest quantity you can form using k repetitions of d in base using only the explicit operator #. Then, obviously, f(1, d, b, #) = d, regardless of what b and # are. For k = 2, there are clearly only two possibilities: dd, or d#d. This is important, because if, for example, d#d than dd, then we already know immediately that (d#d)#d > (dd)#d and d#(d#d) > d#(dd). So then, for k = 3, we only care about operations involving d#d. But clearly, that again only leaves two options: d#(d#d) and (d#d)#d. And whichever one is bigger, say, for example, d#(d#d), it means that d#(d#(d#d)) > d#((d#d)#d) immediately, as well as (d#(d#d))#d > ((d#d)#d)#d. So again, for k = 4, you only care about operations involving d#(d#d), the largest quantity for k = 3. But this only leaves two options, yet again: d#(d#(d#d)) or (d#(d#d))#d. And this goes on forever. If you know what f(k, d, b, #) is, then necessarily, for k > 2, f(k + 1, d, b, #) = f(k, d, b, #)#d or f(k + 1, d, b, #) = d#f(k, d, b, #). That is it. Those are the only two possibilities. It can never be otherwise. So rather than considering all the possible arrangements with k + 1 repetitions, which is futile and intractable, you only need to compare f(k, d, b, #)#d and d#f(k, d, b, #) to figure out the largest possible quantity f(k + 1, d, b, #). This simplifies everything. There is only value it could possibly be for k = 1, only two values it could possibly be for k = 2 (d#d or dd), and every for k > 2, there are only two possibilities based on the previous k.

If # is not monotonic in both of its inputs (or antitonic), then you do actually have to consider all possible arrangements for every k. But exponentiation is monotonic on both inputs (when it comes to positive integers, anyway), so there is no need to do this.

[nayutaito]

I like how you explicitly said "exponential" to prevent googologists from using tetration.

[TheEternalVortex]

It’s pretty easy to solve this in general since for n>2 2s it’s always better to exponentiate with another 2 as the base than any other action. For n=2 since 22>2^2 that’s the best. Thus we get 2^2^2^…^22 as the general solution.

[Ntivat]

At some point the "2" starts looking like an "a"

[MichaelRothwell]

Nice problem and solution.
One way to guess the answer is to solve the problem for three 2s - getting 2²² - then for four 2s raise 2 to that power. The proof is, of course, best done as in the video, "converting" to powers of 2 with appropriate inequalities.
Now, if we could only use Knuth's up-arrow or Conway chained arrow notation...

[owlsmath]

Great video! It's like a "which is bigger" problem but with 8 choices.

[LadyViolet]

I thought this was going to be a different question before I opened the video. Originally I thought it'd be using 2, 2, 2, 2 with any notation that you can think of to make it larger, but I guess finding the largest exponential number would be much more reasonable.

[italixgaming]

My solution (of course way faster). Let's split our candidates into four categories: 2222, 222^something, 22^something and 2^sometthing and let's find the "champion" of each family.
2222 is alone is his family, it's below 2^11.
222^2 is also alone in his family and it is below 256²=2^16.
We have two members of the 22^something family: 22^22 and 22^(2^2) and the first one is obviously the greatest. And we can see that it beats the two other champions because 22^22>2^22>2^16.
So now we have the family of 2^something. How to find the greatest number of this family? The problem is the same that finding the greatest possible "something". So we have the same problem but with only three "2" used.

Again we have three families: 222, 22^something and 2^something. 222 is alone, 22^2 is also alone and obvisously greater than the first one but below 32²=2^10. So our local champion is beaten by 2^22, who also beats the other member of the last family, 2^(2^2)=2^4. So the greatest possible "something" in our original problem is 2^22.

Now we have the final of our global championship: 22^22 vs. 2^(2^22).
22^22<32^22=2^110 so 2^(2^22) is obviously greater.

[perplexedon]

If you allow logs and sqrts:

= 2^n, where n is the number of square roots. It can therefore grow as large as you like, though it does feel like you are sneaking a 0.5 in for every square root. It does technical find the largest exponential number though, as taking roots is exponentiation

Taking a root is technically a binary operation 2^0.5, so this solution amounts to defining 2# = 2+1 and then saying 2##...# can be arbitrarily large, or by saying 2!!!...! Can be arbitrarily large.

[iamjimgroth]

Recreational maths... It sounds insane... But yea, here I am, enjoying it. 😁

[snejpu]

You might have uesd factorials. 2222! has to be massive. : )

[purbitasaha]

You've got 94K subs now! I've been watching you from 9.4k lol

[Qermaq]

Can we use Graham's arrow notation? ;) If not, I suggest 2^2^22.

[GameJam]

I mean, there's really no limit to the size of the number you can make when you consider how repeated operations work. Addition is simply repeated incrementation, multiplication repeated addition, exponentation repeated multiplication, and tetration repeated exponentation. Who says this can't keep going? Who says we can't make repeated tetration? It's not like tetration is a particularly useful operation, so why not keep going and make even less useful operations? I see no reason we can't.

[_Elliot]

A really fun trick about dealing with big power of two, is that every 10th power of two is (approximately) 3 zeros added.
i.e. 2^10 ≈ 1, 000, 2^20 ≈ 1, 000, 000, 2^30 ≈ 1, 000, 000, 000 etc.
There for, 2^22 ≈ 1, 000, 000 * 4 ≈ 4, 000, 000.
Great video!

[shophaune]

Without using factorials, hyperoperations or logarithms:

Let f(n) = 2^n, and f^x(n) = f(f(f(...(n)...))) with x applications of f.

Then, I believe f^22(2) is the largest number possible without using more than four twos in the definition of the number or any non-exponential operations, with a value of 2^^23

[jasonzervos]

I was so focused and didn't realized that 9 minutes had past. I felt like 3!

[davidseed]

sometimes a simpler prroblem givrs sn insight.
so with three 2s. we quickly see that the largest number is 2^22.
now look to see the best place to add a fourth 2.

22^22, 2^2^22 , 2^222, 2^22^2

[brickviking]

It took me about 90 seconds of testing to see the answer - most of the rest can be handled by FX-82 and the like, but for the last one, I needed bc (a program with Unix heritage) to cough up that many digits.

[MudakTheMultiplier]

Ooh, this is really similar with a sample file/puzzle that came with my TI-Nspire calculator. The goal was to use only the digit 2 four times in order to make every number up to 100 (I don't remember if 0 was included) I think I got through about 70 of them through my time in high school.