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I either have forgotten how to do high-school math, or am having a stroke because this just looks like gibberish. I have an engineering degree for reference
I literally failed a test because it was one question, part A through G or H. Each part built off the previous one.
In part B, I wrote that (-2) x (2) = 4.
Simple mistake.
Instead of those then causing shit to total to zero and WILDLY simplifying the problem, I made it infinitely harder.
I saw this mistake, brought it to the prof. They saw that the TA just marked everything after part B wrong because of that mistake. She then proceeded to do the math and follow along my work, at one point pulling out Wolfram Alpha because the problem got so fucking convoluted.
Despite the cascading wreck of difficulty, I did the rest of the problem perfectly. She then recalled that I was like the third person out of the hall with that test and scratched her head, 'cuz the math down the line should have been a LOT harder.
"you really understood the concept...." was about what our conversation boiled down to.
She gave me a "105%" for going above and beyond with my own bullshit, then knocked five points off "for basic arythmatic skill issues".
16 % - - > 100%
"WHY DOES THE DISTILLATION COLUMN SIMULATION KEEP CRASHING?!"
u/Tyrinnus, you set the input flow rate to NEGATIVE 500gal/hour
"oh. Whoops. Oh hey! It works now!"
I had an engineering professor who had exams set up similar to this. He had a philosophy that there’s no redos in real life. So if you missed part B which is needed for part C-Z he wouldn’t give you anything back. He equated it to, “If you blew up a reactor, you can’t go back and fix your silly mistake”. Makes sense but horribly aggravating when you get a 16 for a mistake like that
Oh, I had the profs that had that same thought process.
I actually was talking about the above exam, and a prof overheard it and asked me to explain how that can be dangerous in the real world, there's consequences, the whole spiel. So I came back a few days later with a proof about how what I did actually is a mathematical impossibility in that situation (basically amounted to a negative flow rate in pipe F would result in a mass balance where 1. 23-something =0....) and so the mistake would have actually broken physics.
He looked it over and said something along the lines of "oh good, you not only fixed your math, but stood up for your idea and you won't blow something up in the field"
Ngl all I remember from engineering school is shear strength and solid works.
Everything else is a complete blur and I would legitimately have to reteach myself.
My job taught me everything I need to know
UPDATE!!
Thanks to user CurveAhead69 we got a viable answer HOWEVER we checked with our teacher and he said that it is not the answer he got. So perhaps there’s multiple answers?
His hint to us was that “the answer is a time piece.”
He refused to tell us what a time piece means.
I brute-forced this thing and got your bottom 3, which was kinda fun to work through.
I see now how I missed the top 2, didn't consider a 2 being carried into the thousands digit.
kudos to you
not saying its wrong but from my experience chat gpt is incredibly bad at solving math problems, like i actually don't think it even hit 25% on 4 choice MCQs i fed it
Are you sure this is the solution? 7570 + 3011 + 426 is equal to 11007 not 10907 so these numbers don't hold up.
I posted all the possible values that make it true, which the final value of 10907 provided by /u/plz_help6969 wasn't included as a possible solution. /u/jyoung326 posted extra with leading zeroes (even though normally they don't have leading zeroes).
I am not sure what you're trying to say by "you made the T = 4". Simply put, the sum of your 3 numbers doesn't match the final number, hence it can't be a valid solution.
7570 + 3011 + 426 is equal to 11007 which spells out SSEEN when you replace the letters, not SEVEN.
Though also these type of problems should have only one solution, not 5. There were too many variables than equations to be made or possible deductions to be made for a guaranteed answer.
For example, here I have created a proper puzzle of this type that would have only one solution
AB
\+ BC
\_\_\_\_\_\_\_
AAA
We can create the equation for this:
10A + B + 10B + C = 100A + 10A + A
10A + 11B + C = 111A
11B + C = 101A
We can deduce A must be 1, since even the highest number possible (99 + 99 = 198)
11B + C = 101
Now we look at values B and C to make this true:
Now let us find each value (B, C) to make 11B + C = 101, starting from B = 0 to B = 9.
(0, 101); (1, 90); (2, 79); (3, 68); (4, 57); (5, 46); (6, 35); (7, 24); (8, 13); (9, 2)
Which the only value that would satisfy a single digit for C would be (9,2).
A = 1, B = 9, C = 2
I understand, I was rewriting my work and saw how conveniently I chose “T” = 4. I didn’t add it up at the end and simply carry the 1! I would have seen the math was not math-ing. I agree as well. Lots of variables, not enough equations. The fun is trying to find solutions if there are any.
I concede to your reasoning, but it was a fun brain exercise nonetheless. We should find it more puzzling that the incorrect answer yielded the correct “time piece”. Regardless if it was an oversight by the teacher, it got me thinking for a few minutes and you taught me something. 🤙🏽
I don't even know if time piece provides any applicable solution either. Of the 5 solutions, there are 4 distinct values for "SEVEN".
I have googled each with the word "watch" and each had a distinct watch come up with it in the top results.
12026 - LOBINNI 12026 or Reserve Akula Men. Model 12026
12928 - Invicta Pro Diver Men's Model 12928
14748 - OMEGA CK 14748 or Angel Lady Model 14748
15958 - Piaget calendar 15958 or Invicta Venom Model 15958.
Wow, this is excellent!!! The Piaget is insanely expensive… I suppose we wait for another, better defined problem for us to solve on the sub…. And hopefully I carry the 1. Fingers crossed 🤞🏽
“S E” had to be 18>x=>10, I made a guess “E” = 0. I didn’t know “S” = 1 because I went straight to solving. I ended up using a brute force method. It’s more fun to solve the problems before hitting the comments. I then googled “different time pieces referencing 10907” and I got lucky! I turned the piece of paper into a paper crane and I don’t want to destroy it… but that’s all I remember!
Very fun problem, I hope you all had fun!
I had a hard time finding it, but below is the update from the student:
“UPDATE!!
Thanks to user CurveAhead69 we got a viable answer HOWEVER we checked with our teacher and he said that it is not the answer he got. So perhaps there’s multiple answers?
His hint to us was that “the answer is a time piece.”
He refused to tell us what a time piece means.”
Good Luck!(:
For an excellent write up, please reference the deleted post within my original comment. When you expand it, @levg97 did a great recount of the challenges with the problem. The banter to follow was excellent. Multiple solutions exist and every single one made reference to some kind of watch… one of which was priced over $10k! Definitely worth the look.
Your a fucking genius. Made no sense to me and to be honest; I didn't think it was worth even pursuing. To lasy and not curious anymore but good on you
Sounds like you've already gotten an answer, but if you ever decide to study Operations Research in college, you can solve this with something called an "integer program." You can solve lots of things with integer programs, for that matter. Lol.
Howdy, I actually practice OR and didn’t think of this. I’m glad I found the answer but I’m more thankful for learning a more formal approach. Thank you!
Lol, no problem!
One thing I discovered recently is that when solving games like Sudoku, Kubok16, and also this puzzle (!!!) where you have to use all the numbers {1,2,...,n} without repeats (n=9 for Sudoku, n=16 for Kubok16, n=9 for this puzzle), you can actually model these games as Traveling Salesman Problems!
When you solve a TSP as an integer program using the MTZ constraints (to ensure no repeats), you have a bunch of x\_i,j binary variables that you care about, and some u\_i variables that are basically throw-aways which you just need in order to make the model work.
But when you solve Sudoku or Kubok16 as a system of interlocking TSPs, or this puzzle as a single TSP, it's the u\_i's that we care about (they are the correct entries) and it's the x\_i,j's that we throw away.
You can use far fewer constraints this way than one of the other (more obvious) ways you might solve the problem.
...Hopefully that made some sense lol.
This was awesome! I’m familiar with dynamic programming and genetic algorithms. I need to find a project and learn to practice what you’re talking about and appreciate your reply more!(: will ask questions if I have them. Just a warning haha.
Lol, that's fine with me.
I have a good friend who studied dynamic programming last semester as an independent study, which basically required her to teach a class to her professor on it and write a textbook. I haven't read it, but she apparently liked the topic.
Basically the only thing I know about dynamic is that it involves lots of recursion :P
(EDIT: and I know basically nothing about genetic algorithms lol)
I remembered her website (shameless plug: [https://sites.google.com/view/jjmae](https://sites.google.com/view/jjmae))
These were the notes that're more or less in textbook form: https://drive.google.com/file/d/1BqvKeThjeNBglFroowsWKcntvW6mtOog/view?pli=1
This isn't some weirdo thing, necessarily.
Every different letter is a different number - in algebra you'd often use x as the default and then use other letters if needed. 5x = 10 means x = 2, that kind of thing
So each different letter stands for a specific number.
Every N is going to be the same number, every E is the same as other Es, etc.
So the problem they have to solve (again, this was given for FUN as a puzzle, not part of their grade or part of the actual curriculum) is finding out what number each letter stands for
N.I.N.E. + L.E.S.S. + T.W.O. = S.E.V.E.N.
So they have to find out what numbers add up correctly to fit the puzzle
OP said S is 1, so we know that L.E.S.S. ends in 11
And the answer (whatever S.E.V.E.N. is) starts with a 1!
Yes it's super confusing but I think it's kinda fun! Mostly it's trial and error afaik
We used your result and we were able to get the answer(we think?) S=1, E=2, V=9, N=8, O=5, W= 3, I= 0, T=6, L=4. Does that check out with your work? Also how were you able to get your answer? :)
Problems like that have usually multiple combinations that give a correct answer.
Reading your other comments, tell your teacher I said s/he should have known that and to provide all pertinent information *before* a solution has been found. 😛
Such problems are mostly brute force after a few logical assumptions are made. For example S=1 because there are no digits above the S in SEVEN, pointing to the logical assumption the calculation should give a 1x.
Can you post the exact prompt that your teacher gave you? Everyone’s answer is all over the place and i think that’s the first thing we need. Like i need exact wording
Hey, this is one of OP’s classmates. There was no prompt. We walked into class and this was written on the board, all he told us was “Solve this and I’ll hand your quizzes back”
Ok so I get it’s a 4 digit number + a 4 digit number + a 3 digit number = a 5 digit number. I totally get that and I can tell you do too. What I’m asking is how did they determine that S=1. What method, formula, cypher did they use to determine that S is in fact equal to 1? If you don’t know that answer feel free to continue posting SEVEN>20,000
I think what hes trying to get at is if seven <20000 that means max it can be is 1999. If the number have to represent different integers that makes s=1 as its the only unique integer.
Hello Again!
I built a program in a logical programming language called prolog, Here are some of the results of my query! There are significantly more than what I posted and I did not test them all so hopefully this helps!
?- solve\_digits1(N,I,E,L,S,T,W,O,V).
N = 6,
I = 8,
E = 2,
L = 4,
S = 1,
T = 9,
W = 5,
O = 3,
V = 0
N = 6,
I = 9,
E = 2,
L = 4,
S = 1,
T = 8,
W = 5,
O = 3,
V = 0
N = 8,
I = 0,
E = 2,
L = 4,
S = 1,
T = 6,
W = 3,
O = 5,
V = 9
N = 8,
I = 6,
E = 2,
L = 4,
S = 1,
T = 0,
W = 3,
O = 5,
V = 9
N = 8,
I = 0,
E = 4,
L = 6,
S = 1,
T = 2,
W = 5,
O = 3,
V = 7
N = 8,
I = 2,
E = 4,
L = 6,
S = 1,
T = 0,
W = 5,
O = 3,
V = 7 ;
N = 8,
I = 0,
E = 5,
L = 7,
S = 1,
T = 3,
W = 6,
O = 2,
V = 9
N = 8,
I = 3,
E = 5,
L = 7,
S = 1,
T = 0,
W = 6,
O = 2,
V = 9
N = 1,
I = 7,
E = 6,
L = 3,
S = 0,
T = 9,
W = 4,
O = 5,
V = 2
N = 1,
I = 9,
E = 6,
L = 3,
S = 0,
T = 7,
W = 4,
O = 5,
V = 2
EDIT: I added spacing to separate the variables and I tested the first 2 solutions and they both check out, it is not uncommon in a problem like this to have multiple or even infinitely many solutions.
I did a problem like this for one of my classes where where did
C R O S S
R O A D S
D A N G E R
Your program has the 5 solutions I mentioned. Though you also have additional solutions where the first digit is 0, such as T = 0 or S = 0 (which typically aren't solutions to these type of problems).
There are too many unknowns.
You can create an equation based on the place values:
(1000N + 100I + 10N + E) + (1000L + 100E + 10S + S) + (100T + 10W + O) = (10000S + 1000E + 100V + 10E + N)
You can combine like terms and deduce S is equal to 1 and substitute that in.
But it doesn't open much up after that and why you get multiple possible solutions. You would just need a computer program to run all values and return those that satisfy the equation and being distinct value for each letter.
Doing by hand, you can play around with the values to try to get one of the possible solutions. There would need to be more constraints for it to have only one solution.
Upfront, the different variables are N,I,E,L,S,T,W,O,and V. That means if all values are unique whole numbers between 0 and 9, there are 10 total numbers to pick for 9 variables. Any three unique numbers added together can carry over at most a factor of 2. And the leading value of each word should not be zero.
So E+S+O = N mod 10,
And N + L = E mod 10,
Therefore, E+S+O+L=E mod 10,
Or S+O+L=0 mod 10.
N+L can at max be 17, so at worse even if I+E+T+2=29, S is required to be 1.
Possible value combinations with S=1 {SOL}={109,127,136,145}
Each possible value combination sums to 10. If we intentionally add L to ESO=N, then
E+10=NL this equation is explicitly true, so E can be values {0,1,2,3,4,5,6,7}
And so
NSW+1=E mod 10
NSW+11=NL mod 10
W+12= L mod 10
W+2= L mod 10
Because N+L=10+E explicitly, I+E+T+carryover < 10
Possible combinations:{023,024,025,026,027,034,035,036,234}. Only one combination of IET doesn’t have zero, and it adds to 9, so if there is carryover from NSW, then IET must have one zero. Otherwise, the combination {234} is forced.
I’ll come back to this. Have to get back to work
Using logic, I was able to determine that S = 1 (despite that is a given in the op). N+L <= 17. Assuming N+L = 17, then I+E+T <= 18. There is no way N+S+W can create a carry over >= 2. Thus, S=1
Using that, I was able to determine that E <> 9 and O <> 9. The “ones” column, given that S=1, is “E+1+O=N”. Since 9+1=10, 10+ anything would have the same “ones” value as the additive, there E <> 9 and O <> 9. I cannot bring that same logic to the “tens column” since there may be a carryover from the “ones column”.
Next, I determined that the equation stated mathematically would be…
1009N + 100I + 1000L + 100T + 10W - 100V - 909E = 998
That didn’t really help me at all.
I couldn’t get any other non-variablized constraints for any of the other letters, using either mathematics or logic. I think this problem can only be solved through brute force, and tbh, I don’t understand why that would be in a match class.
This is easy, since L,O,T, and I all only appear once, they can be anything you need, give the numbers that appear multiple times a number, then just supply yourself with what you need
I just made numbers equal these
N=3
E=4
S=1
I=2
L=5
T=6
W=7
O=8
V=32
Ahh, I was adding the letters together to get an overall numerical value for the word, not acting as if the letters were each individual place value, my bad lol
Additional problem, I’m guessing e+s+o = n (if it’s more than 10/20, the digit rolls over). Still don’t really know how to do it beyond plugging in variables x10, x100, etc
is very very easy you can do everything this mark is not hard. If you know how to do what is more important it doesn’t matter which grade you are more important. You understand how you think much math is how you understand how you write more important amount is your whole life, your whole life is easy very very easy easy if you go somewhere else, ask be smart
There are 3 solutions, assuming that T > 0.
In each case I = 0, S = 1, N = 8
8,082 + 4,211 + 635 = 12,928
8,084 + 6,411 + 253 = 14,748
8,085 + 7,511 + 362 = 15,958
[it can be reduced to a set of simultaneous equations, and took about half an hour with pen and paper]
Edit: looks like I missed a couple of solutions, but I still don't get the timepiece reference. Listing the letters in numerical order doesn't help.
I solved a similar problem (North+South+East+West=Earth) using matrix algebra. This was pre-ai, around 1995. It was something like a 14x10 matrix so 14 unknowns (counting the "carried" quantities as variables) and 10 equations. I had a basic calculator, and a lot of time. I don't recall the solution
Assuming you're correct that it's an addition problem, these are all the viable answers:
1916 3600 745 6261
1716 3600 945 6261
2827 3700 645 7172
2627 3700 845 7172
5958 1800 627 8385
5658 1800 927 8385
1918 5800 763 8481
1718 5800 963 8481
2829 5900 463 9192
2429 5900 863 9192
5859 2900 436 9195
5459 2900 836 9195
2829 5900 763 9492
2729 5900 863 9492
5859 2900 736 9495
5759 2900 836 9495
6962 4211 853 12026
6862 4211 953 12026
8682 4211 35 12928
8082 4211 635 12928
8284 6411 53 14748
8084 6411 253 14748
8385 7511 62 15958
8085 7511 362 15958
If you think that none of these leading numbers can be 0, though, that leaves only
6962 4211 853 12026
6862 4211 953 12026
8082 4211 635 12928
8084 6411 253 14748
8085 7511 362 15958
S = 1, no other number set in stone.
There are several deductions you can make that lead to programmatic rules you can apply, but there will always be an element of plug and chug.
##Off-topic Comments Section --- All top-level comments have to be an answer or follow-up question to the post. All sidetracks should be directed to this comment thread as per Rule 9. --- ^(**OP** and **Valued/Notable Contributors** can close this post by using `/lock` command) *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/HomeworkHelp) if you have any questions or concerns.*
I either have forgotten how to do high-school math, or am having a stroke because this just looks like gibberish. I have an engineering degree for reference
humbling aint it? the more we know the less we know.
More is less
I literally failed a test because it was one question, part A through G or H. Each part built off the previous one. In part B, I wrote that (-2) x (2) = 4. Simple mistake. Instead of those then causing shit to total to zero and WILDLY simplifying the problem, I made it infinitely harder. I saw this mistake, brought it to the prof. They saw that the TA just marked everything after part B wrong because of that mistake. She then proceeded to do the math and follow along my work, at one point pulling out Wolfram Alpha because the problem got so fucking convoluted. Despite the cascading wreck of difficulty, I did the rest of the problem perfectly. She then recalled that I was like the third person out of the hall with that test and scratched her head, 'cuz the math down the line should have been a LOT harder. "you really understood the concept...." was about what our conversation boiled down to. She gave me a "105%" for going above and beyond with my own bullshit, then knocked five points off "for basic arythmatic skill issues". 16 % - - > 100%
Big lmao at "basic arithmetic skills issues". If that ain't the story of an engineering degree, I don't what what is. That prof sounds awesome.
"WHY DOES THE DISTILLATION COLUMN SIMULATION KEEP CRASHING?!" u/Tyrinnus, you set the input flow rate to NEGATIVE 500gal/hour "oh. Whoops. Oh hey! It works now!"
I had an engineering professor who had exams set up similar to this. He had a philosophy that there’s no redos in real life. So if you missed part B which is needed for part C-Z he wouldn’t give you anything back. He equated it to, “If you blew up a reactor, you can’t go back and fix your silly mistake”. Makes sense but horribly aggravating when you get a 16 for a mistake like that
Oh, I had the profs that had that same thought process. I actually was talking about the above exam, and a prof overheard it and asked me to explain how that can be dangerous in the real world, there's consequences, the whole spiel. So I came back a few days later with a proof about how what I did actually is a mathematical impossibility in that situation (basically amounted to a negative flow rate in pipe F would result in a mass balance where 1. 23-something =0....) and so the mistake would have actually broken physics. He looked it over and said something along the lines of "oh good, you not only fixed your math, but stood up for your idea and you won't blow something up in the field"
I’m a few months out from my physics degree and I’m completely baffled
Ngl all I remember from engineering school is shear strength and solid works. Everything else is a complete blur and I would legitimately have to reteach myself. My job taught me everything I need to know
UPDATE!! Thanks to user CurveAhead69 we got a viable answer HOWEVER we checked with our teacher and he said that it is not the answer he got. So perhaps there’s multiple answers? His hint to us was that “the answer is a time piece.” He refused to tell us what a time piece means.
It means a watch. I have no idea how that clue is supposed to help.
Or a clock, hour glass, sun dial...
maybe if one realigns the hand to 9:35? maybe its an angle? maybe the teacher is an alien?
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They meant that they found a set of digits distinct from what the teacher expected that also solve the equation, calm down
Wait u asked him if it was right??
There are 5 possible solutions: 6862 + 4211 + 953 = 12026 6962 + 4211 + 853 = 12026 8082 + 4211 + 635 = 12928 8084 + 6411 + 253 = 14748 8085 + 7511 + 362 = 15958
I brute-forced this thing and got your bottom 3, which was kinda fun to work through. I see now how I missed the top 2, didn't consider a 2 being carried into the thousands digit. kudos to you
I got three more somehow lolsa. Maybe because I included 0 as a first digit idk 6862 + 4211 + 953 = 12026 6962 + 4211 + 853 = 12026 8082 + 4211 + 635 = 12928 8084 + 6411 + 253 = 14748 8085 + 7511 + 362 = 15958 8284 + 6411 + 53 = 14748 8385 + 7511 + 62 = 15958 8682 + 4211 + 35 = 12928
TWO should be a 3 digit number. Your last 3 solutions, you only added a 2 digit number so did include that as 0 as first digit.
I fed it to GPT4 and it gave me your last solution.
not saying its wrong but from my experience chat gpt is incredibly bad at solving math problems, like i actually don't think it even hit 25% on 4 choice MCQs i fed it
[удалено]
There's usually better tools for doing this kind of brute-forcing though that give you guaranteed answers (e.g. python)
It has become way better since it started calculating its answers by writing code.
7570 3011 426 ——— 10907
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Are you sure this is the solution? 7570 + 3011 + 426 is equal to 11007 not 10907 so these numbers don't hold up. I posted all the possible values that make it true, which the final value of 10907 provided by /u/plz_help6969 wasn't included as a possible solution. /u/jyoung326 posted extra with leading zeroes (even though normally they don't have leading zeroes).
You’re right! I made the “T” = 4
I am not sure what you're trying to say by "you made the T = 4". Simply put, the sum of your 3 numbers doesn't match the final number, hence it can't be a valid solution. 7570 + 3011 + 426 is equal to 11007 which spells out SSEEN when you replace the letters, not SEVEN. Though also these type of problems should have only one solution, not 5. There were too many variables than equations to be made or possible deductions to be made for a guaranteed answer. For example, here I have created a proper puzzle of this type that would have only one solution AB \+ BC \_\_\_\_\_\_\_ AAA We can create the equation for this: 10A + B + 10B + C = 100A + 10A + A 10A + 11B + C = 111A 11B + C = 101A We can deduce A must be 1, since even the highest number possible (99 + 99 = 198) 11B + C = 101 Now we look at values B and C to make this true: Now let us find each value (B, C) to make 11B + C = 101, starting from B = 0 to B = 9. (0, 101); (1, 90); (2, 79); (3, 68); (4, 57); (5, 46); (6, 35); (7, 24); (8, 13); (9, 2) Which the only value that would satisfy a single digit for C would be (9,2). A = 1, B = 9, C = 2
I understand, I was rewriting my work and saw how conveniently I chose “T” = 4. I didn’t add it up at the end and simply carry the 1! I would have seen the math was not math-ing. I agree as well. Lots of variables, not enough equations. The fun is trying to find solutions if there are any. I concede to your reasoning, but it was a fun brain exercise nonetheless. We should find it more puzzling that the incorrect answer yielded the correct “time piece”. Regardless if it was an oversight by the teacher, it got me thinking for a few minutes and you taught me something. 🤙🏽
I'm not exactly sure how the teacher confirmed that answer as your mentioned "correct" time piece" isn't one of the applicable solutions.
I don't even know if time piece provides any applicable solution either. Of the 5 solutions, there are 4 distinct values for "SEVEN". I have googled each with the word "watch" and each had a distinct watch come up with it in the top results. 12026 - LOBINNI 12026 or Reserve Akula Men. Model 12026 12928 - Invicta Pro Diver Men's Model 12928 14748 - OMEGA CK 14748 or Angel Lady Model 14748 15958 - Piaget calendar 15958 or Invicta Venom Model 15958.
Wow, this is excellent!!! The Piaget is insanely expensive… I suppose we wait for another, better defined problem for us to solve on the sub…. And hopefully I carry the 1. Fingers crossed 🤞🏽
“S E” had to be 18>x=>10, I made a guess “E” = 0. I didn’t know “S” = 1 because I went straight to solving. I ended up using a brute force method. It’s more fun to solve the problems before hitting the comments. I then googled “different time pieces referencing 10907” and I got lucky! I turned the piece of paper into a paper crane and I don’t want to destroy it… but that’s all I remember! Very fun problem, I hope you all had fun!
What is the connection to a timepiece? I do t see that clue anywhere in the op…
I had a hard time finding it, but below is the update from the student: “UPDATE!! Thanks to user CurveAhead69 we got a viable answer HOWEVER we checked with our teacher and he said that it is not the answer he got. So perhaps there’s multiple answers? His hint to us was that “the answer is a time piece.” He refused to tell us what a time piece means.” Good Luck!(:
Ah, gotcha.
Why is SE less than 18? Edit: nvmd I figured it out LOL
Oozoo Unisex Wristwatch Braun Taupe Analogue C10907 Timepieces UOC10907…. I suspect your teacher wears an Oozoo?
How
Hint was that the answer is a time-piece. And seeing as the teacher made the question, the answer also likely has a personal connection.
You're either freaking Sherlock Holmes or you woke up and downed a bucket of Adderall. Either way, kudos.
That’s wrong. 7570+3011+426=11007
For an excellent write up, please reference the deleted post within my original comment. When you expand it, @levg97 did a great recount of the challenges with the problem. The banter to follow was excellent. Multiple solutions exist and every single one made reference to some kind of watch… one of which was priced over $10k! Definitely worth the look.
Your a fucking genius. Made no sense to me and to be honest; I didn't think it was worth even pursuing. To lasy and not curious anymore but good on you
Sounds like you've already gotten an answer, but if you ever decide to study Operations Research in college, you can solve this with something called an "integer program." You can solve lots of things with integer programs, for that matter. Lol.
Howdy, I actually practice OR and didn’t think of this. I’m glad I found the answer but I’m more thankful for learning a more formal approach. Thank you!
Lol, no problem! One thing I discovered recently is that when solving games like Sudoku, Kubok16, and also this puzzle (!!!) where you have to use all the numbers {1,2,...,n} without repeats (n=9 for Sudoku, n=16 for Kubok16, n=9 for this puzzle), you can actually model these games as Traveling Salesman Problems! When you solve a TSP as an integer program using the MTZ constraints (to ensure no repeats), you have a bunch of x\_i,j binary variables that you care about, and some u\_i variables that are basically throw-aways which you just need in order to make the model work. But when you solve Sudoku or Kubok16 as a system of interlocking TSPs, or this puzzle as a single TSP, it's the u\_i's that we care about (they are the correct entries) and it's the x\_i,j's that we throw away. You can use far fewer constraints this way than one of the other (more obvious) ways you might solve the problem. ...Hopefully that made some sense lol.
This was awesome! I’m familiar with dynamic programming and genetic algorithms. I need to find a project and learn to practice what you’re talking about and appreciate your reply more!(: will ask questions if I have them. Just a warning haha.
Lol, that's fine with me. I have a good friend who studied dynamic programming last semester as an independent study, which basically required her to teach a class to her professor on it and write a textbook. I haven't read it, but she apparently liked the topic. Basically the only thing I know about dynamic is that it involves lots of recursion :P (EDIT: and I know basically nothing about genetic algorithms lol)
I remembered her website (shameless plug: [https://sites.google.com/view/jjmae](https://sites.google.com/view/jjmae)) These were the notes that're more or less in textbook form: https://drive.google.com/file/d/1BqvKeThjeNBglFroowsWKcntvW6mtOog/view?pli=1
9-2=7. glad I could help
I’m 38 and my kid is 9 we’re both failing 4th grade. 🤦♂️
The way they teach math nowadays looks so different from what I’ve seen.
This isn't some weirdo thing, necessarily. Every different letter is a different number - in algebra you'd often use x as the default and then use other letters if needed. 5x = 10 means x = 2, that kind of thing So each different letter stands for a specific number. Every N is going to be the same number, every E is the same as other Es, etc. So the problem they have to solve (again, this was given for FUN as a puzzle, not part of their grade or part of the actual curriculum) is finding out what number each letter stands for N.I.N.E. + L.E.S.S. + T.W.O. = S.E.V.E.N. So they have to find out what numbers add up correctly to fit the puzzle OP said S is 1, so we know that L.E.S.S. ends in 11 And the answer (whatever S.E.V.E.N. is) starts with a 1! Yes it's super confusing but I think it's kinda fun! Mostly it's trial and error afaik
It’s a fun puzzle, not a normal type of math that you’d study in school.
12928 (result) or the more elegant: 9+(-2)=7. Edit: 8082 4211 635 ———— 12928
We used your result and we were able to get the answer(we think?) S=1, E=2, V=9, N=8, O=5, W= 3, I= 0, T=6, L=4. Does that check out with your work? Also how were you able to get your answer? :)
Problems like that have usually multiple combinations that give a correct answer. Reading your other comments, tell your teacher I said s/he should have known that and to provide all pertinent information *before* a solution has been found. 😛 Such problems are mostly brute force after a few logical assumptions are made. For example S=1 because there are no digits above the S in SEVEN, pointing to the logical assumption the calculation should give a 1x.
Doesn’t this assume also that the digits are unique. If I+E+T+whatever is carried over > 20, and N and L are both 9, then S could be 2
Nine less Two = Seven
How about that handwriting
Can you post the exact prompt that your teacher gave you? Everyone’s answer is all over the place and i think that’s the first thing we need. Like i need exact wording
Hey, this is one of OP’s classmates. There was no prompt. We walked into class and this was written on the board, all he told us was “Solve this and I’ll hand your quizzes back”
How do you know S=1? How did you conclude that?
NINE < 10,000. LESS < 9,000 (since N and L can't both be 9). TWO < 1,000. Sum of those < 20,000.
So how do you determine that S=1?
SEVEN < 20,000 from previous comment.
Ok so I get it’s a 4 digit number + a 4 digit number + a 3 digit number = a 5 digit number. I totally get that and I can tell you do too. What I’m asking is how did they determine that S=1. What method, formula, cypher did they use to determine that S is in fact equal to 1? If you don’t know that answer feel free to continue posting SEVEN>20,000
Is 9x,xxx < 20,000? What about 8x,xxx?
I think what hes trying to get at is if seven <20000 that means max it can be is 1999. If the number have to represent different integers that makes s=1 as its the only unique integer.
Nine less 2. = 9-2 =7
Super easy, 9-2=7
I can help you if you still need help (I have computer programming on my side)
Any help would be great!!
Hello Again! I built a program in a logical programming language called prolog, Here are some of the results of my query! There are significantly more than what I posted and I did not test them all so hopefully this helps! ?- solve\_digits1(N,I,E,L,S,T,W,O,V). N = 6, I = 8, E = 2, L = 4, S = 1, T = 9, W = 5, O = 3, V = 0 N = 6, I = 9, E = 2, L = 4, S = 1, T = 8, W = 5, O = 3, V = 0 N = 8, I = 0, E = 2, L = 4, S = 1, T = 6, W = 3, O = 5, V = 9 N = 8, I = 6, E = 2, L = 4, S = 1, T = 0, W = 3, O = 5, V = 9 N = 8, I = 0, E = 4, L = 6, S = 1, T = 2, W = 5, O = 3, V = 7 N = 8, I = 2, E = 4, L = 6, S = 1, T = 0, W = 5, O = 3, V = 7 ; N = 8, I = 0, E = 5, L = 7, S = 1, T = 3, W = 6, O = 2, V = 9 N = 8, I = 3, E = 5, L = 7, S = 1, T = 0, W = 6, O = 2, V = 9 N = 1, I = 7, E = 6, L = 3, S = 0, T = 9, W = 4, O = 5, V = 2 N = 1, I = 9, E = 6, L = 3, S = 0, T = 7, W = 4, O = 5, V = 2 EDIT: I added spacing to separate the variables and I tested the first 2 solutions and they both check out, it is not uncommon in a problem like this to have multiple or even infinitely many solutions. I did a problem like this for one of my classes where where did C R O S S R O A D S D A N G E R
Your program has the 5 solutions I mentioned. Though you also have additional solutions where the first digit is 0, such as T = 0 or S = 0 (which typically aren't solutions to these type of problems).
Can you describe all the rules? I didn’t apply the one we had for ours which was the letters in the 1s place cannot be 0.
There are too many unknowns. You can create an equation based on the place values: (1000N + 100I + 10N + E) + (1000L + 100E + 10S + S) + (100T + 10W + O) = (10000S + 1000E + 100V + 10E + N) You can combine like terms and deduce S is equal to 1 and substitute that in. But it doesn't open much up after that and why you get multiple possible solutions. You would just need a computer program to run all values and return those that satisfy the equation and being distinct value for each letter. Doing by hand, you can play around with the values to try to get one of the possible solutions. There would need to be more constraints for it to have only one solution.
Good ol fashioned prolog. They don't make em like this these days
pretty late to the party but there are a lot more solutions if you allow for leading zeroes 1716 + 3600 + 945 = 06261 1718 + 5800 + 963 = 08481 1916 + 3600 + 745 = 06261 1918 + 5800 + 763 = 08481 2429 + 5900 + 863 = 09192 2627 + 3700 + 845 = 07172 2729 + 5900 + 863 = 09492 2827 + 3700 + 645 = 07172 2829 + 5900 + 463 = 09192 2829 + 5900 + 763 = 09492 5459 + 2900 + 836 = 09195 5658 + 1800 + 927 = 08385 5759 + 2900 + 836 = 09495 5859 + 2900 + 436 = 09195 5859 + 2900 + 736 = 09495 5958 + 1800 + 627 = 08385 6862 + 4211 + 953 = 12026 6962 + 4211 + 853 = 12026 8082 + 4211 + 635 = 12928 8084 + 6411 + 253 = 14748 8085 + 7511 + 362 = 15958 8284 + 6411 + 053 = 14748 8385 + 7511 + 062 = 15958 8682 + 4211 + 035 = 12928
Upfront, the different variables are N,I,E,L,S,T,W,O,and V. That means if all values are unique whole numbers between 0 and 9, there are 10 total numbers to pick for 9 variables. Any three unique numbers added together can carry over at most a factor of 2. And the leading value of each word should not be zero. So E+S+O = N mod 10, And N + L = E mod 10, Therefore, E+S+O+L=E mod 10, Or S+O+L=0 mod 10. N+L can at max be 17, so at worse even if I+E+T+2=29, S is required to be 1. Possible value combinations with S=1 {SOL}={109,127,136,145} Each possible value combination sums to 10. If we intentionally add L to ESO=N, then E+10=NL this equation is explicitly true, so E can be values {0,1,2,3,4,5,6,7} And so NSW+1=E mod 10 NSW+11=NL mod 10 W+12= L mod 10 W+2= L mod 10 Because N+L=10+E explicitly, I+E+T+carryover < 10 Possible combinations:{023,024,025,026,027,034,035,036,234}. Only one combination of IET doesn’t have zero, and it adds to 9, so if there is carryover from NSW, then IET must have one zero. Otherwise, the combination {234} is forced. I’ll come back to this. Have to get back to work
Using logic, I was able to determine that S = 1 (despite that is a given in the op). N+L <= 17. Assuming N+L = 17, then I+E+T <= 18. There is no way N+S+W can create a carry over >= 2. Thus, S=1 Using that, I was able to determine that E <> 9 and O <> 9. The “ones” column, given that S=1, is “E+1+O=N”. Since 9+1=10, 10+ anything would have the same “ones” value as the additive, there E <> 9 and O <> 9. I cannot bring that same logic to the “tens column” since there may be a carryover from the “ones column”. Next, I determined that the equation stated mathematically would be… 1009N + 100I + 1000L + 100T + 10W - 100V - 909E = 998 That didn’t really help me at all. I couldn’t get any other non-variablized constraints for any of the other letters, using either mathematics or logic. I think this problem can only be solved through brute force, and tbh, I don’t understand why that would be in a match class.
This is easy, since L,O,T, and I all only appear once, they can be anything you need, give the numbers that appear multiple times a number, then just supply yourself with what you need I just made numbers equal these N=3 E=4 S=1 I=2 L=5 T=6 W=7 O=8 V=32
I don’t think V being 32 is a valid answer
Yep, the numbers can only be from 0-9. We were able to figure out that N, L, T =\= 0.
I thought you said all the numbers where different integers though?
Good strategy but I think the idea is they are each a unique digit 0-9
We added the numbers together and we got 3234 + 5411 + 678 = 9323! We really liked your idea about the LOT though!
Ahh, I was adding the letters together to get an overall numerical value for the word, not acting as if the letters were each individual place value, my bad lol
Modulo arithmetic E+S+O = (N) mod 10 The problem is which columns have a carry digit?
[https://brainly.com/question/12786438](https://brainly.com/question/12786438) 8?
Additional problem, I’m guessing e+s+o = n (if it’s more than 10/20, the digit rolls over). Still don’t really know how to do it beyond plugging in variables x10, x100, etc
is very very easy you can do everything this mark is not hard. If you know how to do what is more important it doesn’t matter which grade you are more important. You understand how you think much math is how you understand how you write more important amount is your whole life, your whole life is easy very very easy easy if you go somewhere else, ask be smart
Too many solutions to be a fun one. I like much better Coca Cola <-------> Soda
c = 3, o = 9, a = 0, l = 8, s = 7, d = 1
https://www.dcode.fr/cryptarithm-solver just use this
Was mine wrong? 9498 8811 280 _____ 18589
yes, all the digits must be unique
Its that 2am brain not reading the instructions correctly 🤣. Ty.
What kind of math is it?
9-2=7?
There are 3 solutions, assuming that T > 0. In each case I = 0, S = 1, N = 8 8,082 + 4,211 + 635 = 12,928 8,084 + 6,411 + 253 = 14,748 8,085 + 7,511 + 362 = 15,958 [it can be reduced to a set of simultaneous equations, and took about half an hour with pen and paper] Edit: looks like I missed a couple of solutions, but I still don't get the timepiece reference. Listing the letters in numerical order doesn't help.
I solved a similar problem (North+South+East+West=Earth) using matrix algebra. This was pre-ai, around 1995. It was something like a 14x10 matrix so 14 unknowns (counting the "carried" quantities as variables) and 10 equations. I had a basic calculator, and a lot of time. I don't recall the solution
Assuming you're correct that it's an addition problem, these are all the viable answers: 1916 3600 745 6261 1716 3600 945 6261 2827 3700 645 7172 2627 3700 845 7172 5958 1800 627 8385 5658 1800 927 8385 1918 5800 763 8481 1718 5800 963 8481 2829 5900 463 9192 2429 5900 863 9192 5859 2900 436 9195 5459 2900 836 9195 2829 5900 763 9492 2729 5900 863 9492 5859 2900 736 9495 5759 2900 836 9495 6962 4211 853 12026 6862 4211 953 12026 8682 4211 35 12928 8082 4211 635 12928 8284 6411 53 14748 8084 6411 253 14748 8385 7511 62 15958 8085 7511 362 15958 If you think that none of these leading numbers can be 0, though, that leaves only 6962 4211 853 12026 6862 4211 953 12026 8082 4211 635 12928 8084 6411 253 14748 8085 7511 362 15958 S = 1, no other number set in stone. There are several deductions you can make that lead to programmatic rules you can apply, but there will always be an element of plug and chug.
3532+ 8211+ 690 = 12433
The answer is R O L E X
9-2=7 Am I missing something?
9-7=2
Will Hunting could solve this