• Member Since 28th May, 2016
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A collection of short-story logical-deduction puzzles, where the reader is encouraged to find the solution to a given scenario.

Those interested in posting their solutions should use black-out text when commenting.

Currently, only a single puzzle is provided; over time, additional puzzles will be added to this collection.

(Formerly titled "Pony Prediction Puzzles")

Chapters (12)
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Comments ( 58 )

I LOVE these kind of puzzles. I drew out a matrix to solve it, and this is what I got.

It must be Yona and Smolder being correct. I knew Yona had to be pinkie, otherwise she would be Twi, which is a contradiction with the rules. So that means one of the others has to be Twi. Next, Smolder or Gallus has to be correct, or it's a contradiction. (If Smolder is right, Gallus is wrong and vice versa.) It's easier to check if Smolder is right, cuz she has to be Twi. After that, go down the list for the others. Oscelus can't be with Rainbow. Silverstream needs to be with Rarity, and then Sandbar doesn't get Rainbow or Twilight. That leaves Gallus with Rainbow, Sandbar with AJ, and Oscelus with Fluttershy.

Did I miss anything?

I think I got it. This is a fun idea.

Smolder: Twilight
Yona: Pinkie
Ocellus: Fluttershy
Sandbar: Applejack
Gallus: Rainbow Dash
Silverstream: Rarity

Ok so we have these predictions
Yona: she wont get twilight
Sandbar: Ocellus gets dash
Smolder: she gets twilight or dash
Gallus: smolder is wrong
Ocellus: Silverstream gets anyone BUT rarity
Silverstream: she gets dash or flutters
Afterwords we have twilight say the 2 who guessed right get her and pinkie
So lets start deducting
gallus and smolder must be 1 right 1 wrong as they directly contradict
If yona is wrong she will get twilight,but since the 2 right creatures get twi and pinkie, that wont work, so yona is right and gets pinkie
Since I've narrowed down the 2 correct predictions that make all other certainly wrong
So ocellus gets AJ
Sliver gets rarity
If sandbar gets dash, smolder and gallus can have twi and shy in some way
If sandbar gets shy, smolder will get twilight and gallus gets dash

So in conclusion yona gets pinkie
Ocellus gets AJ
Silverstream gets rarity
sandbar get dash
And both smolder and gallus get twi and shy in some way

yona gets pinkie
Ocellus gets AJ
Silverstream gets rarity
Sandbar gets Fluttershy
Smolder gets twilight and gallus gets dash

I tried, but I think I must have missed something.

If Yona is wrong, then she will get Twilight, but since Twilight goes with a pony who is right, Yona cannot get her and also be wrong. Therefore, Yona must be right, and since she cannot be with Twilight and be right at the same time, she must be with Pinkie Pie. After that, either Gallus or Smolder must be Twilight--since they have a contradiction, one of them must be right, and the other must be wrong.

But Twilight said that we had enough information to figure out exactly what student goes with which professor, so we've got a while to go still. We now know that Ocellus, Silverstream, and Sandbar are all wrong, which means that Silverstream is Rarity and Ocellus is not Rainbow Dash. But from there... there's nothing. Even trying to assume that either Smolder or Gallus is correct and move from there gets me nowhere. The closest I got to was Gallus being with Twilight, which allowed me to find the conclusion that Sandbar was Rainbow Dash, but Smolder and Ocellus were tied between Applejack and Fluttershy, with no clues either way. If Smolder was correct, I didn't even get that far. Twilight said we have all the information we needed, but I'm not sure we do. I'll check the other comments to see what I'm missing, but I'll go ahead and post this first, so that I'm not stealing their answers.

You're very close to a complete solution. I will post the next chapter (i.e., the solution to this first problem) tomorrow.

I liked that the solution was written as its own little story, as a conversation.

I’m not sure if there’s enough information to solve this one. Especially since determining whether or not one of the predictions is true or not relies on what one of the other scores is which seems to require more information than you’ve given us. There’s also the fact that we don’t have any information that can be used to determine Sandbar, Ocellus, and Silverstream’s scores.

Well i dont have an answer but i found a few facts
yona must be false and thus 95 or 99
Smolder must be true thus 96 or 98
If Ocellus is top 3 Silverstream is even, if not Silverstream is odd
If Gallus beats yona, yona is 95 and gallus is 96 or 98, if yona beats gallus, yona is 99 and gallus is 95 or 97

Witch with the fact that the 96 will solve it means
smolder is known no matter who it is
if its gallus, yona is known
If its Ocellus, Silverstream must be odd
If its Silverstream, ocellus is top 3
It CANT be yona
If its sandbar we get nothing

So either i missed something or its impossible

Known Assumptions (Unknown if true):
Yona: 97 (Predicted by Yona)
Gallus: better than Yona (Predicted by Gallus)
Smolder: NOT 94 (Predicted by Smolder)
Ocellus: 97 OR 98 OR 99 (Predicted by Silverstream)

Predictions may not be right.
Anyone who made a true prediction had an Even score
Anyone who made a false prediction had an Odd score

Conclusions drawn so far: Yona CANNOT be true, as she would have to be 97, then. Which is Odd. She, therefore, must be 99 OR 95.

Acting under the assumption that Gallus. Smolder, AND Silverstream are correct AND Oscellus is 99.
Yona is 95
Gallus must be an Even number above 95 (96 OR 98)
Smolder CANNOT be 94, but IS an Even number (96 OR 98)
Silverstream MUST be 94 for Smolder and Gallus to be correct.

Solution if this is the case:
Silverstream: 94
Yona: 95
Gallus: 96 OR 98
Sandbar: 97
Smolder: 98 OR 96
Oscellus: 99

Acting under the assumption that Gallus. Smolder, AND Silverstream are correct AND Oscellus is 98.
Yona is 95
Gallus must be an Even number above 95 (96 OR 98)
Smolder CANNOT be 94, but IS an Even number (96 OR 98)
Silverstream MUST be 94 for Smolder and Gallus to be correct.


Acting under the assumption that Gallus. Smolder, AND Silverstream are correct AND Oscellus is 97.
Yona is 95
Gallus must be an Even number above 95 (96 OR 98)
Smolder CANNOT be 94, but IS an Even number (96 OR 98)
Silverstream MUST be 94 for Smolder and Gallus to be correct.

Silverstream: 94
Yona: 95
Gallus: 96 OR 98
Sandbar: 99
Smolder: 98 OR 96
Oscellus: 97

Let me know if I missed something!

While everything seems right i must add
if smolder is wrong, she gets 94,but wrong means odd so smolder is right (either 96 or 98)
none of these solutions distinguish who of the 2 is the 96&98, and twilight said if we know the 96, we can solve it(no word on who it is) since we dont know we cant assume either one
This means smolder is definitely right and either Silverstream or gallus is wrong

We know that Gallus or Smolder would be either of the two. So, if Twi told us which of them got 96, we know what everyone got. Once we knew everyone's score BUT the 96 or 98, we have solved it as much as possible.

Did you account for Silverstream being wrong?

I think I got it? Had to guess and check a few times, but this answer seems to work, and if there is only one solution, then hopefully I got it.

Silverstream: 96
Smolder: 98
Gallus: 95
Yona: 99
Sandbar: 94
Ocellus: 97

I hope to post the next chapter (containing the solution to this problem) tomorrow morning, and I need to double-check my work, but I believe this is the solution I have in mind.

Yeah you really didn’t do a good job with this problem. Like I said before, you didn’t give us enough information to solve this with and most of it relied on guesswork as a result. My advice would be to try to avoid that in the future.

Dang it! :derpyderp2: I went a completely different direction. I had:

Well, there was enough information to solve it, you just had to guess who got the the 96 and check if the assumption had a definite solution. I liked this puzzle, myself.

I said it involved guesswork which is something that, as far as I know, shouldn’t be involved in a logic puzzle..

Sorry … shortly after posting, I realized I erred. I need to retool some parts, and will repost shortly.

UPDATE: fixed the issue; now there should be a unique solution.

That was my original intent … until I realized there was a 2nd solution. Sorry for the confusion.

Really? Because by my figures, I still managed to get only one answer. And I'm having a hard time finding where any others might be.

I'm fairly confident that what I now have is correct, and I intend to post as much tomorrow afternoon.

Ok so lets list the numbers that are if only 1 correct guess from yona
perfect square: 100,121,144,169,196,225,256,289,324,361,400,441,484,529,576,625,676,729,784,841,900,961
Perfect cube: 125, 216, 343, 512, 729
The others are too long to list but you get the point

Now for if 2 were correct
square and cube:729
Square and 1st&last digit: 121,484,676
Square and above 200: see above
Cube and 1st&last digit:343
Cube and above 200: see above
1st&last digit and above 200:too long to list

Now for 1 wrong(or other 3 right) its just not numbers that fall into the first spoiler groups
with gallus saying its over 600, that means yona thought it wasnt over 200, so its either squares and 1st&last digit or cubes and 1st&last digit, witch leads to either: 121,484,676 OR just 343, since yona didnt it then she must have thought it WAS a square and not a cube, and below 200 means she thought it was 121
Now for Ocellus
the number is not square, is cube, 1st&last digit are different and its above 200, so we have 216, and 512 witch is what gallus had when he said 2 options, problem is neither of these is above 600 you still have an error

Your lists with only one correct guess are too long, if I'm reading your work correctly.

I have double-checked my work on this one, and remain confident that there IS a unique solution (after changing the 500 to 600, that is). I plan on posting once I get back from work later today.

Much appreciation from all who are taking an interest in this one (as well as the ones before).

Your final comment illustrates what was one of the solutions that would have worked … but there IS another which still works after changing 500 to 600.

I believe I have the answer.

Yona: 125
Gallus' 2 numbers: 484, 676
Ocellus' number: 676

Gallus' answers (Yona's are opposite):
Square? Yes.
Cube? No.
Palindrome? Yes.
Over 200? Yes.

125 is a cube under 200, is not a palindrome or a square.
484 and 676 are the only 3 digit palindrome squares above 200, and not cubes. 676 is the only one greater than 600.

Yay, I got it. That was a fun one.

Ok lets do this
right away i see smolder statement to yona is a paradox, so she must be an error
And since both gallus and yona say they are opposite things to ocellus, one of them is another error
So now that ive narrowed down the errors i can solve this
Sandbar says smolder is a squirrel, if thats true then they are both squirrels, if its false than smolder is a chipmunk and sandbar is a squirrel,so sandbar MUST be squirrel
Similarly, Silverstream must chipmunk due to her first Statement to gallus
Her second statement however must be false as its to sandbar, so yona isnt squirrel and smolder isnt chipmunk
So sandbar & smolder are squirrel and yona & Silverstream are chipmunk
Now Ocellus is squirrel due to yona's statement so gallus must be chipmunk

Think I got it. Not sure if it was intended to be a guess and check, but that's how I ended up solving it. At least, I think I solved it. Another fun puzzle here.

Gallus: Squirrel
Ocellus: Chipmunk
Yona: Chipmunk
Smolder: Squirrel
Sandbar: Squirrel
Silverstream: Chipmunk

Gallus correctly lies to Ocellus.
Yona incorrectly lies to Ocellus (should have said she is a chipmunk).
Smolder incorrectly tells truth to Yona (should have lied and said we are the same).
Sandbar correctly tells the truth to Smolder.
Silverstream correctly lies to both Gallus and Sandbar.

Found the solution. Same method and answer as AstralMouse
(Yeah I read the comments after solving the puzzle and before putting my answer in my own comments)

Comment posted by Coyotek4 deleted January 12th

I'm getting a little burned out about doing these, but I don't want to disappoint if there's enough demand. My plan going forward is to have the number of puzzles I create here match the number of positive thumbs. As of now, that's at equilibrium … but if the mood strikes me, I might do another sooner rather than later.

Regardless, thx to all who are actually solving these as I create them; that part makes the endeavor worthwhile to me.

Yeah, don't burn yourself out on this. I like them, but take a break if you need. It was nice seeing the logical process Twilight used to find the solution here. Thanks for writing it! You have a gift for making these, they've been a fun challenge. Take care of yourself though.

What in the world


Am I truly your first thumbs up?
Never had that happen before.

1. B doesn't play against D since there are 3 students saying the same statement. This point eliminates possibility of Sandbar, Yona, and Smolder being either team B or D
2. Sandbar and Smolder states that team A won against team E, making both of them the only candidates for team A and E
3. Sandbar and Yona have contradicting statements about team A and C, which means one of them is lying
4. In accordance to point (1), Silverstream and Ocellus are the only ones available for team B and D
5. In accordance to point (2) and (4), Yona is team C
6. Point (3) and (5) proves Sandbar is team E
7. Point (2) and (6) results in Smolder being team A
8. Conclusion up to point 7:
Sandbar = E, Yona = C, Smolder = A, Silverstream = ?, Ocellus = ?
game result : A beat E, C beat A, D beat C, D beat A
9. Since one team only plays either two or three games, Silverstream can't be in team B, making her member of team D, and Ocellus team B
10. In accordance to point (9), team B doesn't play against team A since team A would be playing four times
11. Final result:
Sandbar = E, Yona = C, Smolder = A, Silverstream = D, Ocellus = B
game result : A beat E, C beat A, D beat C, D beat A, E beat D, B beat E, C beat B

Comment posted by Os deleted Last Tuesday

Ok lets do this
minimum 3 of 4 gems are used
C is sapphire if D isnt ruby
D is emerald if B&E are different
If E is sapphire A is diamond
If E is emerald A is ruby
C is ruby if B isnt sapphire or emerald
If B is sapphire C is diamond
If B is emerald C is the same

Now that ive clarified the clues lets solve this
if B&E are different, D is emerald, witch makes C sapphire, but C cannot be sapphire as it wont work with whatever i make B, so B&E are the same
D cant be emerald then
If B&E are emerald, C is as well and A is ruby, if D is ruby,only 2 gems are used, if D isnt ruby C must be sapphire, since C cant be sapphire and emerald, it leads to B&E being sapphire, while A&C are both diamond and D is ruby

I had missed this one, so I'm only now getting around to it.

This one seemed hard at first glance, but started coming together really satisfyingly as I peeled away bits of it. Anyway, I think I solved it here.

Ocellus: B
Silverstream: D
Sandbar: E
Yona: C
Smolder: A

A beat E
C beat A
D beat A
B beat E
C beat B
D beat C
E beat D

Final counts of games played are:
A-3, B-2, C-3, D-3, E-3

Also, I love that you used the word "sportscreatureship" hehe. It looks so silly but I could totally see it being a thing.


E must wear either Sapphire or Emerald
If E wears Sapphire
A wears Diamond
-End of Chain-
If E wears Emerald
A wears Ruby
-End of Chain-
If D wears Ruby
and B wears Sapphire
Then C wears Diamond
Since 3 gem types have to be used E cannot be Emerald as that makes A and D wearing Ruby and E, C and, B wearing Emerald
D wears Ruby
B wears Sapphire
C wears Diamond
E wears Sapphire (Due to provision stating D must be emerald unless E the same gem type as B)
A wears Diamond

Okay, this one was tough at first until I got over one hurdle. I believe I have it, though.

A: Diamond
B: Sapphire
C: Diamond
D: Ruby
E: Sapphire

The difficulty was in finally noticing that if D wasn't Ruby, then C had to be two gems. So, D had to be Ruby. Then it was a matter of figuring out the rest from there.

Another fun one. It looked intimidating but I am glad I kept at it. Thanks for another puzzle!

I think I got it.

This first one is a gimme. If Yona is wrong, she gets Twilight. The two correct students get Twilight and Pinkie Pie. Therefore, Yona is correct. Therefore, Yona gets Pinkie Pie.
If Sandbar is correct, he gets either Twilight or Rainbow Dash. Assuming he is correct, Sandbar gets Twilight.
From here, the rest fall into place.
Ocellus is wrong, so Silverstream gets Rarity.
Both Sandbar and Silverstream are wrong, so neither Ocellus nor Sandbar gets Rainbow Dash. The only one remaining is Gallus, so Gallus gets Rainbow Dash.
Silverstream is wrong, so Sandbar does not get Fluttershy. Therefore, Sandbar gets Applejack.
Therefore, Ocellus gets Fluttershy.

This was fun. Now to see if I'm right!

Yes, I got it! Though, I got a bit confused for a second when Yona referred to Sandbar as a seapony.

Very well done so far. This is fun! Let's see what the next puzzle is.

Okay, this one took some trial and error, but I think I found a unique solution.

This first part was the trial and error. Assume Silverstream got the 96.
Smolder can't be wrong because the lowest score is even, so Smolder got either 96 or 98. Therefore, Smolder got the 98.
Yona cannot be correct because 97 is odd, so she got either 95 or 99. If Gallus is correct, he got 96 or 98, so Gallus must have scored lower than Yona. The only way this is possible is if Yona got the 99.
If Silverstream got 96, her prediction is correct and Ocellus is in the top three. 98 and 99 are taken, so Ocellus got the 97.
Gallus is wrong, so he got 95 or 97. 97 is taken, so Gallus got the 95.
There is only one score left. Sandbar got the 94.

I tried several other possibilities for the 96, and they all had two indeterminate solutions. This was the only one I tried that yielded a unique solution, as per Twilight's statement. Let's see if I was right.

Two for two! That took me a bit longer than I'd care to admit. Still, this one was also fun! NEXT!

Just because it involves guesswork doesn't mean it's not logic. Twilight's statement about a unique solution is the key.

Finally got it. I think.

Either Yona's or Gallus's number must be a perfect cube. If Gallus's number is a perfect cube, it must be 729, which is also a perfect square and greater than 200. Thus, Yona's number would be neither a square nor a cube, but it would be a palindrome. That still leaves eight possibilities, so the cube must belong to Yona. Since Gallus's number is greater than 600 (and therefore 200), Yona's number must be the only perfect cube between 100 and 200. Yona thought the number was 125.
125 is a perfect cube, not a perfect square, not a palindrome, and less than 200. Therefore, Gallus's number is a perfect square, not a perfect cube, a palindrome, and greater than 200. There are two possibilities, 484 and 676. Gallus's final clue, that the number is greater than 600, eliminates the former. Gallus's number is 676.

Huzzah, three in a row! Sadly, I must take a break now. I shall finish the rest later this evening.

I wish my test scores were this high. :fluttershysad:

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