Matthew sent me this tweet, which has people’s knickers in a twist. It looks easy, right? Three equations, three unknowns, and you don’t even have to combine them. But if you look at the thread after the tweet, the answers are all over the map.

Actually. there are two correct answers, depending on where you put the parentheses in the last equation. But I advise you to look carefully, and then answer below.

Mate this is hilarious

GO! pic.twitter.com/fK3av04Zj4

— Dr. Jessica Taylor (@DrJessTaylor) March 28, 2020

1st line: each shoe is 5

2nd line: each person is 5

3rd line:each cone is 2

4th: group x before +, answer: 15

Of course because a pair of shoes is worth 10 needn’t mean a single shoe is worth 5. Try selling a single shoe.🙃

What indicates that it is a pair of shoes, rather than just two shoes?

And the correct answer is 15.

5+11×3=38

Order of operations allows for only one answer, because there ARE no parentheses.

Correct, however (parentheses) 4th line:

A. 5 + (5 x 2) = 15

B. (5 + 5) x 2 = 20

There are no parentheses so we go with order of operations. 15.

English grammar school maths c.1960 took a different view!

Besides, our host warned about (parentheses).

Further, looking further down the thread, I am probably wrong anyway…

I think it was a red herring since there are no parens!

No parentheses, but still order to operations…

Order OF operations…

Order OF operations…

I really dislike these kinds of “puzzles”, because the symbols are too ripe for misunderstanding.

In your case, you failed to notice that the person on the final line is holding two cones. So what does a person holding two cones stand for? We don’t know, if we treat the equations formally. But we’re probably meant to think it’s a person (5) plus two cones (4).

That means the final equation is 5 + ((5+4) * 2) = 23

In reality, it means nothing, because the only symbols reliably defined are the pair of shoes, the person, and the pair of cones.

He is also wearing the shoes so he might be worth 19 and the answer is 43. Without guidance as how to treat the combinations, the puzzle is meaningless.

If you multiply items in combination, the answer is 405. But who knows? Maybe the shoes and cones in combination with the man raise his value to a power.

I did not see the shoes.

*Sigh* Gets out a pencil again…

Agreed, the symbols and perhaps operations are undefined in the last line.

It could be, for example, that two shoes is a shoe times a shoe. So one shoe is sqr root 10.

But, assuming that a composite picture is addition you get…

two shoes = 10 (thus one shoe = 5)

two boys + two shoes = 20 (thus one boy = 5)

four cones + boy = 13, (thus one cone = 2)

and thus boy wearing two shoes and with two cones = 5 + 10 + 4 = 19

You get

5 + (19 * 2) = 43

But without the assumptions, which are lacking in foundation, you get nowhere.

Also wearing a pair of shoes.

I thought along the same lines as you but included some addition ideas. Here’s what I posted to my smug facebook friends who always thought they had the right answer:

1. In general math, two symbols that are written next to each other are multiplied.. so take your vanilla quadratic equation:

ax^2+bx+c it’s properly read as “a” times “x squared” added to “b” times “x” added to “c”. Since the first line included the “+” operator, it seems to be written in that style.

2. If the two shoes are actually multiplied the value of each shoe could be different. The left shoe might be 2 and the right shoe 5 or the left shoe might be 1 and the right shoe 10 (and visa versa)… that changes the value of the final equation in both cases. The value pairs would be [(1,10),(2,5)] and that’s if there are no rational number.

3. The same condition could exist for the cones (I have no idea what they are, looks like black caviar)… the cone on the left could be 1,2 or 4 as could the cone on the right so long as the product is 4. The value pairs would be [(1,4),(2,2)].

4. In the fourth and final “equation” the middle “symbol” is a conglomeration of the boy, both shoes and both cones. The “trick” in the meme is that an observant person would note all three symbols together (which I admit I failed multiple times). However, there is no mathematical standard to derive what the operation should be if they are all combined into one symbol… If I were writing a computer program to try and solve this, the middle symbol would be a completely new variable. In addition, the multiplication and addition in the final formula is explicitly stated so it’s really impossible to assume anything (maybe writing them on top of one another should be exponentiation?).

5. The “correct” answer is 43, which assumes addition for all objects shown together. In addition, it assumes both objects are the same value despite there being a way to differentiate one object from the other (there is a left shoe and a right shoe, a “front” cone tilted differently from the back cone…etc)… If this is the were the case, even assuming addition, again the shoes could take on the value pairs [(0,10),(1,9),(2,8),(3,7),(4,6),(5,5)] and cones could be [(0,4),(1,3),(2,2)].

In conclusion, the only way to derive the “correct” answer is to know it apriori… because if you wrote down all the possibilities that fit with the values shown you would already have multiple solutions. As far as I’m concerned, logically and mathematically any answers that utilize the assumptions consistently are as “correct” as 43… and besides, everyone knows that 42 is the actual answer.

But how many cones is the person in the 4th line carrying?

Nope: 43! Just noticed the person in last line is wearing shoes and holding two cones!

28.125?

Sorry, 48. I misvalued the cones.

If you go by standard convention of “multiply first, the answer is 20.

If you don’t (do the addition first), it is 60

This is really:

3x = 30

2y + x = 20

2z + y = 13

Then, using conventional rules:

x + yz = ?

OR, unconventionally

(x+y)z = ?

ack..didn’t pay attention..KD 33 is right:

6x = 30

2y + 2x = 20 (so x = 5, y = 5

4z + y = 13 so z = 2

NOW x + yz = 5+ 5(2) = 15 OR

(x+y)z = 10(2) = 20

49

76

23?

This boy is naughty 🙂

Uh, what about the cones the person is holding in the last equation?

43

OK… so where do we put our answers?

I guess here.

48 or 43.

Ha ha… ha.? What’s I do wrong?

What’s the joke? I don’t get it.

There are lots of subtle things that are easy to miss here. We have to start with an assumption about what two shoes mean. Are they just added? Given that the problem writes out a multiplication symbol, this is a reasonable assumption.

The first line, then gives us shoe = 5. The second line, we substitute that (2m + 2s = 20, 2m + 10 = 20, m = 5).

The third line is then (4r + m = 13, 4r + 5 = 13, r = 2) (r for rice).

Then the last line tries to trip us up. Notice that the man is also holding two things of rice. I interpreted this as (s + (m + 2r)*r, or 5 + (5+4)*2) or 23.

And, I of course, also missed that the man is also wearing a pair of the red shoes. That would be (s + (m + 2r + 2s)*r) or 43.

There is also a sense where multiplying by count of a specific object doesn’t really make sense.

And, I suspect I’m still missing something in the picture.

I didn’t notice he also wore a pair of the red shoes!!!

I contemplate that the red-green color blind math Ph.D. person (male? or is that a social construction?) may be at a disadvantage.

I get either 28 or 23

1 shoe = 5

1 kid = 5

1 bag = 2

(5+(5+4))x2=28

or

5+((5+4)x2)=23

PEDMAS suggests multiply first so my best guess is 23

oh damn, he’s wearing sneakers. Forget what I said.

1 shoe = 5

1 kid = 5

1 bag = 2

(5+(5+4+10))x2= 48

or

5+((5+4+10)x2)= 43

So 43?

An unpaired shoe is worthless, and so are the snow cones once that kid gets his paws on them. On the other hand the kid is nicer when he’s got two treats in hand, so +2 points for that. Applying KD33’s math for the rest, I get 7 x 2 = 14.

25

35

81

23

One shoe = 5

One Trump = 5

One cone full of turds = 2

One shoe + (Trump with two cones) * one cone

43

(I don’t like that pair of dorky shoes.)

I got 30. Because there are no parentheses given, and following PEMDAS, the multiplication should occur prior to the addition.

3x=30 -> x=10

2y+10=20 -> y=5

2z+5=13 -> z=4

10+5*4=30

Upon a closer examination, I can see my previous effort is incorrect. I didn’t pay close enough attention to the number of items depicted in the cartoon. Reassessing now!

I now think the answer is 43. I agree with Sue above.

There is a rule in math that without parentheses you do a multiplication operation first and an addition operation second. The answer is 15.

Yes. There is only one correct answer to this puzzle.

The answer is 23

Each shoe = 5

Each person = 5

Each cone = 2

The final equation is:

1 shoe + (one person + 2 cones) * one cone

= 5 + (5 + 2*2) * 2

= 5 + (5 + 4) * 2

= 5 + 9 * 2

= 5 + 18

= 23

There are not two answers, for standard mathematics always multiplies before adding.

Well shiver me timbers! He IS wearing shoes on the last line!

The answer is 43

Each shoe = 5

Each person = 5

Each cone = 2

The final equation is:

1 shoe + (one person + 2 cones + 2 shoes) * one cone

= 5 + (5 + 2*2 + 2*5) * 2

= 5 + (5 + 4 + 10) * 2

= 5 + 19 * 2

= 5 + 38

= 43

There are not two answers, for standard mathematics always multiplies before adding.

Don’t forget about the kid wearing the shoes!

I also agree with KD33, it’s 15. However, that assumes that left and right shoes are of equal value, for which we have no direct evidence. If I claim that one shoe is worth 2 and the other is eight, I could satisfy all of the given equations and arrive at 12 or 18 as correct answers. Also, I think the claim that two answers are possible depending on how we place parentheses is wrong. We only use parentheses when we want to force an execution order other than the convention of multiply first, then add.

Now I see my mistake. I failed to notice that the boy is holding two cones. That gives us 2X9 + 5 = 23. However, now we are assuming that the little cones the boy holds are equal in value to the giant, free-standing cones.

And you are failing to notice he is wearing the shoes.

Indeed – I failed to spot what the kid was wearing and holding in the final row, despite the instruction to look carefully. Epic fail!

Where is the official explanation of this joke?

Sorry, posted my answer in the wrong place.

Pretty sure it’s 42.

What are those cones? Paper cones with raisins? Who would even

wantthat?I think it’s caviar.

I traced their outlines, then colored them with crayons. I like the smell of crayons. I’ll probably go take a nap now. 😎

Yum. Crayons.

Sub

The correct answer is 793. How do I know? As I was contemplating the fact that just about any number could be the “answer” to this stupid puzzle, that number popped into my head. Obviously a revelation from God. Don’t bother to challenge me with any facts or logic; they are useless against God.

Clearly it’s 42.

You are listening to a false god. My tru god spoke to me from a burning trash pile to tell me that the answer is 666.

This is really a “Spot the …” problem! I’m so unobservant that I initially didn’t even notice that the paired objects in the first three lines were single in the last line. I then belatedly saw that the kid was holding a pair of one of the other objects (whatever it is). I only realized he was wearing the shoes until I started reading the comments.

Making reasonable “mathematical” assumptions (not entirely realistic ones — I agree that one shoe is not worth half the pair in real life) I concur with the answer of 43.

I find my failing to notice the things above interesting. It’s due to assumptions I automatically made of what a system of equations will look like—and of course the problem playfully sabotages those assumptions.

5 + 19 x 2 = 5 + 38 = 43

assuming we can just make the sum of that naughty boy and the items he is carrying/wearing.

I’m less interested in the answer and more interested in knowing which cuisine or culture serves black beans in a newspaper cone? That’s the weirdest street food ever.

I don’t get the logic behind this. It’s simply an open problem with an infinite number of possible interpretations or values (or varying values) for the items.

I’d simply concur that the answer is a red question mark.

I may have fucked up royally, but without looking at anyone else’s answers I came up with:

either 41 or 48 depending on where the brackets are placed for the final sum.

I’m usually shit at these things so I expect I’ve missed something glaring.

…Sorry, 43 not 41.

Still not checked though. Tenterhooks

2(SHOE) + 2(SHOE) + 2(SHOE) = 30 <- add SHOE

6(SHOE) = 30 <- divide both sides by 6

SHOE = 5

BOY + BOY + 2(SHOE) = 20 <– sub 5 for SHOE

BOY + BOY + 2(5) = 20 <- simplify

2(BOY) + 10 = 20 <– subtract 10 from both sides

2(BOY) = 10 <– divide both sides by 2

BOY = 5

2(CONE) + 2(CONE) + BOY = 13 <- Sub 5 for BOY

2(CONE) + 2(CONE) + 5 = 13 <- subtract 5 from both sides

2(CONE) + 2(CONE) = 8 <- simplify

4(CONE) = 8 <- divide both sides by 4

CONE = 2

SHOE + ((BOY + 2(SHOE) + 2(CONE))CONE <- sub #s

5 + ((5 + 2(5) + 2(2))2 <- Simplify

5 + (5 + 10 + 4)2 <- Simplify Parenthesis first

5 + (19)2) <- multiply first

5 + 38 <- add

Answer is 43

30

NO! 15! This is a damn trick question in more ways than 1!

Damn again! Answer is 43.

I had to squint to see.

I’m going 43.

It’s 43 if you look closely at the boy in the last line: he’s wearing a pair of red snickers and holding two cones of what I called blueberries. And if you assume the order of operations, as we were taught in our (Polish) Maths classes from the kindergarten through the Uni. If you don’t assume that multiplication “trumps” (pardon my French) addition, you get 48.

But it’s a pretty tricky question. The first line, which consists only of the red shoes, could be written as 3*R+3*L=30, to make it more complicated. The right and the left shoes don’t have to have the same value, I think, but it would make the puzzle much harder.

All of the terms in the last equation are undefined, and so is the answer. Don’t argue.

28 is the answer is functions are simple sequential ones.

2(5)+ 2(5)+ 2(5) = 30

5+5 + 2(5) = 20

2+2 + 2+2 + 5 = 13

5+(5+ 2+2) x 2 = 28 OR

5+ (5+2+2)x 2 = 23 OR

The small cones held by the boy are meant to signify some other math function than addition – exponential factors, division?

WRONG! see corrected response.

38 is the answer if functions are simple sequential ones.

2(5)+ 2(5)+ 2(5) = 30

5+5 + 2(5) = 20

2+2 + 2+2 + 5 = 13

5+(5+2+2+5+5)x 2 = 38

The kid is holding cones and wearing shoes

I would argue that multiplication takes precedence. But 38 is wrong whatever. The kid with the shoes and cones is worth 19 by himself. So with multiplication taking precedence that’s 5 + 19 x 2 = 43. If you simply do the operations left to right, it’s (5 + 19) x 2 = 48

43 is correct for simple addition. I had forgotten to add in the first kid (5) before.

It’s 74.

One shoe = 10

One man = 5

One licorice popcorn = 1

The man is holding two licorice popcorns, is wearing two shoes, and is five himself. The man at the end is 27. Add 10 for one shoe. That’s 37 x 2. 37 x 2 = 74. It’s 74.

It’s 43. My previous answer of 42 was a smart-ass

comment without even looking at the problem. Hitchhikers guide…

42 might be the meaning of life, but 43 is a fun number. It is the 14th smallest prime number twinned with 41. It is also the largest non-McNugget number.

https://mathworld.wolfram.com/McNuggetNumber.html

🤓

My sister gave this to me two days ago with different numbers and it took me three times to get it wrong. I thought I had it on this one but overlooked the shoe being five instead of 10. I also added the man’s total everything (19) to the five for the shoe instead of multiplying x 2 first. It’s 19 (total man with licorice popcorn and shoes) x 2 = 38. 38 + 5 = 43. Order of operations. Please excuse my dear Aunt Sally.

Looking over the answers here takes me back to my last year of college when I was a grader for a math prof. I thought it would be easy because in math the answer is right or wrong, right? But the prof insisted that I go through the work (which they had to show) of anyone with a wrong answer and show them where they went wrong. My god that was a miserable job.

At least he didn’t make you go through the work of anybody with a right answer to show them both places they went wrong.

😀

what if we let a shoe be s, a man be m and cone be c

ss +ss + ss = 30 => 3 s^2 = 10 => s = sqrt(10) …

+- sqrt(10)

Nobody yet did the following, so before the old man goes to bed ,

CLAIM: There is no unique solution, all sorts of possibilities, nothing to do with order of operations at the end. But just double the number of solutions if you want to fuss about that and allow addition sometimes to precede multiplication without using the customary brackets.

So we have 3 equations to solve, no dispute there.

Except in ‘unusual cases’, 3 variables should yield a single solution when 3 equations, any different number of variables normally would yield none or more than one solution. Not always, but here that’s not an issue.

At first sight it looks like there are 3 variables, namely Twerp (T, I dislike his eyes), Cone (C, the pair of cones we’ll just treat as two of them), and pair of Red shoes (R; we’ll treat the single red shoe as having value 2 x R, okay, thinking the numbers are counting.) You can alter how you do pairs if preferred—talk about red herrings!

Now clearly the red shoes on T at the end must be significant; or else why do that? It would be utterly inconsistent therefore not to have a 4th variable, (B), namely THE BLACK SHOES ON T in the equations.

So now we have 4 variables, not 3, namely (B, R, C, T). And without checking, it seems likely there will be infinitely many solutions. Now as long as you are asking for real numbers as solutions (or even rational numbers, or integers, possibly negative) there will actually be infinitely many solutions. Do it! If you require only non-negative integers, it won’t be infinitely many, but still not unique I think. Do that!

If you don’t like the two colours of shoes, there’s always two, now different, Twerps wth shoes disagreeing, again 4 variables.

At risk of being labelled as a Pecksniff by a sometimes over the Topp respondent, I’d say the annoying thing about this type of puzzle to a mathematician, if ever used in a classroom, is that a much brighter than average student may well become confused. Do it often enough and she (or he) will decide she is no good at mathematics, when the opposite is probably true. But there may be no one there to re-encourage her. All there might be is a teacher who insists that deviation from whatever that teacher has seen before will result in a shitty mark.

A brief explanation: variables must be named as very clear ‘marks’ so that each new occurrence is clearly the same variable, and each different variable is a clearly sufficiently different mark, so as to not be mistaken for the earlier one. Obviously; otherwise use whatever you like as variables; e.g. nothing to it, just use x, y, z, etc. There’s nothing sacred about them, just names for things which you wish to ultimately know in the usual form, like 3.14 (not=\pi!), so you give them some such names, x,y, etc. so that you can manipulate with them to hopefully get an answer. Elementary algebra is exactly that, no more and no less, as a step up from arithmetic and the beginning of actual mathematics.

But it’s fun for people uninterested in education at that moment (like maybe getting their mind off 7 deaths per million of population, and rocketing up very fast). So I enjoyed it as much as anybody else. But jeez, don’t use something like that when you’re trying to get a young person interested in science and mathematics.

Imo the BLACK shoes (like the Norwegian Blue’s plumage) don’t enter into it.

But, for you, the red shoes do enter into it because they are explicitly in the three equations, I assume?

Fair enough, but counting what’s on Twerp’s feet in the question to be answered seems to be just as questionable, I’d say.

But the name of one (or is it two?) of the variables in the whole thing has suddenly changed if you look closely enough at his feet. So suppose you’ve got some algebra text with a question in which ‘x’ occurs a few times as one of the unknowns. But then “X” seems to occur a bit later; as another one? or the same one?. And actually it’s a lot smaller than the “X” I could print, it’s just about the size of “x”. But “x” is just a bit curly compared to what will likely show up. As a textbook, I’d dump it, if stuff like that was not just a single screw-up.

The point is that, however entertaining, and it is and it is welcome any time, but particularly now as a diversion, the thing is just full of red herrings, diversionary tactics, ambiguities.

It ought not to be called a “MATH teaser”, to quote our gracious host, who is probably far from the first person to call it that. It’s a ‘let’s see how many ways we can make this as ambiguous as possible teaser’; but fun.

I should have added that this makes reading peoples’ answers very boring in all but one aspect: when somebody comes up with a new ambiguity, or a new interpretation of an old one. Then it gets even more amusing to me, seeing that creativity and sharp-eyed observation.

Red- and black- herrings😬

I would agree with you. I have been calculating the death rate of the coronavirus globally for about a week or so and when I saw this problem (with different numbers and sunglasses instead of licorice popcorn, I took a welcome break. Percentages aren’t really that difficult but I saw this as a relaxing activity during everything else and literally saw the multiplication sign as an addition sign in my haste to have the correct answer over my family members. As some of my family members are on the front line and didn’t participate, my little sister “bragged” to her co-worker that I would get the answer in less than a minute because I was a “savant.” Whatever the hell that is. I am not. I got it wrong many times and I was using the wrong order of operations at the end anyway (despite all of the nonsense confusion of the man holding or wearing these other items). Do you remember the continental math league tests? There were two people in my grade who always got 5s and 6s and I was with them with 4s, 5s, and 6s. My public school system was always competitive and I was always at the top. In eighth grade my mom took me off the track with all of the smart people I was competing with and put me in the dumb math class. I went from straight As to straight Cs because of the average fails of doing nothing. I never recovered from that. It took me over ten years to figure out what went wrong. My older sister had a hand in convincing my mom to put me in the average math class. My mom said, “Your sister is math/science and you at English/social studies.” Years later I found out my older sister abused my mom in sixth grade for putting her in the not most advanced class. I found this all out maybe three years ago. I stopped trying in 8th grade because I was with the dumbest people on the planet for math and science and so I knew someone had to have thought I was dumb. My mom told me once that the difference between my older sister and me was that she would crawl over her own dead mother to get what she wants and that I have a heart. Well, my sister is in on the front lines now and let’s just not hope that we reap what we sow.

The black shoes are actually socks.

He’s wearing the red-laced (Converse possibly) sneaker shoes in the last picture.

I know.

Now you’ve made it even more interesting:

There’s a famous explanation of the Axiom of Choice in axiomatic set theory.

With shoes, if you had an infinite sequence of pairwise distinct pairs of shoes, you could always find a set with exactly one of the shoes in each pair as an element of your choice set; just choose, say, the left one. Instead, more interestingly, you could choose the left one from pairs 1st,3rd,5th,7th, etc., but choose the right shoe from pairs 2nd,4th,6th, etc.

With socks however, even though it is known that the black hole at the middle of the

Milky Way contains all the many missing socks, the lost one from each pair which came out of the clothes dryer as only one (now useless) sock, this joke doesn’t help, but it does explain why the black hole has the mass of millions of suns.

It’s only with the help of the Axiom of Choice that you can assert the existence of a set of socks which contains exactly one sock from each pair.

Well, I’m sort of doing a diversionary tactic here, but surely in this problem you now realize that the BLACK socks can be counted, but the left RED shoes must be counted separately from the right ones (and the wrong ones must be ignored as a black herring).

So everybody, back to work to fix your solution, no time for navel-gazing!

This answer is certified to have done no harm to any living numbers, nor to variables, either alive or deceased–passed away–gone to meet their maker, …,… cf. Merilee’s shrewd remark about the connection of this problem to the Norwegian Blue parrot within whose plumage was found a nail.

✔️+ for philosophy of mathematics!

(I do think so but am not completely qualified to make that judgement.)

I thought that the single socks made up the rings of Saturn??

Well, in today’s NYTimes we have

“Infinite Visions Were Hiding in the First Black Hole Image’s Rings”

So maybe there’s hope for all those bereft socks in my drawer missing their spouses! I knew a drawer with infinite capacity was needed. Maybe uncountable infinity.

Disclaimer: Just in case, this is a lame joke. There is no evidence connecting Saturn to Saggitarius. But recovering those socks is serious business.

Each sock is worth on hole.

I get 43.

Which is obviously wrong, since the answer to everything is 42.

Also, dunno about mentioning parenthesis since multiplication goes first.

I sadly confess that I haven’t a clue what any of this is about.

If you are going to comment in response to this post, it needs to be in a numerical format.

|x| ly

The absolute value of x and what does ly mean? I was in the retarded math class since 8th grade. I lived and loved for school until I was put with the retarded people. I didn’t do anything since then. I’m so sorry but I have no idea what you mean.

Absolute-ly. As per your request for numerical format, I commented in math symbols. I’m just being a smart ass. 😎

I’m sorry for your educational background difficulties. I was labeled somewhat suspect (a daydreamer) but decided to apply myself in the 7th grade. Everyone has a history of pushes and pulls in our schooling. I look forward to improvements so that people are not pigeonholed and misdirected.

“I was labeled somewhat suspect (a daydreamer) but decided to apply myself in the 7th grade.”

Just congenially curious – did you one day have a pedagogical epiphany and realize that larnin’ was good for its own sake and would create opportunities? Did anyone have to appeal to and make academic subject matter (e.g., science, math) connections to your pop culture interests in order to make academics “relevant” to you?

I see too many teachers give material rewards to students for doing well. I should think that doing well/outstandingly would be its own reward. Are they following some educational/psychological/management guru’s received wisdom?

I’m not sure what made me more attentive. It could have been a sudden rise in testosterone levels. I think it starts to kick in at about that age. I remember being attracted initially by geometry. I loved the way you could “prove” some relationship by assembling the right theorems in the right order. I seem to remember becoming a little more competitive and wanting to be seen as a smart kid.

I think you’re right that kids can be self motivated in the right environment. But, as we know, there are many personalities to motivate. There probably needs to be many kinds of motivators.

When I was four and my older sister was six, she bet me $50 of money I didn’t know existed that I would fail at violin lessons my heart desperately wanted to take. It was scary. She won. She would win again and again. When I was excited to join the high school lacrosse team eagerly, she told me that if I joined, she’d make my life a living hell. See post above. She cares about her successes and not about people’s feeling. It is what it is. At her neuroseogeory graduation we sat there and watched her colleagues “make fun” of her for making the nurses cry. Is that something to be admired? I suppose. Best of luck to her and everyone like her.

She’s worn me down so much at this point that I really don’t care if she dies of the coronavirus on the front lines. The scary thing that my family knows, though, is that she would take a healthy mask off of a nurse to save herself and let that nurse die. They live among us.

Sounds like your sister is a psychopath. Sorry about that. I hope you can get outside all that and live your own life.

That is the “nicest” and/or “kindest” thing I think you could say. I hope it’s not accurate maybe. Of course I would have always protected her to the death even though she’s always been such a terrible bitch to me in all other ways. After I learned of the math thing, though, and how she is in general, I really just have hope for her that she has some love or whatever in her heart. Some people maybe don’t have that. The rest of my family are all loving and caring so maybe she’s the black sheep.

I had a younger brother, black sheep type, who was always angry growing up – from the time he was a baby. He ended up dead, unable to adapt to the demands of societal life. Almost every family I know of has a spectrum of personalities. It’s a biological fact of life.

@rickflick I am so sorry about your brother. I feel awful now about saying this about my sister even further. Okay. So in High School we had to take Home Economics. It was the late ’90s. In her class and group they made an Angel Hair Food Cake. I guess they all made one and there was a girl with whom my sister grew up and competed with in her class. She had made an Angel Hair Food Cake for her family. My older sister, for no apparent reason, was sitting there in the class and reached over and just started mushing this girl’s angle hair food cake until it was mushed all the way down to almost nothing! My sister had to then bake a new angel hair food cake to replace it at home. My mom lightly scolded her but we were all just thinking, “What is the matter with you?” I mean she’s okay. Maybe she’s not terrible terrible but maybe just severely autistic and undiagnosed. Ha just like so many others. Who knows. I don’t know but she mushed another girl’s angel hair food cake in her own grade who was and a current friend since at least second grade. Who does that?

I’m no psychologist, but it sounds like some kind of nephropathy. A missing component in her mental machinery. It’s probably not unlike a congenital loss of hearing or vision. I suppose it did not produce enough impact to have her diagnosed, but it could be a marginal case. We’ve all met people with something like this. It’s probably just a bit on the tail of the curve. A bit out there near the end of the spectrum of personality. I’m sure it’s sometimes hard to live with and creates a lot of stress for people around her. This is life.

That should be

neuropathy. Some physical anomaly in the brain.That’s possible. I really appreciate your response. I’ve never been able to really articulate a lot of that. I am grateful for all of my experiences in life and what I have learned from them. I am very lucky in general and am always learning (new things and new things about myself also) as saccharine as that may sound. Thank you, though, very much.

Thanks. I’m glad my comments were helpful, and wish you all the best.

15

Good effort.

I have not read any answers yet but…

Unless I assume that a left shoe and a right shoe have the same value, I cannot compute an answer. But, if I do make that assumption, then my answer is:

Line 1: 1 shoe (left or right) = 5

Line 2: 1 person = 5

Line 3: 1 cone = 2

Line 4:

(Note: When there are no brackets, multiplication takes preference)

1 shoe + (1 person wearing a pair of shoes and holding 2 cones X 1 cone) = 5 + (19 X 2) = 43

I am also assuming:

1) that the small pair of shoes in line 4 have the same value as the large pair of shoes in line 1.

2) that each of the small cones in line 4 have the same value as the large cone in line 3.

But, if I am forced to give a numerical answer, it would be: 43

But it’s a pity the answer wasn’t 42!

Okay, I added the red shoes in line 4, but I forgot to subtract the black shoes.

🙁

Conclusion: The result is not computable.

(This is worse than the Monty Hall problem when people almost invariably misstate the problem).

Stuff like this is so silly. It’s wrong to call it “math”. There is no area of math that involves one variable holding another (kid holding two ice cream cones) and what that means is completely up to the reader.

Math is a lot more precise than this… if it weren’t, you wouldn’t be reading this right now (ie, the Internet wouldn’t work).

I got 15.

It’s easy to overlook the fact that the boy in the last equation is holding two cones of black stuff (whatever it is).

It’s easy to overlook the fact that the boy in the last equation is holding two cones of black stuff (whatever it is). And wearing red sneakers.

Oh, that’s amusing!

I find these sort of things a bit frustrating, although it’s fun to see how people infer rules.

I think I have to agree with 43. And no, no free choice of operation order!

Unless, that is, everything in the same cell needs to be multiplied?

That means a shoe = sqrt(10), but cones and boy are unchanged. So a boy with 2 cones and 2 shoes would be worth 10*4*5=200.

Giving: 400+sqrt(10)

Easy!

I think I should have mentioned that the boy is clearly still wearing the black SOCKS under the shoes.

And added a smiley!

Certain assumptions have to be be made and for clarity! I enlarged the pictures of the objects. I assume the pictures represent what they are:-

the pictures are not to scale

the red sneakers are red sneakers

the twerp is an child who, amongst other things, is wearing black shoes with socks (the faint grey lines on the legs above the shoes) except in its last appearance where it appears to be wearing sneakers with black toecaps, black sides, a black line around the bottom of the soles, black tongues and are reddish between the laces; clearly different to the other representations of red sneakers.

the twisted paper cones (called pokes in my childhood) filled with what appear to be seeds of some description. the two paper cones the twerp is holding have a different pattern to the other cones.

Different values can be ascribed to these differences but on moral grounds the child has to be taken out of the equations as it would be morally degenerate to put a material value on a child in this day and age and even if a child was to have a material value it would be worth more than a pair of sneakers.

So sharpen your pencils and start again!

Just noticed that he is wearing the shoes:

5+21×3=68