• Member Since 26th Apr, 2013
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Don't mind me, I'll just sit here, quietly writing bad stories. And songs. And stories with songs.


This story is a sequel to Life is a Test: A Series of Pony Logic Puzzles

Come back to cradle your cerebral cores in a new series of puzzles throughout Ponyville, Canterlot, and beyond Equestria's borders!

As Canterlot Castle's new Royal Dispute Settler, there will be many odd problems to solve with Twilight Sparkle and her friends, some of which may or may not involve the Magic of Friendship, but all are fiendish in their solutions.

NOTE: I invite readers to try solving these puzzles for themselves! Try not to spoil too much in the comments, though.

Answers are uploaded as soon as a reader gets it right. Or I get bored waiting.

Chapters (80)
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Comments ( 190 )

I can't do square root on my phone so exact numbers I can't present. But it's pythagoras. A package 1.5x1.5 will definitely contain the pool cue if it's placed diagonally.

This probably doesn't count, but the story pic's answer is 13.

I say the answer to the story image is 14

Apples equal 10 as the Apple + Apple + Apple = 30 = 3 x 10
The banana is 3 as unlike the other bananas in the image that have 4 in a bundle, this one is missing a banana.
The single coconut is 1 as the previous coconut has two halves, this one has only half.

Coconut = 2
Apple = 10
Banana = 4 (or a Minion) :twilightoops:
Total = 16

Awesome more riddles
make the box as big as possible on each side, then place the cue so one end is in the bottom left front corner,and the other in the top right back corner, should have just enough room to close the box


Stick it in the box diagonally.

He knows the amount he values it at right? He just wont say? If thats the case bid 1 bit less than that, if he get the painting then its 1 bit profit, and if not the other pony over spent
If he isnt sure of the price then im not sure

To give numbers to this
1.42×1.42×however thick the cue is

I remember this exercise. Optimal strategy for this kind of auction is to bid 1/30 off the expected value, or 29/30 of his expected value. Actual proof is long and can be found here.

Fic picture - the missing value is [undef].

Both the Coconut and Banana emblems are different from those used in the earlier calculations. Therefore, both count as completely new variables, the values of which we have not been given. As such, the summation of those, plus the 10 from the apple, is an unknown value - undefined.


I was right!

but analyzing the images for both. in both cases the final additive each the coconut and the bananas is missing one from it's previous iteration. If we consider the values we calculated before, it can be considered that it's simply those minus one.

I see 2 options yes or undecided, if they are in a line, jet set is behind quiet time and thus looking at them, or undecided as they are next to eachother

My Answer:
If Charity Case is not married, then Jet Set (married) is looking at Charity Case (not married)
If Charity Case is married, then Charity Case (married) is looking at Quiet Time (not married)

Now thats a good one
based on the 2 left most collums, we know E+E+(possible carry)=10+k and based on the right most collum, k+k=E, so by trial and error, i get K=3 and E=6 and that possible carry is now definite
E+k=9 so thats that blank space
E+e=X so X=2 and there is a carry
Since r+r+1=an odd number, r can be 7 with the blank being 5
M+3+1=h and 2+h>10, so h=8 and m=4
So the final equation is 6247663+6837633=13085296

They never said that the cue couldn't be designed to unscrew and be placed in the box with both parts side-by-side. She just said it couldn't be damaged. So logically at the handle would be the ideal place to put where the cue unscrews and screws back in. Additionally, it can be hollowed out this way so the package doesn't exceed the weight limit (assuming there's that unspoken requirement) and the package won't exceed the length or width of the box.

So E is either 6 or 8, which means that K must be 3 or 4.

K can also be 8 or 9 in the cases where E is 6 or 8 respectively. This step in the solution has not ruled out any of these two possibilities.

It can't be, that would create carryover, which is impossible.

If K is 8 when E is 6, then the tens digit wouldn't have both K and E, because a carryover of 1 would throw that off.

Thank you, why does nobody think of colorblindness when it comes to this
ok, so if azure is colorblind, he would say no first,and then Big looks and sees minimum 1 black, then says no(if both Azure and Cirrus have white, then Big knows they have black), cirrus now knows that either their hat and/or Azure is black,if azure is black then cirrus says no, and since cirrus sees a black on him he'd then call black, so Azure is not colorblind
If big is colorblind, azure would see minimum 1 black hat and say no, big then says no as he cant tell if cirrus wears black, cirrus determines either him and/or big is wearing black, sees black on big and says no, then azure says no as he saw the black on big and cant determine thier hat (so here we have big is colorblind and wears black)
If cirrus is colorblind, azure sees minimum 1 black and says no,big knows him and/or cirrus is black,sees black on cirrus and say no, cirrus knows big saw they had black and would call black
All in all, only the 2nd path works

I have no idea who you are, why I'm here, or if anyone has answered your riddles, but I ACCEPT YOUR CHALLENGE!!!

Aren't pool cues design to screw apart, or am I just uncultured?

Okay, this one is harder. I'm going to go with my Monopoly bidding strategy and say one dollar (or bit in this case) less than his estimated amount. That way he forces his unknowing opponents to over bid, or he gets one bit in return!

Ouch, you got me there. I feel embarrassed

I sware I was planning on solving this tomorrow because I don't think I can handle the caliber of this question before I call it a night in about a hour. But when I went check so notifications, I came back and hit continue reading and it sent me here since I technically finished reading the last chapter...oops.

Despite not solving these exactly like how you want, I'm enjoining myself and am learning a thing or two.

I'm confused. Why is color-blindness being treated as an issue? Color-blindness only affects what colors someone can see among the RGB spectrum - even full achromatopsia still allows for grayscale vision. With only black and white (dark and not-dark) hats in play, the color-blindness shouldn't affect the ability of anypony to determine their hat color.

Wait, the Puzzle was stated as

PUZZLE: Is an unmarried pony looking at a married pony?

The answer should be B: NO. The reasoning is very different. Since we're looking for whether an "unmarried pony is looking...", we can always ignore who Jet Set is looking at since he's married and nopony is looking at him. Since we need to determines if the watched pony is married, we only need to worry about Quiet Time being observed. Since he is not married, the answer must always be no.

Was something changed or just a mistake?

No big. Also, love what you're doing with the MLP flavors of these puzzles. Very fun ^^

since we are considering worst case scenario, that means going through every combination available which equals to 24 tries

Worst case?
we get 0 right, then 1 then 2,then all 4, no matter what its 4 trys

Since we're imagining worse case logical scenario, we will have to ask the builder a maximum of 6 times.

Her first answer will be 0, 1, 2, or 4. If it's four, great! We're done. But worst case scenario, it's not four.

Let's say her first answer is 2. The means out of the 24 possible combinations, only 6 could be correct. Of these six combinations, e pick one and ask again She can answer 0, 1, or 4. Is she answers 4, we're done. If she answers 0, there is only one combination that can then be possible and we don't need to consult her a third time.If she answers 1, this leaves 4 possibilities where asking her will result in a 4, 0, or 1. Again, if her reply is 0 or 4, we will not need to consult her. If it's 1, we'll have to build again from the remaining 2 combinations and ask one last time. Either way, after her fourth answer, we know the exact order, whether our current build is correct or not.

Now let's say her first answer is 0. This reduces our pool to 9. For our second build, her response will only be 0, 2, or 4. 0 or 2 will reduce the pool down to 4 possibilities. In those pools, the third build will get answer of 4, 0, or other (it's not the same for the 2 pools). If it's 0 or 4, we know the exact build. If it's a different number, the pools are reduce to 2, meaning we need to do one more build, and depending on whether she answers 4 or 0, we know the correct outcome.

However, the worse case scenario can occur if her first answer is 1. This reduces our pool to 8. We select a build, and, if the response is not four, each build after this only eliminates 1 possibility. If her second response is 0, this means there are 3 possibilities and we need to only 2 more times at most. Once will eliminate a possibility, leaving us with only 2 after, which will need a confirmation.

Unfortunately, if her second response is 1, this narrows our choices to 4. Each possibility will either net us a 4 or a 0, with zero only eliminating 1 possibility each time. The first zero narrows it to three, the second to 2, and then we need to ask one last time between 2 choices.

So most you'll have to consult the builder is 5 times to know the correct order. Keep in mind, you don't need an answer of four in some cases to know what the correct order is. Since we will have to confirm the final is correct, we will have to ask one last time giving a maximum possible of 6. I hope you're ready for the long haul.

I worked this out on a spreadsheet. If you care to look, you can see it at https://docs.google.com/spreadsheets/d/1JcrFRA2RoZv8TM15YGUO-RLBwhLO5hExkD4C-GKMtIE/edit?usp=sharing
Ignore the link if you don't want a spoiler for this.

You only go through 24 tries if you don't use any of the information the builder gives you.

While this sounds right i need to ask

Now let's say her first answer is 0. This reduces our pool to 9. For our second build, her response will only be 0, 2, or 4. 0 or 2 will reduce the pool down to 4 possibilities. In those pools, the third build will get answer of 4, 0, or other (it's not the same for the 2 pools). If it's 0 or 4, we know the exact build. If it's a different number, the pools are reduce to 2, meaning we need to do one more build, and depending on whether she answers 4 or 0, we know the correct outcome.

Now the problem comes about halfway when you say other the only numbers that show up at all is 0,1,2 or 4 since thats how many choices we deal with, so how does this play out?

Actually, if the new picture has a different value, there is still a solution.
3 apples=30, so 1=10
1 apple +2 bunches of 4 bananas= 18
2 bunches of 4 bananas= 8
a bunch of 4 bananas=4- logic dictates that three bananas is 3
a bunch of 4 bananas- 2 coconut halves=2
2 coconut halves=2... 1 coconut half is 1
1 coconut half+3 banana+1 apple= 14

Depending on which pool you get, her answer will be one of 4, 2, or 0 or 4, 1, or 0. In either case, an answer of 4 or 0 gives us the exact build. If it's not one of those, you will have to decide between 2 remaining builds. I used other since the process would be the same. Perhaps I simplified a bit much. Then again, it felt like my explanation was a bit long winded to start.

You can't say that - you're not dealing with two coconut halves, or four bananas - you're dealing with VAR[IMG:2HalfCoconut] or VAR[IMG:4Banana]. There is no guarantee that the value of VAR[IMG:1HalfCoconut] or VAR[IMG:3Banana] has any relation to the previously-used variables.


You watch Ted-Ed.
I approve.


Wait, if he's color blind, doesn't that mean he shouldn't be allowed to fly?


This is obviously not the real answer, but it's a practical one.

Assuming it takes 1 hour to build the structure, it should take only 2 hours.
While they are building it the first time around, we make a list of all 24 combinations of color order.
If we are lucky, the 1st time is correct and it only took one hour.
If not, we show the builder the list and he points to the right one.
Structure goes up, and it only took us 2 hours.

Once again, this is not the right answer. Just common sense here.


Here's a picture for you.
Basic, but it does the job.
= and || bridges
~ water
----------- shores


Thanks to that picture that make this pretty easy
its 1 in 64 all bridges from the shores to their nearby islands are down, so at first glance it seems thats it but its not
we have .5×.5×.5=1in8 chance of NOT going from north shore to A,B or C, and an equal chance of NOT going from D,E or F to south shore, so minimum 2 in 8 or 25% chance we cant cross


Here's a hint for you guys.
The max number of bridges that we need to cross is 7
The minimum is 3.

place one at the corner, making a right triangle, and the other from the corner of the island to our board

Wait a minute, it can be this easy but...
I got this from adding the 25% of fail i had before and another 25% from the chances of all other combos that prevents crossing

Ok so this got me on the right track hopefully
lets say we try abcd and get 1 right,out of the 24 combos in total, none get eliminated
Then i try bcad and get 1 right
Next try will be dabc, witch gets a 0
Then cdab gets 1
Then adcb gets 2
The solution is then acdb in 5 trys

Comment posted by Fluxxdog deleted Nov 19th, 2018


lets say we try abcd and get 1 right,out of the 24 combos in total, none get eliminated

If none get eliminated, we are ignoring the information the builder gives us, and that wouldn't do. After all, we can at the very least eliminate ABCD :pinkiehappy:
When we build our pool of remaining combos, we should be using ones that share the matching number of traits with our guesses. For your example, when the builder responds with 1, we should keep any combos that match exactly one of the criteria:
First color is A
Second color is B
Third color is C
Fourth color is D

So if our first guess is ABCD, ACDB would be acceptable since it only matches on the A. However, ABDC would be useless since it has 2 matches. Likewise, DCBA is out since it has 0 matches.

Remember, our job isn't to get the tower built correctly. That task belongs to that pecan muffin muncher. Our task is to find the worse case scenario and determine the number of times we need to consult the builder.

I updated the link so comments could be posted (thought I did that the first time!) but here it is again if you want a look. Spreadsheet link

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