2 nice logic puzzles - spy, liar, truthteller and chocolate wrappers

[@InsaneSynbad]

Wait, if Charlie is a liar… how do we know if Charlie actually has $1000?

[@MatthewMorris6148]

Charlie never said he had $1000. The problem says it, so we’re just assuming the problem is a truth-teller. So unless Charlie is the problem writer, then it’s probably fine. Although I suppose Bob could be the problem writer, and therefore may or may not be lying? Hmm, this is getting tricky. You know what? I’m just gonna assume Derrick is the one writing the problems, and that he’s another truth-teller. If there’s any contradictions, then I’ll know my original assumption was wrong.

[@jimmyh2137]

He's a liar, not a thief. With 1000$ he can buy 1333 of those chocolate bars, regardless if he has that much money or not.

[@yurenchu]

We asked Charlie "Would you say that you have $1000 ?", and he said "Yes".

[@jaydentplays7485]

Charlie said he didn’t have $1000 so that he didn’t have to buy me any of the chocolate so he has it secured.

[@gotafarmyet4691]

@@jaydentplays7485 Charles did not unwrap the bars and then after getting 1333 of them returned them and turn 1000 to 1333 dollars. He did this every day for 10 years and became a millionaire. Then he posted what he did on tik tok got sued for a billion dollars and when to prison for 20 years.

[@DinJaevel]

Diabetes, Charlie gets diabetes.

[@Mysteryspy]

The correct answer right here folks

[@WicherBos]

Only if he eats them…

[@samueldeandrade8535]

Nope. He gets a mansion next to Easter Bunny's mansion.

[@alanclarke4646]

Liar! 😂

[@yurenchu]

​@@WicherBos Somebody probably eats them, otherwise Charllie can't turn in the wrappers...

[@Species1571]

Charlie's whole scheme is in ruins when he discovers that when he hands over the 4 wrappers, he gets a free bar that doesn't have a promotional wrapper. Then discovers the non-promotional bars are half price in the shop next door.

[@alexmashkin863]

You're too much real world for those problems :-)))

[@omg_look_behind_you]

Also he needs to pay taxes on the free bars

[@general_paul]

Scam 9999

[@scottmcshannon6821]

ive noticed that happens a lot, before the mark goods off 40% THEY DOUBLE THE PRICE AND THEY COME OUT AHEAD.

[@berndbeispielmensch]

Also, Charlie is getting diabetes soon. It is a good thing that he likes maths a lot, because from now on, he must count his carbs every day.

[@harryh536]

1250 bars easily. The gifted bars come in different wrappers and the store claims they cannot be used for redeeming.

[@DawnDavidson]

😂 Sadly, likely true!

[@__christopher__]

The store has a policy that you can only buy a maximum of 10 bars per day, the offer is for 5 ays only, and only wrappers from the offer period can be redeemed. Of course, as already noted above, wrappers of free chocolate bars cannot be redeemed. So Charlie can get at most 62 chocolate bars.

[@akale2620]

I see someone works for Amazon

[@MrEscape314]

​@@__christopher__wouldn't he still be about to get at least 1000 bars over time? Even if the promotion ends, surely they still sell the chocolate..

[@__christopher__]

@@MrEscape314 well, over time Charlie will surely also get new money (or else he surely wouldn't use all his money on chocolate), so the amount of chocolate he can buy from the shop will be limited by other factors like how long he lives near the shop.

[@mattpburgess]

😂 I love when the “easier” way of working something out is VASTLY more complicated and confusing!! 😂

[@LibertyMonk]

It's not, he just explained it much more thoroughly to leave no holes in the word problem. "Every three bars means one extra free bar" so "1000 * 1 & 1/3, or 1,333" is pretty simple, but it's a minor leap of logic to get from "four wrappers is a free bar" to "three bars, plus the free bar, is four wrappers, paying for the free bar" so he walked through an example of it. It's fair to be confused by his walking through, or confused by the "slow" way, but try to understand them. Seeing both (or more) ways to get to the right answer is a valuable skill.

[@whiteninjaplus5]

I don't think it's up to you if it's more complicated. What's simple for one person may not be for the next.

[@karolinaj5045]

@@LibertyMonk I like your explanation better

[@extensivity29]

I love when adults forget how to do basic algebra 😂

[@MiccaPhone]

It is easier because it can be generalized for any number N other than 1000, so you can get the right number with quick calculation even if N=1234e56 instead of 1000.

[@willchurch8376]

"Here's the long way." *Shows the basic arithmetic. "Here's a shorter method." *Shows a convoluted scheme in which Charlie, a known liar, over a span of three years, eats a candy bar every day and slowly goes broke.

[@bentonrp]

Lolol! Oh yeah! It was the same Charlie! 🤣 Conflict of interest...! That's true that it was a longer method. Also, I wonder why he made this complex graph in the first problem, when a simple line drawing to two separate columns would have done the trick. :)

[@treyk9007]

It's been a few months, but I had to tell you that you just made my day with this response. I laughed and laughed and laughed. Thank you.

[@Chase3141]

My way of solving the second problem felt easier than either method you used by, counting bars and wrappers separately. At first he spends all $1000 to get 1000 bars and 1000 wrappers. Every time he trades in 4 wrappers, he gets 1 bar and 1 wrapper, so his total number or wrappers decreases by 3. This means he can essentially trade 3 wrappers for 1 bar without a wrapper as long as he has at least 4. So I take the 1000 wrappers, divide by 3 to get 333 more bars with a remainder of 1 wrapper (this remainder also means he never had less than 4 until the final trade). Since he can’t do anything with 1 extra wrapper, you add the 333 bars to the first 1000 to get 1333

[@vxllabh09]

I completely agree with your logic, however there is a small caveat... You've written that he never had less than 4 until the final trade and this is not wholely correct. "The final trade" is him trading the last 4 wrappers for a bar and a wrapper. The wrapper that is left over isn't actually left over, it's just the wrapper of the chocolate he bought and that's why he has a remainder. The wonderful thing about this is that no matter how this problem is set up, you'll always end up with a remainder!

[@daviniarobbins9298]

Ignoring the fact that the the promotion would have some limiting factors like you can only have so many free bars of chocolate. No store would let you buy 1333 bars each costing 1 dollar with 1000 dollars, well not if they are smart and don't want to go out of business.

[@hermanphilips4617]

@@daviniarobbins9298 Hahaha, if a store did have the profit margin on the chocolate bars to do this, and the inventory to support it, you could only get the 1000 on the first trip. After that, you need to go unwrap the chocolate bars and come back with just the wrappers. So, it would take time to get them all... because yeah, no store would let you walk in and get all the future bars at the same time. Because you then have 1333 wrappers while they are running a 'trade in wrappers' promotion. But, that does presume the store in question is making enough profit that on $1 chocolate bars that they can afford to do this. Basically, it means they would still have enough profit selling them for $0.75 to be able to just list them at that price. Of course, more likely... they accidentally ordered too many so just want to clear them out before they expire. Or they decided to break even or take a loss using the promotion, in order to bring in more customers who might purchase other products at the same time. But yeah, the business would only go under if they didn't consider their profit margin or didn't put a time limit on it. Because honestly, if they are selling the bar for $1 each, they probably are buying it for 50¢. So effectively selling it for 75¢ is probably not going to sink them.

[@johncox7169]

@@daviniarobbins9298 Depends what it costs the store to get them. If it cost the store anything less than $.75 per bar, then the store is still making a profit no matter how many free bars they are giving out (Each bar costs $1, and gives $.25 towards another bar, hence $.75)

[@MiccaPhone]

I think this is the same as his 2nd method, just differently explained. In either case you end up with a generalized formula for for the number B of chocolate bars for N Dollars and a bonus scheme of 1 extra bar per each W wrappers: B = N + floor((N-1)/(W-1))

[@xnick_uy]

I approached the 2nd problem similarly two times, as in the video. But for the second try, I just used directly the fact that the effective cost of a chocolate bar is 3/4 $ = 0.75 $. This gives the answer that 1333.33... bars can be afforded with 1000$. Round down and presto!

[@mohammadfawwaz4804]

Could you tell me how it would be 0.75? 1 dollar = 1 chocolate + 1 wrapper(1/4 chocolate) So that should be 0.8 per bar ye?

[@AllyWheels]

Only the very first bar costs $1. The effective cost of all subsequent bars is then $0.75 (buy 3 get 4). So how many $0.75 bars can one buy with $999?

[@tylerbrown4483]

@@mohammadfawwaz4804if you view the wrapper as worth $.25 because 4 of them can be redeemed for a chocolate bar valued at $1, then each time you spend $1 on a chocolate bar, you get $.25 back in wrapper credit, making the effective cost of the bar $.75.

[@tylerbrown4483]

⁠@@davidwalton3604but what does work for every amount is 1+(N-1)/.75 So 1+999/.75=1333 for $1000 And 1+14/.75=19.67 rounds down to 19 for 15 And 1+1001/.75=1335.66 rounds down to 1335 for 1002

[@JamieFoundWaldo]

I just think of it as x = 1000 + x/4 (total amount is 1000 + one quarter of the final amount of wrappers), rounded down

[@_orko]

For the second one I just used the knowledge that four wrappers gets you 1 bar so the wrapper itself is worth 25 cents. So the chocolate is only really .75 cents. 75 cents into 1000 gets you 1333.333

[@NeKrokodilu]

But the correct answer would be 1333.25, surely? Round 1: take $1000 cash and buy 1000 chocolate bars. Round 2: use 1000 wrappers to get 250 chocolate bars (running total: 1250 bars) Round 3: use 248 wrappers to get 62 chocolate bars (Rtot: 1312 bars + 2 wrappers) Round 4: use 64 wrappers to get 16 chocolate bars (Rtot: 1328 bars) Round 5: use 16 wrappers to get 4 chocolate bars (Rtot: 1332 bars) Round 6: use 4 wrappers to get 1 chocolate bar (Rtot: 1333 bars + 1 wrapper, which would be worth 1/4 of a chocolate bar if that was how the promo worked)

[@RexxSchneider]

@@NeKrokodilu Yes, but that 1/4 of a chocolate bar would have 1/4 of a wrapper which would be worth 1/16 of a bar, and so on ... Add up the infinite sum and you get 1/3 as the OP said.

[@NeKrokodilu]

@@RexxSchneider interesting. So either the wrappers have no value until you have 4 of them (making the answer 1333.0), or you have to keep counting the fractional wrappers (making the answer 1333.333...). Thanks for the insight!

[@ADudeWhoDo]

The way I did it was spend 4$ for 5 bars then the rest is 4 bars per 3$. I end up with the answer in exact

[@olerask2457]

Your argument only work because 1000 is NOT divisible by 3!

[@TyphoidBryan]

Yeah, I did it the long way with the chocolate: 1000 + 250 + 62 + 16 + 4 + 1 = 1333. I liked the part where I had to remember I had some extra wrappers hanging around. Sure enough, I needed them again. :)

[@Rammkard]

I did the first one with a shortcut. I noticed they were all saying what they weren't, and the liar can only say that they aren't the liar.

[@ReaGool09]

Unfortunately, I only realized this after solving it the more complicated way. I didn't even consider that I could solve it w/o considering what anyone else said.

[@soundsoflife9549]

Very easy to see that Alice is a spy then the rest is obvious.

[@michaeledwards2251]

I noticed everyone was saying what they weren't, it didn't occur to me the liar could only say they weren't a liar.

[@Aname-l-]

Couldn’t the truth-teller tell the truth and say that they aren’t the liar,since they,in fact,aren’t the liar? Actually,doesn’t that same logic apply to the spy as well?

[@michaeledwards2251]

@@Aname-l- Only 1 individual said they weren't a liar. Since the liar had to say they weren't a liar, they couldn't not say it, the one individual who said they weren't a liar, had to be the liar.

[@brunofreitasrego9182]

The sum of a geometric progression is calculated using the formula: amount / (1 - multiplier). For this scenario, the boy starts with 1000 bars, then exchanges the wrappers for 250 more bars, and repeats this process until he can't buy anymore. With a multiplier of 0.25 and an initial amount of 1000, the sum of all terms will be 1000 / (1 - 0.25) = 1333.333. The integer part, 1333, represents the total bars the boy can acquire.

[@chaos.corner]

That doesn't work with 78 though. 78/(1-0.25) = 104 and, in fact, it works out to be 103 bars with 3 wrappers left over. It's a shame because it would have been a nice answer. Credit to davidwalton3604 above.

[@leif1075]

I don't see why ANYONE would solve this problem the way he does the second way with supposed "clever" trick...does anyone??

[@soundsoflife9549]

We use the type of calculations to determine dilutions for obtaining particular concentrations when diluting or mixing solutions.

[@icarus877]

Actually that method can be flawed and potentially not correct under all circumstances.

[@TimothyRE99]

Its flawed because it ends up counting the wrappers from bars you have not yet gotten. In any situation where you end up with an integer from the formula (i.e., you bought an initial number of bars that is divisible by three), simply subtract one off. Otherwise, take the floor.

[@sweepingtime]

I was delighted by the 2nd puzzle. I'm sure that the long way occurs to everyone, but the short way is an example of how you'd try to simplify a tedious calculation, which is important in everyday life.

[@sfcs3743]

I failed it :( but how is it important in everyday life?

[@PSiOO2]

I had a similar thought process - you take 4 dollars to get 4 chocolate bars, then get a fifth one for free. You now have 1 wrapper, 5 chocolate bars, and 996 dollars. For now we forget about these 5 chocolate bars. For the next free chocolate you are buying 3 more chocolate bars. From now on it will always be 4 chocolate bars for 3 dollars. 996/3= 332. That's the amount of "transactions", that all end in us getting 4 bars for 3 dollars. We multiply that by 4 to get the number of chocolate bars, and get 1328. Now add the 5 chocolate bars we got at the start. 1333, answer is the same

[@LibertyMonk]

I really don't understand how the "long" way is significantly more steps than the "shortcut". The "long" way is just long division, or multiplication in this case. The alternative method is shorter, if we complete ignore the cost of checking that our addition & translation of the word problem is right, then doing the small-scale test (check 4, then that 3 more gives a spare), then the possibility of whatever shortcut you tested not working and needing to try a different one (which doesn't happen in this case), then multiply by the shortcut 1,000 * 1+(1/3) rounded down. They're the same duration. It it was a billion dollars, the shortcut is more likely to save time, but going the "long" way you'll notice the pattern pretty quickly and that the sum is 1,33... as you're going along.

[@elpme16]

Here is a fun insight into the second problem. In a single trade, Charlie can spend $1 in exchange for a chocolate bar and a wrapper. The promotion allows 4 wrappers to be traded for an additional bar and wrapper. Since 4 wrappers can be used to get another bar, that means 4 wrappers are equivalent in value to $1, meaning 1 wrapper is equivalent to $0.25. So if we go back to the original trade, $1 is being exchanged for a chocolate bar plus a wrapper worth $0.25. So in effect Charlie is trading $0.75 to get one chocolate bar. So to find the amount of bars Charlie can get for $1000, we just need to divide 1000 dollars by 0.75 dollars per bar to get 1333.3333... bars. But we can't get a fractional number of bars, so 1000 dollars will get Charlie a maximum of 1333 bars.

[@XJWill1]

Yes, that is the way I solved it. I think it is easier and clearer than either of the methods shown in the video. For the general case with D dollars, the number of candy bars is: floor(4*D/3) where floor(x) is the greatest integer strictly less than x.

[@atary86]

This is not a correct solution. Assume he had $1002 to start with. Your method gives the result of 1336 bars, whereas the correct answer is 1335.

[@Nick-M1]

​@atary86 is correct. Since you cannot end with 0 wrappers it isn't as simple as $1000/0.75 Also each wrapper is actually worth 1/3 of a chocolate bar, which is why it would be dividing by 0.75 Since you can't both start and end with 0 wrappers it's simpler to subtract the first one then take the remaining amount and divide by 0.75 For the video that's what Presh Talwalkar did. For the $1002 example that would give (1001/.75)+1 = 1335.67 bars

[@beardymcbeardface69]

@@atary86 You could be a accounting company software engineer and the OP could be a 90's 3D game company software engineer. LOL

[@XJWill1]

@@atary86 If your dollars are a multiple of 3, it will still work but you need to go into the store with your 3 wrappers, grab a candy bar, take off the wrapper, and then give the store the 4 wrappers and hold up the candy and say, "all good! see ya!" (or you can just subtract 1 from the answer if your dollars are a multiple of 3)

[@nickwilson1645]

The explanation for the “quick way” takes longer than working it out the “long way”😂

[@bluerizlagirl]

Yes; but once you know _how_ the quick way works, you can reuse the method in future. Real-life example: Spending four hours writing a program to automate a task that would have taken half an hour to do by hand is worth it, if you are going to be asked to perform that task nine times or more.

[@MiccaPhone]

Yes, but now repeat it with 35745763 dollars instead of 1000 dollars. Which method is faster? I think the 2nd method, which provides the general formula for the number B of chocolate bars for N Dollars and a bonus scheme of 1 extra bar per each W wrappers: B = N + floor((N-1)/(W-1))

[@glowhazel]

I'm not the best at logic puzzles but I paused right when you introduced the 3x3 organizer and got it from there. Second one I wrote a linear equation once I realized (past the first 4 bars) he'd be getting each subsequent group of 4 bars for $3 each.

[@gaboignacio]

Problem 1 is direct if you understand that a liar can not say "I am not a spy" nor "I am not a truth teller" because they would be telling a truth. The only sentence a liar can say is "I am not a liar".

[@dandanlec1996]

For the first question you don't need to know what the other two are. The liar has three can say one of three phrases: they say that they are not a liar, they say that they are a spy and they say that they are a truth teller. The only person who sais any of these is Charlie

[@tarangmendhe7915]

just because a liar has to say any one of these statement dosen't mean that anyone who says them is a liar. for eg" i am not a liar" can be said by a truth teller( he is saying truth in this case) , a liar(he is lying in this case) as well as a spy ( he is telling truth here, which he can )

[@SgtSupaman]

@@tarangmendhe7915 , the point isn't that others can't say it, but that none of the others do say it. Because those statements are required by the liar and only one person says one of those statements, that person is automatically the liar.

[@dandanlec1996]

​@@tarangmendhe7915 sorry if I was not being clear. What I'm saying is that anyone who DOESN'T say one of those three statements is NOT a liar. The liar can't say they are "not a truth teller" because that would be a truthull statement. Thus Alice is not a liar. The liar can't say that they are "not a spy" because that would be a truthull statement. Thus Bob is not a liar. The liar can say that they "are not a liar" because that is an false statement. Charlie COULD be the liar. He could also be a spy. However the only person who COULD POSSIBLY BE A LIAR is Charlie. This solution would also work if there was at least one or more liars in this scenario, while elimination would not work in that case. But, my solution would not work a second person said that they where "a truth teller" or that they were "not a spy" (which anyone can say). (Sorry about the caps, I don't know how to bold on mobile)

[@antonfilipenkov864]

@@SgtSupaman "... but that none of the others do say it. ..." // So? It doesn't mean anything. Others were not obliged to say all the phrases they could say.

[@SgtSupaman]

@@antonfilipenkov864 , it does mean something, because the liar HAD to say it. Instead of these phrases, imagine the liar had to say the word "blue". The other two are allowed to say "blue", but can also say other words. Person 1 says, "red". Person 2 says, "red". Person 3 says, "blue". The others could have said "blue", but they didn't, so you know person 3 is the liar, because he is the only one that said what he had to say (he had no other options).

[@mittfh]

My workings: Alice, Bob and Charlie. Alice says she's not a truth-teller. Given the truth-teller always tells the truth, so by denying she's the truth-teller, Alice can't be the truth-teller. The liar always lies, but as Alice can't be the truth-teller, by denying it, she's telling the truth, so can't be the liar. That means Alice must be the spy. Bob says he's not a spy, which as we've worked out, is correct. Therefore, Bob must be the truth-teller. Charlie says he's not a liar, but there's only one option left, the liar. So by denying being the liar, Charlie confirms he is the liar. Chocolate. The initial $1,000 gets you 1,000 bars, but also 250 free bars (1000 / 4). The wrappers from those 250 free bars can in turn be exchanged for 62 more free bars, but you've also got 2 wrappers left over. Add those to the wrappers from the 62 free bars, and you can exchange the 64 wrappers for 16 free bars. The wrappers from the 16 free bars can be exchanged for 4 more free bars, the wrappers from those 4 can be exchanged for 1. Add them all up, and you've got 1,333 bars. One of which contains a golden ticket to visit the chocolate factory...

[@shykitten55]

It is so refreshing your posting these puzzles. Though I miss the days when I could work them out and not have to "cheat" and watch the answer.

[@shubham_stark]

Second problem is simplifies to 1000 * (1 + 1/4 + 1/16 + 1/256 + ....) this is a geometric series with which has sum = a/1-r, where is first term and r is ratio. so the sum is 4/3 . so total chocolates are 1000*4/3 = 1,333.333 which round offs to 1333 chocolates. :)

[@ianfowler9340]

LIke your method. 1/64

[@YoungGandalf2325]

What happens if Charlie finds the Golden Ticket?

[@DaCostaGuitars]

He gets to move his family into the factory.

[@Woad25]

Then his "bed ridden" grandfather suddenly has enough energy and mobility to go on a factory tour..

[@DaCostaGuitars]

@@Woad25 don't worry, Grampa will have more than just a tour at the end of it. He'll go back to work

[@quigonkenny]

He can write the $1000 off on his taxes.

[@shlatekkin]

You mean Grandpa Joe "found" a golden ticket

[@dorderre]

Second problem, first solution is, for me at least, way more intuitive and straightforward than the second solution. I solved this before you even finished reding it out. My only mistake was that I forgot to add the leftover wrappers and kept rounding down, so I ended up at 1330.

[@beardymcbeardface69]

Same.

[@Visstnok]

So you didn't solve it.

[@itachiuchiha8875]

There is 3rd method that is more easy to me. 4 wrappers are worth same as $1. That means 1 wrapper is worth $0.25 and that means actual cost of bar is $0.75. So 1000/0.75=1333.33333 1333 bars and last bar wrapper is represented by 0.33333 value which is non useable (0.75:0.25). So total bars 1333

[@Mark-vj7zd]

I got 1333 by remembering the leftover wrappers - 1000+250+62+16+4+1

[@TeramanV3]

If it was "way more intuitive and straightforward" then you would have gotten the correct answer.

[@kicorse]

Very nice. I was expecting the clever second method for Q2 to use the formula for the sum of the infinite series: Sigma_(n=0)^(inf) (1/4)^n = 4/3. Congrats for coming up with a more intuitive quick approach that also avoided having to think about rounding.

[@Solrex_the_Sun_King]

I just took the remainder

[@xnick_uy]

It would rather be the sum for a *finite* series in this case...

[@kicorse]

@@xnick_uy Nope. That's just the first method, which is slow. The formula for an infinite series is how I solved the problem, and is very quick. You just need to satisfy yourself that rounding down to 1333 is the correct way of handling the finite case.

[@MrEscape314]

This is the method I used. Each dollar gets you 1 bar, plus a quarter of another bar that also gets a quarter of a bar.. aka, $1 gets your 1 and 1/3 chocolate bars.

[@piol29]

I never heard about promotion that takes more than 1000 days 😂

[@dewhi100]

Replace "day" with "minute", and "Charlie" with "Agustus"

[@easy_s3351]

Charlie can buy 1000 bars with the money he has. He needs 4 wrappers to get a free bar, which gives him 1 extra wrapper. So he needs to buy 4 bars first and afterwards for every 3 bars he buys he'll have 4 wrappers (with the extra one he gets every 4 bars). So 1000-4=996 (first free bar/wrapper) and 996/3=332 other free bars/wrappers. Which gives a total of 1000+1+332=1333 bars.

[@adrianalexandrov7730]

I read it as buy 4 get 1 free, so 1250

[@johnshaw6702]

That's the best solution I've read here. Well done. 🎉

[@PopeVancis]

​@adrianalexandrov7730 That is not how it works. That *may* be the situation if the promotionally collected bars' wrappers cannot be turned in for more promotional bars, but this problem assumes the opposite.

[@stephenj9470]

But if Charlie is a liar, did he ever have $1,000 to begin with?

[@Woad25]

I see you've never worked with consultants before...

[@Doktor47]

Plot twist: Charlie is the spy

[@misshell]

B. Dang, I did it the long way on paper. Nice alternative way to calculate the chocolate bars.

[@PrinceAlberts]

I thought I was being clever, but I didn’t think to account for the wrappers from the free bars.

[@Robbedem]

Neither did I. Didn't really undersand what was meant with wrapper. I thought it was just like a sale thing. ;)

[@donc9260]

The vendor will…lest they go out of business

[@__christopher__]

​@@donc9260no, the vendor will just go out of chocolate bars once the expected amount is sold (and yes, the "free" ones are sold, too, their price is included in the other four). They somewhere in the fine print have a clause "as long as supply is available", so you can't do anything about it.

[@michaeledwards2251]

In practice you would never get a promotional wrapper as they would sell out before you got to the 4th bar, or by the time you did, they would be out of stock.

[@bentonrp]

Me too. 🥴

[@CheekiScrubb]

Vids like these are exactly why i love ur channel more than the other similar content out there. Pls post more puzzles like these every once in a while i love them so much bc it relies on logic n cognitive aptitude instead of memorizing equations. I managed to solve both of these quickly n easily bc its simply fun to do unlike back when i was still at school where quick theorems r drilled in my head so i rejected them n always solved exams w my own methods n refused to use the method taught in class which almost got me expelled.😂 TL;DR : more logic puzzles pls

[@GlobalReach0]

For the second problem we can use the formula m*p/(p-1), where m is total money, and p is wrapper.

[@kungfuduckyreal]

The second way to solve the chocolate bar riddle is actually cool

[@evanrosman9226]

Chocolate? Did you just say "chocolate?"

[@p111SC]

CHOCOLATEEEEEE

[@samueldeandrade8535]

Did you mean "Did you just say 'chocolate'?"? It looks like that's what you meant.

[@fishrocker95]

yes sir. with and without nuts

[@SoDamnMetal]

I'VE BEEN TRYING TO CATCH YOU BOYS ALL DAY!

[@StevenLubick]

YES and I want some

[@Sotanaht0]

For any series 1/X^n where X is an integer >1, the sum will converge to 1/(X-1). So for the wrappers you have 1/4^1 + 1/4^2 etc, which converges to 1/3, hence the 333 as 1/3*1000. The only complication is because we are only using the whole number wrappers, the series will always end early

[@mgancarzjr]

Alice? Bob? I think I get to call my friends Whitfield Diffie and Martin Hellman.

[@Phroggster]

Oh my gosh, someone that actually mentions the first/middle names of Diffie and Hellman exists in the wild. Here I thought that we weren't allowed outside of academia. Don't worry Ralph Merkle, your name was mentioned here, too.

[@Trephining]

I have renamed all my IRL friends to Alice, Bob, Charlie, Dave, Elizabeth, Fred, Greg, and Heather so I can more easily turn all of our social situations into easy-to-label math word problems, lololol. And yes, it helps to only have eight friend when employing this strategy in your life. I made the mistake of making extra friend when I only had four, and had to come up with the E through H names, and wow, that was tough. But it made my life so much easier. My friends hate it though, so hopefully I don’t have to illustrate any ideas needing eight people too many times, because I don’t think I’ll have eight friends for long, 😜🤪🙃.

[@CoolitDC]

The chocolate problem is solved in one step. For $1 you get 1 bar and 1 wrapper, you then turn in the wrapper and get 1/4 bar and 1/4 wrapper, then 1/16 etc. So, for $1 you get: 1+1/4+1/16+1/64… = 4/3 of a bar. The answer then 4/3 * 1000 = 1333.

[@woodysmith2681]

Second solution to Problem 2 is what I call a backwards-answer. It's a method of solving the problem that you'd only come up with after already solving the problem.

[@ffggddss]

Pre-watch 1. Alice-Bob-Charlie; From the first sentence (the 3 of them - A, B, C - are one of each type - T, L, S), I read that as saying that they are all different; no 2 of them are the same type. So take Alice. She says she's not a T. If she were, that statement would be a lie, so her statement is true. Thus she can't be T or L, and must be S. For if she's T, then her claim not to be is a lie, so that's impossible. If she's L, then her claim is true, so that, too, is impossible. So she's S. Now take Bob. He says he's not S. That's true, since Alice is S, and no two of them are of the same type. He can't be L, because he just spoke truth. So he's T. That leaves only L for Charlie, who says he's not L, and that being a lie, is consistent with his being L. Ans: B) Liar 2. The chocolate bar problem: I did this by "slugging it out." 1. $1000 buys 1000 bars (B), including 1000 wrappers (W). 2. Those 1000W get you 1000/4 = 250 more bars, including 250W. 3. Those 250W get you 248/4 = 62B, including 62W; + 2W you didn't use. So at this point you have 64W. 4. Those 64W get you 16B, including 16W. 5. Those 16W get you 4B, including 4W. 6. Those 4W get you 1B, including 1W. So in the end of it all, you have obtained (1000 + 250 + 62 + 16 + 4 + 1)B = 1333B . . . and 1W, which you could use toward another free bar if you get more money, but which isn't getting you anything at this point. And looking at that answer, I'm sure there's a streamlined way to get it. You could do it as a (converging) geometric series if it weren't for those 2 "leftover" wrappers from step 3 . . . Fred

[@ffggddss]

Presh: Great job, except you skipped an essential final step in puzzle 1, verifying that Charlie's statement is consistent with our finding that he's a liar. Without that consistency, the problem would have no solution; none of the 4 multiple choices would be correct.

[@veezhang6988]

For 2nd question, separate chocolate and wrapper from the very beginning. $1 = 1 chocolate + 1 wrapper, 4 wrapper = 1 chocolate + 1 wrapper.

[@finris1]

For the chocolate problem, you get one additional bar for every 4 bars you buy. Then Bars = 1000(1 + 1/4 + 1/16 + 1/64 +...) = 1000(4/3) which is approximately 1333.

[@goatgamer001]

Indeed. If the value of the chocolate function c(x) is the number of bars with x dollars, c(x) = ceil (4/3*x-1). Cents do not count though, only full dollars.

[@MiccaPhone]

1002*(4/3) gives exactly 1336, which is approximately correct! 😂 Correct would be 1335. Beware of populists, you are sensitive to "seemingly simple answers" to difficult questions. Think twice before entering the ballot box.

[@MiccaPhone]

​@@goatgamer001wrong! formula gives wrong results in 33% of all cases (whenever N is divisible by 3)!

[@finris1]

@@MiccaPhone I think you committed an error. 1000*(4/3) = 1333.333333, which is approximately 1333. And 1002*(4/3) = 1334. Where are you getting the extra 2 from?

[@goatgamer001]

@@MiccaPhoneit is easily fixable to reduce the output the function for 1/3 of the values by 1. The solution is nearly as simple.

[@N8570E]

1,250 chocolate bars. Thank you for your efforts. May you and yours stay well and prosper.

[@PoeLemic]

I liked that puzzle. I almost got it, but I was one short. However, these are awesome little puzzles that I like trying to solve.

[@chrish7336]

I almost did the same thing and had to go back and count the extra wrappers not evenly used (remainders) and add them together to get the final one.

[@ReaGool09]

Same! I incorrectly calculated that 50/4=12 with 1 left over because 50/4 is too difficult for me, but 50*2=100 and 100/4=25 and 25/2=12 with one left over and it didn't occur to me that 12*4=48, not 49 and I failed to notice that [even number such as 12]*[even number such as 4]+1=always odd, so definitely not 50! I don't think I explained this very well, but I'm late to this video, so how many people are really going to read this?

[@lerarosalene]

Same. I failed with 250/4, got it to be 62 and remainder 1, instead of 2.

[@tanjavandermeer3522]

I did it a little differently. Charlie converts the first four wrappers into 1 chocolate bar, giving him the first wrapper of the next set. He then only has to buy 3 new ones. So, I divided 996 by 3, which gave me 332 extra bars. Add the one extra bar for the first set of 4, and you have 1333.

[@pavloslav]

My favorite variation of the second problem: $10, 3 wraps for a bar. Charlie buys 10 bars, exchange 9 wraps for 3 more bars, and he's left with 2 wraps... But next he asks the next customer: "can I borrow your wrap for a moment?" So, now he has 3 wraps. He buys one more bar and returns the wrap. Now, he has 15 bars for $10.

[@jonathanstrasner2594]

For the first, I figured out Alice and then somehow figured out Charlie, and then lastly Bob. When he went over the explanation I no longer understood how I got Charlie but I was right anyway lol. For the second, I did the long way and then realized the actual cost of a chocolate bar is .75 since each bar gives you .25 of a new one and divided 1000 by .75 = 1333.33

[@ReaGool09]

You might have realized that the only "I'm not the ____" statement the liar can say is "I'm not the liar," since saying "I'm not the spy" or "I'm not the truth-teller" would be true statements for the liar. So, you didn't even need to solve for Alice at all. Unfortunately, I only realized this after solving for Alice and Bob, because why would I go in reverse-alphabetical order?

[@Hans_Gruber]

You can tell a lot about Presh Talwalkar just from the fact that the "first day" is "Day 0" and not "Day 1"

[@foamheart]

I have a question: How did the liar get the $1000 ?

[@SgtSupaman]

That's the most realistic thing in these problems, unfortunately...

[@Leopoldshark]

Bob is a sucker

[@Kernel15]

You can also just spot that the second puzzle is a geometric progression with starting term 1 and common term 1/4, which sums to 4/3 :D So number of bars bought = (4/3) * 1000 rounded down to the nearest whole number

[@Kernel15]

​@@davidwalton3604 Fair, but that's because you can't actually get to infinity and you have a discrete amount of wrappers, not because there's a problem with using the concept of a sum to infinity. So (4x/3) if x is not a multiple of 3, else (4x/3)-1

[@mtaur4113]

This should be slightly finicky with edge cases in the geometric series. If you need to redeem fractions of quadruples to get the last whole bar, you can't. But in the intermediate steps, you don't throw away partial progress either. One bar left over before you earn three bonus bars gets you another quadruple. It's hard to work it out without paper though.

[@torstenpersson5629]

Being a dementia candidate, I'm happy to boost that I had both right!

[@marcusscience23]

For the chocolate riddle, I was thinking, since on average, each chocolate wrapper gets you another 1/4 bar, we can write it as an infinite series, but take the floor: floor[ 1000*Σ({♾,n=0}, 1/4^n) ] = floor[1333.3_ ] = 1333.

[@ffggddss]

Expression under the ∑ needs to be 4^-n, or (¼)ⁿ not 4ⁿ Fred

[@whosdr]

When you see this kind of series, an infinite sum of (1/x)^n from n=0 to infinity, it simplifies to 1/(1-1/x). That further just simplifies in the case of x=4 to 1/0.75 or 4/3. Multiply it by the starting amount as you did here for a rather simple equation of 1000(4/3) = 1333.3333 (and then round down once again like you did, since we can't buy fractional bars. Bummer, right?) Oddly enough this is something I figured out from playing too many MMORPGs. :P

[@marcusscience23]

@@whosdr Exactly what I was thinking, just couldn't be bothered to include it in comment, so thank you.

[@PopeVancis]

​​@@whosdr Isn't 1/(1-1/x) just x/(x-1)? Reciprocal: 1-(1/x) Common denominator: x/x-1/x Simplify: (x-1)/x Since we took the reciprocal, we must do so again: x/(x-1)

[@whosdr]

@@PopeVancis Yup, just another form of the same equation.

[@franklinturtle9849]

Answer 2: $1000 = 1000 bars. 1000 + 250 from wrappers. 250 wrappers still got value so 250/4 = 62 bars (2 wrappers left) 62/4 = 15 2 wrappers left so go with 16. 16/4 = 4 4/4 = 1 So 1000 + 250 + 62 + 16 + 4 + 1 or 1333 bars in total. Then he is out of money and only has 1 wrapper left.

[@unjugglable]

How much is a wrapper worth? 25 cents. So every bar he wants to buy he spends a dollar and receives a 25 cent "gift card". So his budget goes down by 75 cents per bar. 1000/0.75=1333.3333... With every bar being 75 cents, the 0.3333.... bars he has left is the final wrapper that's "worth 0.3333 of a 75 cent bar". So 1333 bars. Makes total sense 😊

[@SGKdi]

The first problem is very easy : The liar can never say "I'm not a truth-teller or I'm not a spy", otherwise, he won't be a liar. The only sentence that he can say and lie is "I'm not a liar". Charlie said : I'm not a liar ==> He is a liar.

[@nix_]

Well, the liar can also say "I am a truth-teller" or "I am a spy" and still be lying. And, technically, both the truth-teller and spy could say "I am not a liar" and be telling the truth.

[@SGKdi]

@@nix_ As per problem statement, there is only three affirmations : * I'm not a truth-teller. * I'm not a spy. * I'm not a Liar. He can't say the two first ones.

[@nix_]

@@SGKdi There are actually six statements total: I am a truth-teller I am not a truth-teller I am a spy ! am not a spy I am a Liar I am not a Liar The liar can LIE and say three of these: I am a truth-teller I am a spy I am not a liar

[@SGKdi]

@@nix_ The problem statement says exactly : "Alice, Bob and Charlie are one of each type: a truth-teller (always tells truth), a liar (always lies) and a spy (can lie or tell the truth). Alice says she is not a truth-teller, Bob says he is not a spy, and Charlie says he is not a liar : What type is Charlie ? " Where do you see the six statements you are talking about?

[@davidstigant9466]

@@SGKdi Ok, I was going to come on here and say that Charlie DIDN'T say either of the first two statements but that doesn't mean that he CAN'T say them, so your reasoning is wrong. However, I think what you're actually saying is that there is a liar who made one of the three statements. The liar can't have made either of the first two statements, so they must have made the third statement which means that the liar must be the third person (who happens to be Charlie... if that's his real name).

[@mrkeller8000]

The first problem has a much easier solution than the one you presented. Each of the three characters says they are not any particular role. We know from the information that one of them is a liar. Given this, the one that says they're not the liar will always be the liar. This is because anyone claiming to not be any other role cannot be the liar. EDIT: This question could be made more complicated by having some answers be "I am.." rather than "I am not..", which may force a more complicated solution.

[@christheother9088]

(n - n/4) x $1 = $1000 (n is total bars, every 4th bar is free) 3/4 X n = 1000 (units cancel) n =1333

[@GaminghunterYTOfficial]

I did 333÷5 instead of 333÷4 . I had not included the day 1 strategy or smthn

[@krabkrabkrab]

1+k+k^2+k^3+... is a well-known sum, it is 1/(1-k). In this case, k=1/4, so sum is 4/3. Now multiply by 1000.

[@snowfloofcathug]

I like the easy formula of the second, I was doing 1000 + 250 (1000/4) + 62 (250/4) + 15 (62/4) + 3 (15/4) + 2 (10 leftovers / 4) + 1 (4 leftovers / 4) = 1333, not the easiest to keep track of but hey it works

[@SysFan808]

pretty sure for charlie it's either 1250 or 1000+1000/4+1000/16+... repeat until less than 4 bars, depending whether the shop gives wrappers with the promoted bars.

[@sparkyshore3543]

I almost got fooled by the second riddle but I caught my mistake in time. Saying this before watching for the answer: If my math is right, Charlie (who is or whose namesake is the liar) can get 1333 chocolate bars. ETA: Halfway through the video I thought I'd double counted, but I hadn't.

[@Cheezymuffin.]

another fun way of doing the second one is like this: for 1 dollar, you get 1 chocolate bar, which comes with a wrapper 4 wrappers are 1 chocolate bar, so a wrapper is 0.25 chocolate bars, but then you realise, that the chocolate bar from the 4 wrapper would also be 0.25 chocolate bar, meaning you add 1 + 0.25 + 0.25*0.25 + 0.25*0.25*0.25 etc, meaning you could write it as the sum of 0.25^n with n from 0 to infinity, this is actually equal to 1.333......., meaning for 1 dollar, you get 1.333.... chocolate bars, and for 1000 dollars you would get 1333 chocolate bars

[@noahblack914]

Liked your way of solving problem 2. Didn't occur to me that the math is easier if you consider that after the first day, you get chocolate at a rate of $3 for 4 bars, rather than starting with a rate of $4 for 5 bars and then later trying to account for the extra wrappers

[@feynthefallen]

You know, I've lately developed a taste for this kind of problem. Surprising it took me that long, considering how much I enjoyed the numerous IQ tests I went through at school (at least the upper end - they invariably resulted in "above the scale", except when they gave me one for adults in 3rd grade, but that I was deemed unable to complete because I wasn't schooled enough to tackle the math problems)

[@shambhav9534]

I thought of the second one as a series. 1000 + 1000/4 + 1000/4/4 + 1000/4/4/4... That is 1000(1 + 1/4 + 1/4² + 1/4³...) If you solve for 1 / 1/4 + 1/4², you get 4/3 = 1.333... Multiply that by 1000 to get 1333.333, and round it down, making it 1333. As a general solution, n/(n - 1) = 1 + 1/n + 1/n²...

[@Wink-Wright]

I did the counting method to confirm my answer, but initially I i just modeled the equation. Total cost = money invested + money returned $1000 = x + x(-0.25) $1000 = 0.75x 1333.33 = x Rounding down, that's 1333 bars.

[@sagittariusa2008]

After seeing the long calc for chocolate, I noticed if you divide the 1K by 4, then the divide the answer by 4 etc another 5 times you arrive at almost 1. Then sum the results with the 1st 1K you get 1333.008. Coincidentally close enough. Doesn't work so well with 5 or 3 wrappers.

[@aidanbrumsickle]

For the second one: bars = 1 * dollars + ¼ * bars bars - ¼ * bars = 1 * dollars ¾ * bars = dollars bars = 4/3 * dollars = 1333

[@dyslexicswift4007]

For the second one I realized by him receiving a chocolate from wrappers after spending $4 this adds to the total by him receiving another wrapper etc.

[@osmanbadroodin3215]

I done the chocolate one like this (W=wrapper / C=chocolate) 4W = 1C Therefore W=1/4C 1/4+(1/4^2)+(1/4^3)+.... =0.3333.... $1000=1000C 1000C + (1000C×0.3333....)=1333.3333.... Rounded off =1333C Therefore he gets 1333 chocolates with $1000

[@AndreSomers]

Didn't watch the video, but I think 1333 bars. I arrived at it like this: initially, he can buy 1000 bars. Those 1000 bar wrappers yield another 250 bars. Those wrappers can be traded for another 62 bars, keeping 2 wrappers. The 62 bars + 2 leftover wrappers mean another 16 can be had. These 16 yield another 4, and these 4 allow him to get 1 final bar. 1333 in total.

[@tannerarmstrong1496]

My thinking was that each wrapper was essentially worth $0.25. Therefore, being given a wrapper when you pay a dollar was essentially the same as the price being $0.75. $1000.00 / $0.75 = 1333.33 (repeating of course). You can't buy a fraction of a bar so the answer is 1333

[@bledlbledlbledl]

didnt bother with much formula, just took it day-by-day: 1) 1000 dollars -> (1000 bars, 1000 wrappers) 2) 1000 wrappers -> (250 more bars, 250 wrappers) 3) set 2 wrappers aside because you can't use them yet 4) 248 wrappers -> (62 more bars, 62 wrappers) 5) pick up those 2 wrappers you set aside becaues you can use them now 6) 64 wrappers -> (16 more bars, 16 wrappers) 7) 16 wrappers -> (4 more bars, 4 wrappers) 8) 4 wrappers -> (one more bar, one more wrapper) total) 1333 bars, and one wrapper left over.

[@Dragondog-Gaming]

I solved the first problem in exactly the same way. And I solved the second problem slightly differently to make it easier to count without using a calculator.

[@MorpheousXO]

Woo! For once I was able to figure them both out in my head!

[@lethalty6055]

For the first problem, I just said liar before the logic comes into play, and I lost it when logic and process of elimination happens.

[@OrenLikes]

After the first chocolate bar, for every 3 he buys, he gets 4. So, the total number of chocolate bars he can get (g(n)) is floor[(4n-1)/3]. Here, n=1000 (units as 1/1$), so, g(n) = floor[(4n-1)/3] = floor[(4000-1)/3] = floor[3999/3] = floor[1333] = 1333. Additional useless/pointless checks: n=1, g=f[3/3]=1 n=2, g=f[8/3]=2 n=3, g=f[11/3]=3 n=4, g=f[15/3]=5

[@ADudeWhoDo]

The way I did the chocolate bar problem was buy the first four bars for the first promotional bar and and from then on buy three bars at a time using the previous promotional bar wrapper for the last wrapper to get the next promotional bar. This means that the first 5 bars are 4$ and then the rest are 3$ for 4 bars. My math went as follows: 1000$-4$=996$ 996$(4bars/3$)=1328bars 1328bars+5bars=1333bars

[@olerask2457]

You need only buy 1 chocolate first for 1$. Then you buy 3 for 3$ and get one for free. That is 1 + 333*(3+1) = 1333.

[@lycansan]

I got it all right even im just waking up in the morning😂😂😂

[@zushyart]

Each person says they don’t have a role. The liar lies about not having their role, meaning it is theirs, meaning the liar must be the one saying they aren’t the liar. The truth-teller must be telling the truth about not having a different role.

[@Gretchaninov]

Nice problems, but pretty easy. Took me less than a minute to do both.

[@nuclearcommando9729]

Before watching: assuming all wrappers are valid, 1333 1000 gives you 250 250 gives you 62 with 2 leftover 62 gives you 15 with 2 leftover, or 16 with the previous 2 16 gives you 4 4 gives you 1 EDIT: yay I actually got something right

[@SteBar3000]

There is only one statement a liar can make, which is 'I am not a liar.' Therefore, we can be certain that Charlie is the liar.

[@harpoon2445]

I was about to write just that, and then saw your comment already. The solution is indeed much easier as in the video, since the liar can’t say any of the first 2 statements.

[@Zygnity]

His name literally has the word “lie” in it

[@michaeledwards2251]

Due to the fact the problem was composed to be soluble, forcing there to be only 1 person who stated they were not a liar. If the spy or truth teller had said they were not a liar, the problem becomes insoluble. Both the truth teller and the spy could say they were not a spy, Charlie would still be known as the liar, as he would be forced to say he is not a liar.

[@robgrune3284]

re chocolate. the answer imputes a rate of 1 bar/day, which is not mentioned in the problem's text. the problem states a purchase of $1000 total, which in a real-life purchase means a 1x purchase, not a piecemeal daily rate. after 4 bars @ $1/bar, I receive 5 bars. ergo the average price per bar is $4/5=$0.8. $1000/$0.8= 1250 bars.

[@AshaTanwar-b9w]

1250 bars

[@justforthis3208]

For the chocolate bars it's straightforward. Think about them in sets of 4. First set of 4 Charlie buys costs $4. Then he gets a free one. So the next set of 4 cost $3. And he keeps getting a free one. So any set after that will cost $3. So with $996 he can buy 332 three dollar sets of 4. 4*332=1328. Add on the $4 set 1332. Here's the trick on the last set of 4 he buys he gets one final free bar. So 1333.

[@TocoaPuffs]

I did the chocolate bars puzzle from the thumbnail. 1,333 chocolate bars. I just bought a thousand and then kept using wrappers to buy more until I had one wrapper left.

[@TocoaPuffs]

Charlie is the liar as well. Alice needs to be a spy. Since Alice is the spy, Bob needs to be telling the truth with that statement. Leaving Charlie to lie about being a liar. Which is fitting for a liar.

[@TocoaPuffs]

I figured out the "shorter" way to solve the chocolate bar problem first, but couldn't figure out how to put it into a math problem. So I just did the brute force way haha

[@quixadhal]

It depends. Are chocolate bars taxable in your jurisdiction? Also, since you're gonna have to mail in the wrappers and wait 6 to 12 weeks for a debit card or free candy bar, you have to factor in the price of postage. There's also the chance that the store will either put them on sale, or stop carrying them before the "free" credits arrive.

[@PopeVancis]

Yes, but can't you just go to the store and trade them in instead?

[@bentonrp]

Lol! Are they Duy Free? What if the factory only manufactured production at the MOQ of a 1,200 bar limit. ...That would certainly pepper his pourage...

[@Xi-Teo]

I literally remember getting this in year 5! I’m pretty sure got it right!

[@markwhitis]

It is almost a consistent finite series. But you can't get a fractional bar so at one step you have two leftover wrappers which you can use at the next step. 1000+250+62+(15+1)+4+1 = 1333

[@zpyo27]

I think this is the first MYD video I got everything right!!!

[@dustyoldduster6407]

I don’t know, but my cell phone reception would be through the roof.

[@btf_flotsam478]

For the second problem, Charlie can get one extra chocolate bar via trading for every three wrappers he has, provided he has an exra wrapper to start with (which gets 'refunded' by gettingan extra wrapper). After buying 1000 chocolate bars, he has 333 sets of 3 wrappers and the extra wrapper required to trade them in.

[@SilentDecepticon]

I got an ad for Cadbury chocolate during this video 😂

[@fleetadmiralj]

Okay for problem 1: Alice must be the spy because if she were the truth-teller, she can't lie about not being the truth teller. And if she were the liar, that means saying she's not the truth-teller means she is the truth-teller, which is a contradiction. Bob must be the truth-teller because if he is the truth-teller, saying he is not the spy is the truth. But like with alice, if he were the liar, and says he is not the spy, that means he must be the spy, which is a contradiction, and we also know he can't be the spy because Alice is the spy. Charlie is the liar because the liar saying he is not the liar is what the Liar would say. So: Alice - Spy Bob - Truth-teller Charlier - Liar (so I guess answer is B) Problem 2: You can buy 1000 bars for $1000 and get 1000 wrappers But then you can get 1 bar for 4 wrappers, which gets to 250 more bars (1000 / 4) But wait, you can get even *more* bars from those 250 free bars. In this case, you can get 62 bars, with 2 wrappers left over (250 / 4) Repeat the process: 62 bars gets you 15 more bars (62 / 4) with 2 wrappers left over. But now we have 4 wrappers left over which gets you 1 more additional bar for a total of 16 16 bars gives you 4 more free bars (16 / 4) Those 4 bars get you 1 more free bar (4/4) So the total number of bars is 1000 + 250 + 62 +16 + 4 + 1 = 1,333 (with 1 wrapper left over) Now to see if I was right! Woohoo I got both right

[@knrdvmmlbkkn]

At first 1.000 chocolate bars. 1.000 wrappers gives 250 new bars. 248 wrappers gives 62 new bars (with 2 of the 250 wrappers left over). (62 + 2 =) 64 wrappers gives 16 new bars. 16 wrappers gives 4 new bars. 4 wrappers gives 1 new bar. Sum 1.333 bars at an average price of slightly more than 75 cents.

[@q.e.d.9112]

Haven’t looked at the video, but the chocolate bar one is easy. He spends 4 bars, hands over the wrappers and gets his free one. Then he buys three more and uses those wrappers plus his “free” wrapper to get another free one. From here on he can get a free one for every three he buys. He’s still got $993 to spend, with which he can buy 331 more lots of three for another 331 free bars. Add em all up: he’s got 1,333 choc bars for his $1000.

[@vamsiCAM]

You can also solve the 1st one like this Alice says she is not a truth teller. So, if she was either a truth teller or liar, it would create a paradox. So she is a spy. Bob says he isn't a spy, which is true as Alice is a spy. So bob is a truth teller. Since only the liar is left charlie is a liar

[@originalhgc]

I would really be interested to see how to solve the chocolate bar riddle with an integral.

[@LiamHighducheck]

I feel like an easier way to do the last one would be to see it as you get 1 free bar for every 3 wrappers since you spend 4 then get 1 back. Then it’s just 1000/3 for how many extra chocolate bars you can get (round down)

[@LiamHighducheck]

@@davidwalton3604 can you give an example?

[@LiamHighducheck]

Ahh I see what you mean, it only works with numbers that aren’t divisible by 3. So I guess you’d have to subtract 1 first before making the calculation if it is divisible by 3.

[@chrish7336]

you would be close but off by 2. Doing it out the long way, you get 1000 for $1k, from those wrappers you get another 250, From the 250 wrappers you get another 62 with 2 extra wrappers, that 62 gives you another 15 plus 2 extra wrappers, and the 15 give you another 3 plus 3 extra wrappers. So your total is 1331 with 3 extra wrappers. 1331 bars Max for $1000

[@LiamHighducheck]

@@chrish7336 that’s incorrect. The answer for $1000 is 1333 the explanation for which can be seen in the video. Where you went wrong is 1000+250+62+15+3= 1330 but through all of that you had 2+2+3+3 = 10 extra wrappers. That gets you 2 more bars and 2 more wrappers leaving you with 4 wrappers to buy one more. This gives you your 3 extra bars for a total of 1333

[@chrish7336]

@@LiamHighducheck Yeah I saw that later after the post, while I followed the same process I didn't take into account converting the 4 remaining wrappers right away at 15 making it 16. This definitely would have made the answer 1333. I think I actually made that correction in a later post when it was pointed out.