Impossible Logic Puzzle from Indonesia!

This is one of the hardest logic puzzles I have ever come across.

Translated and slightly re-worded from a 2018 math competition in Indonesia by Gadjah Mada University. Question 15 in this pdf:
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[NathanSimonGottemer]

Three logicians walk into a bar. The barkeep asks if they all want beer. The first says, “I don’t know.”The second says, “Me neither.” The third says, “yes.”

[XkcdEsperite]

This can be solved even more simply:

If someone knows whether the numbers are being added or multiplied then they know the solution.

If your number is greater than half the total (1010) then you know for sure that the numbers are being added together, and would know the solution.

Alzim does not know the solution, so his number must be 1010 or less.

Era does not know the solution, so his number must also be 1010 or less. Because he still does not know the answer there must be a way add Alzim and Era's numbers together to reach 2020, and neither one is greater than 1010, this is only possible when Era's number is exactly 1010.

[bgbbft]

1) if Alzim's number isn't a divisor into 2020, then he would know it was a sum, and would know the answer.
Therefore his number must be 1, 2, 4, 5, 10, 20, 101, 202, 404, 505, , 1010, or 2020
2) If Era's number isn't a divisor into 2020, then he would know it was a sum, and would know the answer.
Therefore his number must be 1, 2, 4, 5, 10, 20, 101, 202, 404, 505, , 1010, or 2020
If his number is 1, 2, 4, 5, 10, 20, 101, 202, 404, 505, or 2020 he would know it was a product, and would know the answer.
However, if is number is 1010 it could still be either a sum or product. The answer could be 1010 or 2.
3) Since Era didn't know, Alzim now know's Era's number is 1010.

[deerho]

When we know that Alzim's number is a factor of 2020 at the beginning, we also rule out 2020 as the sum would make Era's number 0, which is impossible.

[WilliamWizer-xm]

you can even generalize it. let N being the number given by the host.
if the answers given by the players are NO, NO, YES then the second player's number will always be N/2.

why?
if Y>N/2 then X=N-Y. (as Y>N/2 is the same as 2Y>N and that prevents multiplication)
if Y<N/2 that implies X>N/2 so first player would had known Y=N-X (as X>N/2 is the same as 2X>N and that prevents multiplication)

in short. first player must be either N/2 or 2 and second player is always N/2.

[violetfactorial]

This one felt good to solve. There's a lot of information given by the fact that they can't immediately tell that it's a sum. The question is more like "can you tell if it was a sum or a product?"

Since the answer was 'no', it was obvious that their numbers must have the potential to both sum to 2020 AND to multiply to 2020. After some thinking, the only pair of factors of 2020 that could ever sum to 2020 are 1010 + 1010, so that must be the answer. It follows that Alzim's number is either 1010 or 2.

[defnotisaiah]

love your videos presh but ai art is meh

[DaveBermanKeys]

Let me just randomly pick some numbers to show you why they don't work. I don't know...how about 420 and 69?

[ARex]

I did it in a pretty similar fashion but arrived at the solution pretty quickly.

Here's my trick-> We need to ask ourselves what will create ambiguous situation even if we had all the information.

An ambiguous situation only arises when 2020 can be formed using both the multiplication and summation of two numbers. This allows us to come down to factors of 2020.

Now again going by the same logic as earlier, what can still cause us to be ambiguous even though we have the information? We can clearly see 2×1010 and 1010+1010 are the only numbers that would still create confusion while solving the puzzle for Alzim and Era.

The process mathematically is the same, its just that logically I find it easier in these types of question to create a forceful ambiguous situation even though we have all the information. Nothing groundbreaking just helps me solve more quickly!

[Indian_Ravioli]

This took me more time than i initially gave it credit for, because I went from thinking it was easy to discovering why it's not that easy, to finally finding the actual answer😮‍💨😌. I'm glad I follow your page.👏👏

[HeirToPendragon]

Quick solution for those that want it: Era's number must be a factor of 2020 such that
Era's number * Another factor or 2020 = 2020
Era's number + Another factor of 2020 = 2020
This is why Era doesn't know Alzim's number. Once Era doesn't know, Alzim just has to find the only numbers that could work, and in this case there is just one: 1010
Because 1010 * 2 = 2020
And 1010 + 1010 = 2020

[ctsmd]

I think you can definitively say that Eras number is also 1010 and here’s why I say that:

When Alzim is first asked what Era’s number is, he only knows that his own number is 1010 and that 2020 is either the sum or product. From that he can say that Era’s number is either 2 or 1010 but not which, so he says “no.”

Next Era is asked. If his number were 2, then he would realize that Alzim’s number could’ve been either 2018 or 1010, but if it had been 2018 then Alzim would have known that Era’s number had to be 2 and he would have answered “yes”. From that Era would know that Alzim’s number had to be 1010 and so Era would’ve said “yes.”

If Era’s number was 1010, though, he would know that Alzim either had 1010 or 2 but not which and so also said “no”.

Therefore the fact that Era didn’t know Alzim’s number tells Alzim that Era’s number had to be 1010.

[dustinbachstein]

Nice puzzle! I found it rather easy, though. A few notes:
1. The same puzzle works for all even numbers instead of 2020. Listing up all factors of 2020 is also unnecessary; instead, it's enough to solve the equation 2020/a + 2020/b = 2020, where the 2020 cancels (that's why every even number works) and a=b=2 is the only solution.
2. The information that Alzim says "yes" in the end is irrelevant because he couldn't have given a different answer anyway.
3. Also, I don't think the extra condition that the chosen numbers must be positive is necessary. The puzzle should work the same way if any whole numbers were allowed.

[QuantumOverlord]

I haven't looked at the solution yet. Here is my solution:


Alzim does not first respond by saying he knows. This rules out him having any number not in the set F (where F is any factor of 2020) because if Alzim's number is not in F then Era's number must be 2000-Alzim's number. So we know from the first 'no' that Alzim has picked a number in F.

Era concludes the above so knows that Alzim has picked a number in F. Therefore we can conclude Era's number must also be in F (if the card is the product) or in F' which is defined as the set of 2020- each element in F.

If Era's number is exclusively in F then Era would answer 'yes' since he could determine Alzim's number is 2020/Era's number. If Era's number is exclusively in F' then by similar reasoning he can answer 'yes' and determine Alzim's number is in 2020- Era's number.

Given Era does not answer 'yes' we know that his number cannot be exclusively in F or F'. But since Alzim's number is in F it can't be in neither either. Therefore the only other option is that it must be in *both* F and F'. The only factor of 2020 that is in F' is 1010 since 1010*2=2020=1010+1010. Therefore Era's number is 1010

Alzim also concludes the above. However there is not enough information to determine whether Alzim has picked 2 or 1010. Alzim's number could be either of those options.

[zermar]

What I find the most interesting in this problem is the insight in hindsight:
As soon as the 2020 card was shown, Era (having picked 1010) immediately knew that Alzim had to have either 2 or 1010 and Era learned absolutely nothing else about Alzim's number throughout all the rounds of questions!
Alzim, having picked 2 or 1010 (we still don't know which), would have also known from the getgo that Era could have one of 2 numbers.
If Alzim has 2, then Era has 1010 or 2018.
If Alzim has 1010, then Era has 2 or 1010.
Here is an alternate reality scenario with an ever so tiny change, that ends up in a very different result:

What if mufti-saab questioned Era first instead?
If Era was asked first if he knew what Alzim's number is, he would say "no", and if Alzim had picked 2, Alzim would immediately know that Era could not have 2018, otherwise Era would have immediately known what Alzim had, and conclude that Era has 1010.
If Alzim picked 1010, Era not knowing what Alzim had would not have sufficiently narrowed it down for Alzim, and Alzim would say "no".
Ironically, this difference in outcome would actually expose to Era what number Alzim had! Alzim would say "yes" if he had 2, or say "no" if he had 1010.
So, if Alzim picked 2, Alzim's declaration that Era picked 1010 would tell Era that Alzim picked 2, thus both of them knowing each other's number in the end!
And if Alzim picked 1010, Alzim's not knowing what Era has would give away that Alzim has 1010. Era then declaring that he knows what Alzim has would confirm to Alzim that Era also has 1010, and this is because if Era had 2, Alzim knows that ERA KNOWS that Alzim could not have had 2018 without knowing what Era has.

Isn't it fascinating that asking fewer questions, but changing the order could have actually revealed MORE information to all parties?
Alzim and Era strictly know more than outside observers like us do. Would there be a scenario in which they both know each other's number, but we don't know one or both? I don't think so?

Well, thank you so much for reading!
I'm actually not 100% confident that my reasoning is actually logical, because, man, I didn't get enough sleep last night! If you somehow managed to parse through all the Alzims and Eras in that, I congratulate you.
I'd love to see someone call me out on being wrong, because that showed they care enough about my rant.

[averagejuveenjoyer]

With person's number n other's number can be only 2020-n or 2020/n.
If a person doesn't know, then both variants are possible, so his number is a divisor of 2020. Both dint know, so both numbers are divisors of 2020.
The only way sum of 2 divisors of 2020 can be 2020 is 1010+1010 (since at least one of them must be not less than 2020/2: 1010 or 2020). This variant fits
Let's consider that 2020 is a product. The second one knows that both numbers are divisors. If his number isn't 1010 then sum isn't possible -> the second knows both numbers at his turn - contradiction.
Era's number is 1010

[janjanssen]

I think the riddle works also if you don’t specify “greater than zero “ in the beginning.

[JohnRandomness]

Once Alzim answers "No", Era knows that Alzim's number is a factor of 2020. If Era's number is greater than half of 2020, and not 2020, 2020 must be the sum -- he figures it out. If Era's number is less than half of 2020 or is 2020 itself, 2020 must be the product -- again, he figures it out. The only ambiguous situation is when Era's number is 1010 -- Alzim's number is either 2 or 1010.

This reminds me of Mark Goodliffe solving a "Genuinely Approachable Sudoku". He kept breaking the puzzle on applying a rule leading to 5 + 7 = 12. The were other rules, and Mark finally broke through saying out loud, that we couldn't have 5 + 5 = 10 -- oh, but we could have 5 * 2 = 10! That was after several trials and crashes, and he got a well-earned dinosaur.

[johnschmidt]

I think you ended up doing it the long way. The key insight for me was if either person's number was over 1010, they would know that it has to be addition, as the product would be above 2020 (or * 1, another easy answer).

But once Era knew that Azeem number was under 1011 he should have been able to divide 2020 by his number to get Azim's number (knowing the result was a product the only way two numbers under 1011 to make 2020), unless his number was exactly 1010, which is the answer!

[taflo]

After watching only the beginning of the video in order to be sure what "number" is supposed to mean in this context:

This works with any even positive integer N (apart from 2) instead of 2020. By saying "No", you reveal that N is both a multiple of your number (vice versa, your number is a factor of N) and a sum of your number and a positive integer (meaning that your number is between 1 and N-1). So when the first person says "No", we know that their number X is a factor of N (but not N itself).
Before the second person answers, they already have this information about X, so when they say "No", we can not only deduce that their number Y is a factor of N (but not N itself), but also that both N/Y and N-Y are factors of N, since those are - from the pov of the second person - the two possible values of X. We can ignore N/Y, since that is a factor of N as soon as Y is. But if Y were strictly smaller than N/2 (but still positive), then N-Y would be strictly larger than N/2 (but still smaller than N) and thus not a factor of N, iplying that the second person's answer would have been "Yes". So we know that Y=N/2. Since N is the sum or the product of X and Y, we have X=N/2 or X=2.

Remark 1: When we look at N/Y for the second person, we immediately see that they couldn't have chosen Y=1, as that would imply that the other person's number was N, which we already excluded. This is why the argument fails for N=2. If you allow X, Y to be 0, then it also works for N=2. You just have to observe that if one of them chose 0, they would immediately have known that the other one chose N, as N is not zero and thus not a multiple of 0.

Remark 2: "Number" could in fact just refer to "integer" (i.e. including negative values) and the argument would still work with a few slight changes. Indeed, if the first person answers "No", then X has to be any factor (positive or negative) of N. So if Y were a negative number, the second person would immediately know that N can't be X+Y, since otherwise |X|>N and the first person would have answered "Yes".

Remark 3: If "number" were just to refer to rational or real numbers, the whole argument would break down and the only things we could deduce is that neither of them chose 0 as their number and that Y is not N (as otherwise the second person would have known that N = X*Y, since they would have already known that X is not 0).