Can you solve this 2nd grade problem that has baffled adults?
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- čas přidán 6. 05. 2024
- A question assigned to 7 year old students completely stumped a mother who posted it online asking for help.
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looking at it, I genuinely have no idea what the problem even is.
English is also not my mother tongue and I understand why you don't understand.
They throw in word words the kids should know, 99% of the time I get stuck on soemthing helping my kid, I ask, “how did you learn this” and they’ll show me a quick easy thing and it’ll click in my head, there’s always a random word for me that I have no idea what they mean lol. Like in this it was “addend” or whatever that word was, remove that word and I’m good hahahah
They never taught me the word 'addend' in school. They just say 'one of the two numbers which get added'.
@@Moses_VIIsame
@@bebektoxic2136 Even with English tonguers, nobody knows what an Addend is.
I need to go back to 2nd grade English because I had no idea what the question even asked
Addend is a word none of us were taught
@@Moses_VIIYeah, I'm pretty good at math, and I truly do not think I was ever taught the words "addend," "aguend," nor "summand." I know "term" and "sum," but not the others.
The wording of the question is confusing.
Haha, right. "Have a go at solving it?" I don't even know what the question is!
That's what my problem with the question was also. 😂 @@Moses_VII
The whole reason this is difficult is because you can overcomplicate the question
exactly, the equation can be written as my first answer (which is 6 + 20 + 800 = 826 aka I only used the digits listed in the question). The thing is, you can TECHNICALLY add more digits to your equation since the question never stated that you should ONLY use the digits listed in the question (for example: 6 + 20 + 800 + 826 = 1452, aka my second answer, nobody said I couldn't add a "1452" to my equation). This makes the question more complicated than what it's probably initially intended for (aka ONLY using the numbers listed to make an addition equation).
edit: I saw another viable solution that makes sense which also bends the supposed "rules" of the question, as nowhere in the question states that you ONLY need a 1-, 2-, and 3-digit addend, resulting in this addition equation: 8 + 68 + 206 + 0 + 0 = 828
The thing that stumps most people the most, however, is the fact that they have no idea what an addend is, and since English is a second language to most people in my country, I get it (lucky for me, I grew up speaking English AND we were taught about "addends", "subtractor/subtrahend", "factor", "dividend/divisor", etc. where I live so it's just an advantage for people like me).
And it's easy to overcomplicate because they used digits that add up to each other. Makes you think the two has to add to the six to make eight, or 2+8 to make a 0.
Honestly this feels more like a language comprehension question than a math question
If i were to construct the question i would write:
Use the following numbers to construct a valid equation by filling in the boxes below
2,2,6,6,8,8,0,0,0
[] [] [] + [] [] + [] = [] [] []
I will go with this... The original question was horribly constructed!
@@hillaryclinton1314 the original question used terms no 2nd grader ever heard.
@@scottmcshannon6821
Says who?
I'm sure that digit, equation, addition, and addend were all mentioned in math classes before.
@@jnharton Strangely I had never heard "addend" and had to look it up as an adult when I was writing some documentation.
Ohhh then it’s easy, it’s 800 + 60 + 2 = 862
The 3 zeroes are the key. Once you realize there are only 2 of everything else, you understand that everything that appears in the addends also has to appear in the sum, and there are enough zeroes to make that happen. There are no odd numbers available to do any kind of carry logic. It's actually probably easier for a second grader, who doesn't have a lot of experience carrying digits forward yet, so they see the obvious solution right away.
Exactly. If you avoid the carrying then it is simple.
Simple, but I wonder what practical use it has. Just realizing that adding a 0 with a digit (an integer in the range [0,9]) yields the digit being added?
Exactly! What a neat question!
The question should have stated, "using only this complete set of digits." Nothing in the problem statement prevents you from adding more digits.
I wonder if this is the intention? It makes the math of the problem MUCH easier, and that kind of logic, while not something most of us were taught in second grade, is something that I think a second grader could grasp, especially if they're already being taught what an equation is. I mean, I mostly remember second grade math focusing on doing basic addition quickly, and they probably don't have to spend nearly as much time on that in the "you DO keep a calculator in your pocket at all times" era.
just what I thought, also with the amount of addends so 6+22+680+800=1508 is a solution.
exactly.
This is only possible if you also have an additional addend (which is also not specified, so the question is still underspecified).
If you have 1 more digit, it has to go somewhere - and if your addends are defined, you'll need a 4 digit sum (so you can fit all the given digits and your new one).
A+BB+CCC=DDDD requires the 4 digit sum to start with a 1, since a 1 digit number + a 2 digit number + a 3 digit number is at most 9+99+999=1107. Since 1 is not part of the given digits, it must be the new digit you added (assuming no leading 0s). So to get this you need to use 6 of the given digits to add together to a 4 digit number.
However, the largest of the given digits is an 8, and 8+88+888 < 1000, so you cannot create a 4 digit sum using the given digits, so the total number of digits matches the number of given digits.
That was my first thought
I really like the compliment to her for trying for an hour before asking for help. And I fully agree! Giving it a proper try *and* asking for help once you’re sure you need it are both amazing habits
Primary school homework should never take more than 30 min. and it should be practice and nothing else. Go have 3 kids and put 3h in every week day. ffs.
@@Fred-yq3fs 1. I agree that her spending half an hour on the question would’ve already been enough
2. One problem taking an hour once isn’t even bad as long as they’re typically much shorter
3. I can commend her habits without thinking it should take that long
@@Fred-yq3fsI'm stumped...ffs?
It is most probable that the question was set as homework in order to reinforce the lesson taught in class that day. Before you help your child with their homework it is a good idea to ask them what they were taught, and to show you the relevant pages in their work books. This will often guide you towards the sort of answers required.
That's a good point. I definitely remember being taught in elementary school how to break numbers down using their place values, such as:
135 = 100 + 30 + 5
So if that was the lesson taught, it would be obvious the answer would be in the form:
a00 + b0 + c = abc, where a, b, c have values 2, 6, 8 in some order.
The books suck ass.
I am 68 and just learnt what addend is. I have never heard that word before. I have o and a level maths. But the problem is not the maths it is the English. How about arrange the above digits to form a 1 digit, a 2 digit and 3 digit number which when added together make a number formed of the remaining digits.. Simple!!
I agree that's much better wording. In fact I believe much of the problem so many people have with math and word problems is not the math, but the way the problems are written.
@@kmbbmj5857 A poor tradesman blames his tools. It's the not the fault of the question writer that you don't understand fundamental mathematics terminology - that obviously would have been taught in this class.
Understanding these kinds of wordings is part of the training.
@@beng4186It maybe now it wasn't then or probably when the mother was at school. Not sure it is a term used in the UK now.
Plain English. No technical pedantic language in primary school. Too obvious I guess.
I get why people might be confused, but probably just because the wording and phrasing is rough for people who have been out of school for a long time and don't do math regularly. It took me a minute to see what the teacher's intent was but I think it's pretty fair for second graders who were just learning it in class that week.
A student in that class would be introduced to the Wording before these questions.
Once they know these words, the question can "make sense" for them.
The poor parents, who learned the "old" language, won't know what is being asked.
@@hrayz That's a problem in and by itself. There is no "new wording" to be had, especially not in primary school. It's basic knowledge. it's thousands of years old. No need to make it new every 2 years... except for the bottom line of text book editors of course.
@Fred-yq3fs This is the old language. I was taught this in grade 2 about 40 years ago.
@@SC-gs8dc Not in the UK. Never heard of 'addend' before. It's not used when teaching numeracy here.
@@yippee8570 English is my second language, but I since i know the word addition I could make it out with no problem.
Took me longer to figure out what the problem wanted me to do than to solve it.
It's asking for us to use those 9 digits to make a valid addition problem: 3 digits to make one of the addends, 2 to make another, 1 to make a 3rd, and then the last 3 to make the total.
Took me 5 minutes to work that out and 1 minute to come up with 600 + 20 + 8 = 628.
I am 58 UK born and I have never heard of or seen or even been told of the word addend before
I am 60. Until today I have never seen the word "addend" used in a math question for 7 year olds.
It is a very poorly formatted question.
2 + 60 + 800 = 862 ?
or any such variation yeah.
Ya, I did 8+20+600=628
There are 3 0s for a reason...
Good one- here's what I did: 0+0+[permutation of 2, 6, and 8]=[3-digit number] (I thought "stretch your thinking" meant "0 doesn't have to be a single digit.")
Yes. The kids are probably learning or reviewing place values. The context makes the intended answer more obvious.
The problem didn’t state that you should only use the given digits, just that they must all be used, so adding extra terms would open up a huge number of additional solutions.
This is not a problem about math this is a problem about definitions. I can guarantee the kids were taught what to do
Solved it in 30 seconds in my head, but I harbor no illusions that it would be within reach of most second graders.
You mean finding out there were 48 solutions?
Yeah, I got a correct answer quickly but most of my problem was seeing if I even understood the problem correctly.
@@sanamite the sufficiency case was one solution, as it would have been for the second graders. I intuited several solutions by digital permutation but didn’t go to the trouble to count them.
That’s always the hype title for these kinds of videos “third grade Chinese math problem stumps physics professors”
All they needed to say is create a equation that adds a 1-digit number, a 2-digit number and a 3-digit number with these digits included
Guessing was never a permitted part of math classes, not even when estimation was taught. We had to learn the teacher’s way of solving problems and be able to duplicate it in any problem we encountered.
Inspection is a very important to learn in math, particularly in Calc II and Differential equations.
The real question is more "what does this question mean?" I thought the sum wasn't supposed to be part of the digits. So my answer was
006+062+822=890 which is a stretch to the definition of an "n digit number".
Same. I never even considered that the answer HAD to be from within the remaining digits and therefore couldn't solve it. Although that is a failure on my part since the question didn't state otherwise either
The key is in the word "equation". With that word, the problem is telling you it's needed to write on both sides of the equal sign
No, a teacher would not accept a number with leading zeros to use up the otherwise unused zeros.
You used 4 zeroes, 3 twos and a number not even in the question. Are you sure you got it?
Every time it's "student's math problem stumps parents" I guarantee it's something the teacher already explained in class but the student wasn't paying attention.
Or the parents haven't bother to ask the question what did the teacher teach you today.
Neither did the parents, and they even lack the basic deductive skills to find that "addend" probably means one of the numbers being *add*ed. I'm a non-native English speaker on the verge of failing English literature, but it only took me about 30 seconds to figure this one out. And no, I haven't seen that word once in my life.
"What's wrong?"
The meta-answer is to look at the child's curriculum.
"Decomposing parts of numbers using the positional system" is the key.
My immediate thought was, “It says I have to use _all_ the digits shown, but not _only_ the digits shown, meaning there is an infinite amount of answers, as I can just add more numbers infinitely.”
The problem with some of these is the way they are worded. They can be ambiguous and confusing and neglect to clarify just what is wanted.
Flipping "6" to "9" is not enough. Why not rotating "8" 90° and turning it into ♾️ ? 😮😨
Or cutting the 8 into two small 0s or two 3s.
this question feels like its easy but has the most absurd wording ever, i have never once heard 'addend' in my entire life
The way I see it, there are 216 separate solutions. You can’t include leading zeros but you can count 200+80+6=286, 80+6+200=286, and 286=200+80+6 as three separate valid responses. Just because the responses are equivalent doesn’t make them redundant.
You missed solutions with a different pattern: 60 + 208 = 260 + 8, 20 + 806 = 820 + 6, etc.
Those aren't really addition equations. They are equations, but from what I can find addition equation means two or more numbers added together with the sum on the other side of the equals sign. If you have anything but just the sum over there it isn't an addition equation..
@@corvididaecorax2991 Whoa, there's an official definition of "addition equation"? >_
@@HarvardHeinous Honestly I just Googled it. Some websites listed that as a definition. I'm not sure how official it is, but I didn't find anything that gave any other definition.
But it is Mathematics. Everything has definitions.
I also considered the possibility of a compound sum on the right hand side. But, second grade, probably not.
Not a solution, since the problem stated "a 1, a 2 and a 3 digits *addend* ".
I honestly found it ridiculously easy once I figured out where to put the zeros. Took me about 5 minutes in total. Hats off to anyone who can find any solutions with two non-zero numbers in the same column with no leading zeros.
I messed with that avenue for a while, but realized there were roadblocks. Can't carry a one, because no odd numbers. Can only carry a two from the first column (6+6+8), but then run out of sixes and nothing else works. Presh's simulation of all possible combos shows the range of answers is all of similar types. Nothing exotic turned up.
I am suprised people found this difficult, but tthis being for grade 2s is insane. Maybe grade 4s or 5s
Since we're supposed to add up 1, 2 and 3-digit numbers together, the sum must have 3 digits too. Also, the most-significant digits of both the 3-digit addend and sum must be the same because all the digits are even numbers.
abc
de
f
agh
c+e+f is either below 10 or above 19. If there's no 8 among them, at least one 0 is used:
0+2+6=8
0+0+6=6
0+0+2=2
If there's an 8:
0+0+8=8
8+8+6=22 (carry 2)
8+6+6=20 (carry 2)
b+d can't exceed 9. d is at least 2, so b
You have a good principled answer. Ruling out the carries first is the way to go if you approach it as a maths problem.
I see 108 solutions, once an initial candidate is found:
- 6 column assignment combinations for the 3 non-zero pairs,
- 6 row arrangements for the 1, 2, and 3 digit numbers, and
- 3 arrangements for the last column (assuming that 0 counts as a stand-alone 1-digit number).
(I did not allow leading zeros beyond the lone zero.)
It took me like a minute to understand what the question is asking but then I got the answer in just 10 seconds
200+60+8, or 200+80+6, or 600+80+2, etc etc. A little bit of thought gets you there and a 7-year old might benefit from the exercise without Mom spreading math-panic
Literally 90s of thinking is enough to get 2 + 60 + 800 = 862
Or it can be any permutation of 2, 6 and 8
Just a silly idea, but if we decide we can rotate digits...
2 + 26 + 600 + 0 + ∞ = ∞
And, of course, lots of variations.
You have replaced the 8's by something that is not a digit
I think this is the most clever answer
that has 5 addends, not 3
@gorak9000 It requires 1, 2, and 3 digit addends, but doesn't specifically limit itself to those.
I think I figured it out- "stretch your thinking" is code: it means, "0 doesn't have to be just one digit!" With that in mind, there's 6 potential solutions: 0+00+[one of 268, 286, 628, 682, 826, or 862] = 3-digit added
those 6 solutions are already included in the answer of 36
@@NihaarB I hadn't even *watched* the video when this solution hit me!
@@wyattstevens8574 👍
the trick of this question is context: i'd say that this exercise is in a chapter in the school textbook probably teaches about "tens, hundreds, thousands", and/or how to write and manipulate summations with "carry over", what to do with leading zeroes, etc
this exercise is training the students'ability to comprehend and interpret the text, but is also to help the _teacher_ discern the individual students' abilities
1) the student that goes for the low effort answer: 0+00+268=268 (or permutations)
2) the student that struggles to find an answer, whether they do it or not - most of them will be in this category
3) the student that finds a couple of answers or possibly a _pattern_ in those answers
and i guess a 4th category would be *you,* the obviously-not-a-second-grader
if they had drawn those boxes representing the equation below the question, i bet people would have gotten it easily
it was just worded horribly
i didnt understand that i needed to use the numbers in the list on the answer too
Someone : i couldn't solve a 2nd grade problem even with an hour help
This guy : here are 48 ways to show you that you failed
i was stunned a first and tried to figure it out while you were explaining, it clicked in my brain as soon as you showed 2:55 the pyramid-like thing. really easy solution that shows how much you have to not overthink at problems like that
IM0, using 0 as an addend is acceptable, but 00 or a number with a leading zero is not as that's not how numbers are typically described. Yes, 00 is 0 and is valid, but I think most teachers would flag that. Also, the answer should not have a leading zero.
eg: 8+60+200=268
This feels like one of those questions the book author thought was so clever but in reality it was just a terrible idea that probably confused all but the smartest kids in the room.
I don’t really get why this was hard for people, you just do 0 + 00 + some combination of 628 and make that equal the same combination
Had never heard the words "addend" and "augend" before -- I got up to multivariable calculus in college. Learned something new today.
The problem doesn't appear to prohibit using decimal separators. Given that we have a 1-digit number as an addend, we can only add decimal separators between the ones and tens places. However, America uses a decimal point while Europe uses a decimal comma. So multiply the final answer by 3.
Only some places in Europe. Please don't group us all together. We're hardly homogenous.
Not quite true. You can put the decimal point as:
60.0 + 2.0 + .8 = 62.8
or
6 + 0.2 + 0.08 = 6.28
or
.6 + .02 + .008 = .628
etc.
And since you're adding decimals, you can also add negative signs and come up with other creative solutions. Here's just one example:
6 + .28 + (-6.28) = 0.00
Having never heard of the term "addend" before, I had to google that. Once I knew that was just the numbers being added together, it took about 10 seconds to see that 600 + 20 + 8 = 628, which uses all the digits and has the 1, 2, and 3-digit numbers required to be added together. Not sure why the 7 yo's mother didn't just look up what "addend" meant, if that was her problem.
I assume that she noticed 6 + 2 = 8 and kept trying to get a solution that incorporated that, without having to use leading 0s.
I'll be honest, I tried this one for a bit and legitimately failed because I never stopped to think that the 1-digit addend might just be the last digit of the sum. Such a simple, obvious step, and I discarded it right from the get-go.
Always impressed. What I learned by watching this channel was that you don't actually need to do it the hard way but the smart way. I thought the equation wording was the catch, but didn't notice the trick to pairing the digits. Now I know. Those little things is what I tried to teach my little brother because the hard way is the pathway of highest electrical potential in my brain's medium. And we may entertain the thought, but isn't the way the scientific notation treats the zeros the standard?
First of all, once I did a search to find out what an "addend" was, it was quite easy for me to find a solution. And, like many others, I had never had a math teacher, professor, etc. ever use the term "addend".
Secondly, using "02", etc. and "002", etc. as two and three-digit numbers is invalid - the only times I've ever seen such a notation was when teachers or textbooks used it to explain the addition of numbers with a different number of digits (as in the problem in this video).
The most difficult part of this problem was knowing what an "addend" was.
I was getting too much carry over originally before playing the video, once I saw your first step I face palmed and figured out the rules for all the solutions.
There are 2 of each number except the 0’s, which there are enough to express a multiple 10 and 100. A number with the digits ABC can be written as A00 + B0 + C. That uses each digit twice and three zeros, there are 6 possible solutions using this formula:
268=200+60+8
286=200+80+6
628=600+20+8
682=600+80+2
826=800+20+6
862=800+60+2
After the step where we figured out that the answer has to be a 3 digit number, I realized the three 0’s meant you could make any 3 digit number and break it into it’s expanded form
I had to look up what an addend was but after that I realized this was a question about understanding digit placements.
I struggled to understand the question as it was written. Once i understood the question, I found an answer quickly
Among the great many previously published math puzzles people have come up with over the years, there are myriad examples containing phrases like "1-digit", "2-digit", "n-digit number", and the like. Notice that in nearly every case, it is implied, if not outright stated, that the numbers referred to are positive integers without leading zeros. Exceptions to this rule are rare. It is a longstanding convention that "00" does not count as a 2-digit number.
The pairs of digits and three zeroes clued me in to the answer, and I'd envisioned the addition in columns.
6+62+288+1000 = 1356. Nowhere is it stated that only the given digits can be used.
Except it does. You're given 10 digits to use.
@@arandombard1197 Firstly, there are 9, not 10 given digits. But secondly, and more importantly, it is not stated that you are not allowed to use other digits, only that the listed gidits have to be used.
@@gabrielgrey2708 It doesn't say you can use other digits. It says you can use these digits. It then gives you the format that those digits must be used in. Pay attention.
@@arandombard1197 The problem states "use all of these digits", and my solution does use all of those digits. The format is just the preferred format of Presh, not a part of the original problem.
Did it in my head before the pause. Thought there'd be 6 possible answers, but didn't consider the )'s alone or in front for a moment.
3:51 Once I saw that the result was included in the digits used, I quickly saw that a valid solution is units, tens, and hundreds, with the result being a permutation of the distinct non-zero digits.
Either the equation uses all the digits, so:
6 + 80 + 800 + 622 = 1508
Or, the solution uses all the digits, which implies that all the *unique* digits must be used, else the question is impossible:
600 + 20 + 8 = 0628 = 628.0
6020 + 8 = 6028 if leading/trailing 0s don't count.
Or both, splitting the digits across the addends and the sum.
*Note: The question never specified that you have to use only those digits, so a possible solution can be easy:*
6 + 80 + 800 + 622 = 1508
or if the specification is required
800 + 60 + 2 = 862
The most difficult part is understanding the question (does a 7-year old know what an "addend" is?). Once understood, it is pretty easy to an answer.
3:28 when it's written like this it is suddenly so painfully simple.
My first thought was "What the hell is an 'addend'?"I couldn't even understand the question. I thought it involved addng up the 9 digits after right-padding some of them with 1, 2, or 3 digits. Maybe it would have made sense if I'd had the benefit of classroom context, but I didn't - so it didn't. Kudos to the mom who at least knew what was expected of her.
It's actually interesting how the question is made intentionally with the zeros in mind.
14 years of math including 2 years of calculus and I don't remember being taught the word addend, but if I replace it with number, I come up with
600 + 20 + 8 = 628.
bruh, it's a trick question. Beautiful puzzle tbh! I started by thinking about the possible places where carry could occur, only to realize there can never be any carry, meaning the pattern is essentially forced by the addends having to start with a nonzero digit (there's technically two other options if the one-digit addend is 0)
I didn't even understand the question until he explained but once I understood it took like 10 minutes to play with the numbers to figure it out then it made so much sense
The problem as stated is easily answered. The tough thing would be convincing the test scorer that the problem didn’t state how many times each digit could be used, agreed that zero is repeated if you wanted to jump to conclusions, or that the problem asked for an equation, not an equality. Greater than and less than feel so left out!
I really like this problem. It’s the kind that seems hard, then when you figure it out it seems simple.
Since the question just said the equation must have a 1-,2, and a 3-digit addend, I'd like to think it doesn't mean we can only have those three numbers, but we can also add more numbers ourselves. We could add some 4-, 5-digit numbers, or two 3-digit numbers as long as a 1-,2-, 3-digit number is added.
Moreover I think we can also use other numbers than 6,2,8,0, as long as we use all those numbers that were required. This would also fulfil the question requirements.
Frankly, took me less than 5 minutes.
Started with the last digits:
8+6+0/2/6 don’t work as carry one doesn’t work (would give an odd number in the next column).
8+8+6 could work (carry 2) but impossible in the next column (no 4 and no more 8 available).
6+2+0 could work for twp columns, but not for a third (missing 0 digits).
Sole solution is to use only digits added to zeros such as 200+60+8=268 or 800+20+6=826 or other variations 😊
Does anyone REALLY believe this was a 2nd grade problem with the word "addend" in it
7:52 Double-zero may not count, but single-zero certain should. - I accidentally got it when I considered making the 1-digit number a 0, then when I wondered if it has multiple solutions, I came up with a template for a bunch of solutions (which is similar to but different from Presh's solution): _0 + xy + z00 = zxy_ Really, the only restriction is to avoid carries.
11:00 The moral of the story is that schools are nasty and children are doomed. Presh's take on this is contrary.
My solution started from noting that that 1) there are no odd digits, so we can never carry a 1, and 2) there are only 2 8's, so if we ever do 2 non-zero digits (e.g. 6+2=8) in one column the other copies of those digits would be unusable (I missed the 22+66=88 angle). I also didn't allow leading 0's (which would also disqualify 0+22+066=088 permutations). That left me with 3 leading non-zero digits and 3 non-zeros in the answer and the 3 zeroes locked in as the non-leading digits in the addends - i.e. of the form a+b0+c00 =cba, yielding 6 permutations. I did neglect 0 as an option for the single digit addend, and allowing this yields answers of the form 0+ba+c00=cba and 0+b0+c0a=cba for a total of 18 permutations when leading 0's not allowed.
2:24 I've lived in the US for most of my life and this is the first time I am hearing of "addend", "augend", or "summand".
I had to google the definition of what an "addend" is.
Answering before I hear the answer but my logic was that there are no odd numbers, so anything that involved carrying a 1 was invalid (8+2=10, etc). This means all columns had to add up to 8 or less. The 1s digit could not be 0 in the final solution since that would require 4 0s since we can't carry digits over. None of the other columns could result in 0 because that would mean one of the numbers on top would be invalid (6+62+020=088 would not have a true 3-digit number). 0s can't start a number and can't be part of the solution, so that left a triangle of A+B0+C00=DEF. If any column used 6+2=8, that would leave an additional 6 and 2 orphaned since each column needed 2 non-0 digits. That leaves any solution where A=F, B=E, and C=D using the notation I used. So 2+60+800, 6+20+800, 2+80+600, 8+20+600, 8+60+200, 6+80+200 are all valid
Once I realised what the poorly-worded question was asking, it took no time: 600 + 20 + 8 = 628
As long as it was presented to the kids as a brain-teaser, I think it's a pretty fun little puzzle. Better than pages of addition problems which is what I got in 2nd grade.
There's no way in hell this was assigned to 2nd graders unless it was extra-credit and nobody was really expected to get it. 2nd grade math is learning to add and subtract through tons and tons of repetition. This video poster does this all the time, he takes these problems that give the average high school math student a lot of problems and says that it was assigned to 8 year olds (usually he says "gifted" 8 year olds). He throws in a strong implication, "A bunch of kiddies got it, what's wrong with you?" Same sh!+, different day.
@@Skank_and_Gutterboy It requires both a knowledge of addition, and a flexibility of mind. I don't think 2nd graders lack either of those things, but they need to learn to use both and that's the great thing about brain-teasers.
For a 2nd grade student this is nuts
it honestly doesn't look too bad. There are so many solutions, just some guesswork and a little thinking should easily stumble you into an answer
I could see some students getting it if they had done similar questions in class and in their homework assignments. If this just came out of the blue, it would be way too hard imo.
In context of their classwork it is probably very simple. I imagine they have been learning to split up a number into hundreds, tens and units, eg 862 = 800 + 60 + 2.
Creating any three digit number from different digits given in the question would then allow it to be split into hundreds, tens and units as an answer to the question.
The only thing that confusing in that question at 0:34 is the word addend which I never come across before.
But the fact there's 9 digits listed to be used and the fact that digits are repeated means they looking for an answer in the form of X+XX+XXX=XXX. So 6+20+800=826 you can swap the 2, 6 and 8 around.
Bro just put salt by finding not 1 but 48 solutions, 😂
All nonzero digits are doubled, and we have just the right amount of zeros to simply decompose a 3 digit number, e.g.
628 = 600 + 20 + 8
It's probably a kind of equation the students have seen many times, although it seems challenging that they would recognise them as a possible solution here. Then again, it says "stretch your thinking" so it seems this is a kind of advanced task.
A lot of people are stumbling over the word addend. For some reason I was taught this in grade school. Here’s another word they taught me: Subtrahend. It’s the number that gets subtracted in a subtraction problem. Both of these have been completely useless words my entire life.
If I understand this right, there are several valid solutions. A key is to notice that there are 3 zeros, and 2 of every other digit. And, this corresponds with growing magnitude of the summing digits. That said: definitely a tough problem for second grade.
General solution I found:
Z + Y0 + X00 = XYZ
Some example solutions:
6 + 20 + 800 = 826. Or, 2 + 60 + 800 = 862. Or 8 + 60 + 200 = 268.
5:00 The problem statement is a bit unclear but I arrived at the same solution you show. The problem statement would be clearer if the problem required to use each provided digit exactly once while forming the addends and the resulting sum. Several variants exist that work the same.
It's pretty easy. You can use any combination you want as long as one side uses one each of the 6, 2 and 8 and all the zeros. For example 600 + 80 + 2 = 682 or 200 + 60 + 8 = 268 or 800 + 20 + 6 = 826. I doubt it was intended for the kids to use 0s in the more unconventional ways such as leading zeros; teachers aren't that smart. That being said, it doesn't say use ONLY these digits so there are an infinite number of possible solutions.
Turning the 8's sideways to make infinity symbols will allow a whole new batch of solutions.
Honestly, once I actually understand the question it was really easy. However, I had no idea what the question even was. It was written extremely poorly for sure.
It looks to me like they are really trying to find the gifted students.
AA BB CC DDD
and we want to consider sums of the form (not necessarily in this order)
A + AB + BCC = DDD
now permute till we get a solution (brute)
Is there anything about the question that prevents you from adding additional digits to the equation? As long as I add a 1-, a 2-, and a 3-digit number and include all of the 9 required digits, I cannot see a problem in e.g. adding another addend or using other digits than 0, 2, 6 and 8 for the sum?
I don’t even know where to start 💀
The problem is that the wording of the assignment is EXTREMELY vague. I am 69, have a bachelor's degree in math, and math was my best subject. I came up with one of the possible solutions (6 + 80 + 200 = 286), but only because I made assumptions as to the meaning of the assignment. Nowhere does it say that some of those digits are used in the sum. And when I was in 2nd grade, I never would have been able to infer the necessary assumptions.
I’m not sure what you mean, it seemed to me like it was implied that you had to use those digits in the sum
we have 4 numbers
A, single-digit
B, double-digit
C, triple digit
D, this equals A+B+C
we have 9 digits to use, specifically 6, 6, 2, 2, 8, 8, 0, 0, 0
lets first establish that if we have a remainder from the tens digit to the hundreds then the 2 addends in the tens digit can be at most 8 and 8, producing 16, which is a remainder of 1. therefore the hundreds cannot recieve a remainder greater than 1. (recieving a remainder from the singles digit does not change this since we would need a remainder of 4; impossible with 3 digits)
lets now explore the hundreds digits. they cannot be 0 and we should have exactly 2 of them. if we had a remainder from the tens digit (as mentioned; at most 1) then the hundreds digits would have to be off by 1, but all our available digits are even and can therefore not be off by 1. therefore the hundreds digits must match each other and the tens digits give no remainder.
next we look at what remainder the singles digits can give us at most, from logic similar to the hundreds digit we can find that sending a remainder of 1 to the tens digits will not be useful for the puzzle, but can we send 2? yes we can, with 8, 8, 6 or 8, 6, 6, in both of these combinations the hundreds digit needs to be 2, so it cannot be 8, 8, 6 since that would need a 2 to be in the sums singles digit.
if we assume that the singles digit then are 8+8+6=20 and the hundred digits are 2=2, then that leaves the tens digits as 0, 0, 6 with a remainder of 2 from the singles. that doesn't work out, therefore, the singles do not contribute any remainder.
thus, since there is no remainders at all to consider, that means we can consider these to be 3 seperate single-digit calculations looking like: A=A, B+C=D, E+F+G=H, where all of these numbers are single digits and A is not 0, at least 1 of B and C are not 0 and at least 1 of E, F and G are not 0, also, neither D nor H is 0.
so where are the 0s?
among B and C at most 1 can be a 0. among E, F and G at most 2 can be 0s. that means we only fit 3 0s, and that's all we have, so that's where they must go.
the solution is then
A=A
B+0=B
C+0+0=C
where A, B and C is 2, 6 and 8, in some order.
all of the possible solutions to the puzzle is then:
800+60+2=862
800+20+6=826
600+80+2=682
600+20+8=628
200+80+6=286
200+60+8=268
Majored in mechanical engineering, minored in math.
I don't think I've ever seen the word "addend" in my life.
The - symbols after 1, 2 and 3 also confused me.
Personally my first tought was to have all 3 addend be 3 digit but with leading 0 so that they can be considered a 1, 2 and 3 digit number and then just do the sum (something like 662+028+008=698). It says to use those numbers but it never tells you to not use any others.
As most of this problems it is hard mostly because it is poorly written
I started by adding 8+8+6=22. (The sum of the units digits can't be from 10 to 19 because all the digits are even.) Then (2)+6+2=10, so that doesn't work. Then I tried 8+6+6=20. This led to 8+66+206=280, but that has three sixes and two zeros, so it doesn't solve the problem. At that time I had to get ready for church. On the way, I figured that the sum of the digits on the two sides of the equation must be equal. It can't differ by 9, since all the digits are even. The obvious way to do that is for the sum to be some permutation of 628.
I don’t even understand what on earth the question is even asking. It’s not even asking for an actual output. Nothing is specific.
I love this question. It has more than one solution. For a second grader. I'm not sure? Possibly an extension question for a period of 1 year.