Can you solve the paper clip question for 2nd grade students?
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- เผยแพร่เมื่อ 7 ก.ย. 2024
- The paper clip problem has stumped many adults after it was posted online. Can you figure it out?
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if someone says they had *some* donuts at home and they didnt have any donuts i would be pretty mad
If they said they had some donuts, and they only had one, I'd be pretty annoyed too.
I think they could even be arrested for that.
Yup. Violation of international law, that is.
What if they said they had some donut holes?
Cake, and grief counseling, will be available at the conclusion of the test.
But it never said that Brian has boxes for all 10 or 100 paper clips. Just some. So the leftover can be a million, because he run out of boxes.
Exactly my thought - after seeing those questions using dirty tricks on me I have zero hesitation to take their questions LITERALLY and assume nothing more.
Here some means enough,
In question like these u do not get ans by trying to out think the question u have to think like the person who made the question to solve it, then u can point out it's flaw
However I do agree that for some people who have not practiced enough word problems it will get confusing when they mention some boxes, so they should've added sufficient, but that much is understood for students. And the word problems are usually in this format,
And clearly it was a question made for people who are trying to pursue math or science at a higher level because it requires u to already have attempted these types of questions
Even if he did have extra boxes, the paper clips may not be in those boxes. He may have more than 9 full ten paperclip boxes. Nowhere does it say he need to break open the ten paperclip boxes and fill an empty hundred paperclip box.
Brian could also have an IQ90 and asume he ran out of boxes.
In fact there is no brian and no boxes, and the answer is 42, because that is the answer to everything.
It also doesn't say that Brian has this kind of obsessive-compulsive disorder that once his spare clips reaches 10 he'll have to find a box for them. He could have just store bought some boxes with 10 clips in it and some with 100 in it, and he has a million leftovers at home.
I think 301 is no solution. That means that Brian has no 10-clip-box, but the problem says "some boxes hold 10 clips", maybe even 412 is no solution, because 1 box is not "some boxes"...
Agreed. He's being too mathematical here. 523 is the lowest possible solution based on the wording of the puzzle.
@@aquamarine99911 I agree that we need minimum 1 box for 10 clips. For me problem with "some" is with translation (english is not my first language). I always treated it as 1 or more.
I'm almost certain that 'some' in a mathematical context always means 1 or more (ie. some number of boxes). I'm possibly mistaken and it means 0 or more, but it definitely in no way means 2 or more.
I agree. The wording of the problem would limit the answers to where the number of boxes of each has to be at least 2. If you accept that "0" is a valid number for "some", why not accept 0 for the number of leftover clips, which would make the number of Ten boxes -1. If "some" can be zero, why can't it also be -1? Why not say you have 0 Hundered boxes, making your total number of clips -32?
@@AverageJoe1999 In deeper though. He has this boxes but he could not use them. So its possible he has 100 boxes for 100 clips, 100 boxes for 10 clips. And only 301 clips
It is an ill-defined problem. A term like "some" must be defined; the idea of zero being "some" is counterintuitive, and even the idea of one being "some" is fuzzy. Additionally, it doesn't specify that leftover clips must be less than 10 or the number of 10-clip boxes must be less than 10. It is within the parameters of the problem that he has 473 leftover clips, thus 475 boxes of a hundred and 470 boxes of ten, and thus a total of 52,673 paper clips.
Yes they really should rule out the possibility of "some" meaning zero or at least consider the possibility of L being equal to zero.
@@londonbobby Suppose you got bit by a snake, and I told you I had "some" antivenom and "some" medical training.
Yes, "some" is definitely more than "none." But the problem doesn't say he has some 10 clip boxes when you read it carefully. It does say that he has some leftovers, so you know that L is non-zero.
@@hyperboloidofonesheet1036 Yes, in normal speech "some" means more than none and generally means more than one also. But following his logic where some can mean none then we have consider that possibility as a value for L to be consistent.
If it wasn't ill defined then nobody would be arguing about it on reddit and twitter and now youtube
-Hey bro, I'm Brian and I have some boxes that hold 10 paper clips.
-Cool, how many do u have?
-0
Dude, I busted out laughing when I read this. Nice one.
@@jdrex02 Hey class, come over and let's eat some bags of chips I have while we play video games. I have 0 bags of chips for us to share!
Not sure I understood which of the info claims that Brian has an empty box to put 10+ leftover clips into..
It doesn’t. There are infinite answers.
It's implied. He has boxes and "some paper clips left over". Not "and some other clips", not "and more besides"; he went through some process involving the clips -- with context, it's probably putting them into boxes.
I 100% agree that it's not clear. Especially in low-grade math, it probably should be made clear. But I'd support this being an extra-credit or higher-grade problem as is because teaching contextual clues is important, too.
@Reignspike It's not implied. In real life, a person could/would buy new boxes of paperclips even if they still have more loose paperclips laying around than would fit in a box. Also, in the context of real life, it's not likely that a person would be putting paperclips _into_ boxes; you take a paperclip _out from a box_ when you need it, but when you don't need it anymore, it usually ends up somewhere in a drawer or something.
@@yurenchu Also, the problem says "stretch your thinking". Assuming that this is a conventional place value question is the opposite of that.
@@TonboIV Exactly! I made the same remark in another thread in this comment section.
2:00 the question never specifies he has extra empty boxes.
No the question specifies that some paper clips are left over that means the no. of paper clips is not a multiple of 10 so the no. Of paper clips left would be 1-9
Therefore those extra clips won't be put in any box or u could say a extra box(which is confusing) so think of it as holding the extra clips in ur hand because they are less than 10
@@ayus0_0 We can interpret "leftover" as "Brian does not have boxes for them". With such interpretation, leftover can be any positive integer.
@@Valentyn007no in these types of questions they usually don't consider the lack of boxes(until the question says so) so if they say they have some boxes means they have infinite boxes
Or
In simple words
Brain will always have enough boxes for the paper clips
Let's say u were brian
And u would report that "i have some boxes that hold 10 paper clips and some left over paper clips" u urself wouldn't leave the information of not having sufficient boxes out even if u were saying that to someone irl,
Like if ur friend says "I have 21 candy, i have boxes that store 10 candy
And have some left over"
U would immediately think he has 2 boxes instead of "naw tf u mean 21 candy what if u don't got no boxes and just lyin, mate I don't trust u enough"
@@ayus0_0 Nowhere does it say in this particular problem that Brian has been sorting his paperclips into boxes of 100 and boxes of 10 . He could just have bought new paperclips in boxes from the store, while still having a bunch of loose (= leftover) paperclips lying around in the house; which is actually more realistic, unless Brian sells self-made paperclips over the internet or something.
By the same logic, where you derive that "some leftover clips" means "at least 1 leftover clip", "some boxes of 10 clips" means "at least 1 box of ten clips". So no, 301 is not a valid answer.
No, when the question says some clips are left over it clearly means that there are clips not in the boxes,
When it says "some boxes hold 10 clips" it means there are boxes that hold 10 clips(which may or may not be used) ,
Think of it this way
Let's say I'm selling candy and have boxes that can store 5 candy, if I have 3 candy I can put it in 1 box and say 3 are leftover(as they are not filling the box), so when we count the no. Of boxes with 5 candy that box wouldn't count
If I have 6 I will fill one box and have 1 left so I will only count 1 box as " I have 1 box that can hold 5 candy" in the question it means the same as this example
I took some as more than 1., so the H, T, and L were all 2 or greater.
So 523 was the smallest number of paperclips.
@@ayus0_0 No. It does not say "boxes that could hold 10 clips", it says "boxes that hold 10 clips". Also, there is no word about empty boxes, so we should not invent them.
Same thing, if u have I tell u I have a bottle that holds 1l of water, would u assume I have water in it, no u won't
It means- ( I have a bottle in which I can put water ranging from 0-1 L)
And there are no empty boxes there are leftovers only, presh only said it to make it easier to think. U can think of there brian putting the leftovers on the floor and only putting clips in the boxes if he can get 10 clips together and put them in one
@@ayus0_0 Please read the first sentence. "Brian has some boxes of paper clips." Brian does not have boxes that are suitable for paper clips but are currently empty.
You could possibly reject 301 and 412 if you want to claim that 'boxes' is plural, so zero boxes or one box of any type does not work. Having said that, there is nothing in the question that limits the number of loose clips to 9 because 10 could be put into a box. Generally a poor question by an obviously poor teacher.
Arguably, "zero" is (grammatically) plural, because in English we say "there _are_ zero paperclipS in the box" and not "there _is_ zero paperclip in the box".
@@yurenchu Aye. You are right my fellow pedant! 🙂
@@Mike-H_UK Cheers! :-)
@@yurenchudid you really expect this thinking from a 2nd grade student ?
@@mohitrawat5225 Why would/should I expect this thinking from a 2nd grade student?
If I were the teacher and a 2nd grade student answered with just one of the infinitely many possible solutions, I would have graded their answer as correct.
If you had no problem with zero being some form of "some", then we should not have problems with zero leftover, because it's also "some"
Following the given conditions in the problem description, 0 leftover would imply (-1) boxes of 10 paper clips; now, how would that be possible? (Would that mean that Brian _owes_ a box of 10 paper clips to someone else?)
A better way to phrase it to clarify "leftover" means "couldn't fit in the box" would be:
Brian is filling paper clip boxes which contain either 100 or 10 paper clips each. After filling the boxes, he had some leftover that couldn't fill a whole box. ...
You're assuming that he is being efficient and every time he has sufficient clips he puts them in the bigger sized box. It would be best if the problem explicitly stated that.
exactly!
Of course we are! It's implied by the word "leftover" that any amount large enough to fill a box would be boxed and not "leftover", and having 93.5 clips leftover from a box of 100 it just silly. Any second grader who has not yet learned the difference between plural and singular should surely already grasp this! :/
This problem requires a couple of assumptions: a) Brian uses as many boxes of 100 as possible and as few loose paper clips as possible, and b) there's no limit to the boxes of 100 that Brian can have. I"m guessing that's what the main source of confusion is.
At about 2:00, you make an assumption which isn't supported by the problem statement. You state that L must be less than 10, because if there were 10 or more, Brian would have put 10 of the left over clips into a box of 10. But that's not necessarily the case. The fact that Brian "has some boxes of paper clips", and that he "has some paper clips left over", doesn't in any way mean that he has less than ten left over. The way the problem is stated, he might not have enough empty boxes to hold all the left over paper clips. Or he might have plenty of empty boxes, but simply opted to not fill them all. So, for example, Brian could have 39 boxes of 100, 36 boxes of 10, and 37 leftover clips, for a total of 3900+360+37=4297 paper clips. THIS ANSWER FULFILLS THE PROBLEM STATEMENT!! In fact, any answer which fits the formula 190 + 111x (where x is any positive integer) is an acceptable answer!!
Exactly My thought
This question is a far better predictor of being a future lawyer than a mathematician. Just reading through some of the comments already and I can see the loopholes in reasoning through the problem; this isn't a math puzzle but a wording one! I'll even add one more. Where does it say he actually FILLS the boxes? Sure, the box MAY hold 100 clips, but that box still holds 100 paperclips even if he only put 17 in there.
👍
Without making some common assumptions, many so-called word math problems are indeterminate propositions and therefore unsolvable. "Contains" is a very common meaning of "holds." You are likely to hear this problem in a math class and unlikely to hear it in a law class. Context is relevant to deciding which assumptions are warranted.
Nowhere in the question does it state that if he has 10 individual paperclips, then he would put them in another box and have a box of 10, or that if he has 10 boxes of 10, then he would put them in a box of 100. He could, for example, have 20 boxes of 10, which would mean 23 boxes of 100 and 21 left over. As such there are an infinite number of solutions to this.
If, however, it had asked what the minimum number of clips were, then it depends on whether you consider zero as a valid instance of "some". If you do, then 301 is the minimum. If you don't then 412 is the minimum.
The statement, " He has some paper clips left over", is my hang up. "Left over" is not a well defined term. Also I wasn't completely convinced that all the boxes that hold 10 or 100 clips were full. Just because a box can hold a number a clips doesn't mean it contains that number of clips. Also it uses clips and paper clips interchangeably. There are many types of clips that aren't paper clips. Assuming you are interpreting the whole question correctly I think a better way to have worded it would be, "Brian has some paper clips that he has put into boxes of 100 paper clips each, boxes of 10 paper clips each, and some left over paper clips that did not fill a box"..... If the question had been worded this way or a better way, the confusion would not have been there.
If the question started by saying "Brian has some apple pies" and then continues to abbreviate apple pies to just "pies", would you also be wondering "Wait, does that mean that Brian could also have some cherry pies?" ?
The verb "hold" can mean "has the capacity of", but it can also mean "is holding"/"contains". It can also mean "stop"/"prevent from going further".
But the phrase "boxes with 100 paper clips" means that they are referring to boxes that are each containing 100 paper clips. You can't say "box with 100 paper clips" when it isn't containing 100 paper clips.
Furthermore, the first sentence says "boxes _of_ paperclips", not "boxes _for_ paperclips"; this suggests that these boxes are "full"/"filled".
In saying that there must be
If he had 5 100 count boxes, he’d need 2 10 count boxes, and he’d need 3 leftover. So 523 is an answer. There are infinite answers.
It never specifies how many he can have left over, or whether he has extra empty boxes. So concluding at 9 extra isn’t necessarily correct.
If you are thirsty and ask me for "some water" and I give you an empty bottle, you will still be thirsty.
Nowhere does the problem state the 10 leftover clips magically get their own box... "I see nothing in the problem that excludes 0 as a number..." Yeah, well, I see nothing the problem limiting the number of leftover clips. You botched this one...
I agree that it's ambiguous: it's not clear whether or not there could be more than 9 "left over" paper clips.
From the thumbnail:
301 + 111k paperclips, for any positive integer k
where k is the number of 10-paperclips boxes.
I did it this way too, and didn't think of this as hundreds, tens, and ones, but as a lower bound and then an infinite series with no upper bound.
My way too! How would the video method work if the paperclips came in boxes of either 49 or 113? The method of solutions to problems should not depend on the specific values in the problem...
Nothing stops me from having 12 left over clips if some of the boxes have been damaged and thrown away. You didn't constrain the problem in the way you phrased it.
nothing stops you from putting 4 clips in a 10-clip box, thus not having any leftover clips.
I don't see the "assume that the number of boxes is always minimised" statement anywhere in this problem, therefore the answer has no upper bound.
The problem never said that all the boxes are full, or that leftover clips that could be put in boxes would be.
He could have plenty of half-filled boxes, and a big pile of leftover clips.
I think the problem I had when first reading the question was understanding the action being performed (filling boxes with paper clips). To me, this is because "boxes of paper clips" implies "boxes FULL OF paper clips." If instead the question had said "boxes FOR paper clips" it would have been more immediately clear what was happening in the question, and then easier to work out the answer.
The problem itself isn't hard, it just sounds really weird. It sounds like you're giving Brian full boxes, stating he already has some loose paper clips, and then asking "alright, how many paper clips does he have?" You could approach that problem that way, but the word "leftover" just sticks out at you and makes it feel wrong.
2nd grade students!? what do these students learn in 3rd grade math...differential geometry?
Maybe it meant 2nd year of high school??
This is such a poorly articulated problem. Presh, you're better than that
Have you been watching for the last year? He is not better than this. Ever hear of Blocktube? Today is the day I add this channel to mine.
The words "boxes OF paper clips" indicate that there are no empty boxes IMHO. Therefore, the minimum is 412. No maximum (?)
'Some boxes' for me means at least two, not zero or one.
"Leftover clips" also is plural.
This problem is ill-defined. Nothing was said about opening the boxes, and "left over" has no unambiguous meaning. Please stick to well-defined problems.
We are told that he has "boxes" of 10 and of 100. If we take "boxes" literally, T must be at least 2.
That is quite vague though - it could be that the 10 and 100 boxes is how they are found in the shop and he has some of these boxes, not necessarily both varieties.
I'm going to assume that there is at least one box of 10. Since we know Brian has three more boxes of 100 than boxes of 10, we know he has at least four boxes of 100. And if he has two fewer left over than there are boxes of 100, he must have 2 left over if there are 4 boxes of 100, meaning that Brain may have 412 paper clips, but no less.
We can increase every quantity by one and still have a valid answer to the problem: 523, 634, 745, 856, 967, etc. In each case, the hundreds digit is the number of 100 boxes, the tens digit is the number of 10 boxes, and the ones place is the number left over. (Assuming there aren't 10 or more boxes of 100).
2nd graders know algebra?
I'd alsp argue that it doesn't state he is filling boxes with any leftovers so it would be possible to have between 1 and 99 clips left over if he has only one of the one hundred clip boxes with clips missing.
Furthermore you could also argue that the answers for one and two clips left over being invalid since in many places "some" means 3 or more, and since it states he has "some" leftover clips, he gas to have at least three leftover.
Some is never 3 or more. I would say that mathematically, some could be as little as 0, but in this context I'd say some means 1 or more. I also think that some is less than 10, otherwise he could fill another box of 10. I take it that he filled as many boxes as he could and THEN realized he had 3 more boxes of 100 than boxes of 10, and 2 less left over than boxes of 100.
2:20 FALSE. It is from 2 to 9 "Some" isn't one, but it's larger than one.
This also means that the minimum number of boxes is larger or equal to 2.
Meaning that the minimum number is: 3 + (3+2)*100 + (5-3)*10 = 523
And the maximum: 9 + (9+2)*100 + (11-3)*10 = 1189
Zero T is impossible, because it is stated explicitly that some boxes hold 10 clips.
"Some paperclip boxes do hold 10; I don't happen to have any right now,"
We get 3 equations
The number of boxes of 10 clips = x
The number if boxes of 100 clips = x+3
The number of left over paper clip = x+1
Let x=1
We get, 10 clips + 400 clips +2 clips
Which is 412
U can have multiple solutions for different values of x
I.e, this problem do not have a definite solution,
(M commenting this before watching the video so I could be wrong)
They're stumped because the system is under-determined; there are infinitely-many solutions.
x,y,z for 100 box, 10 box, and leftover, respectively.
x = y+3, and z = x-2
x must be at least 3 (since we're dealing with non-negative integers), in which case y is at least 0, z is at least 1. So the smallest solution is 301.
Then to get all the other solutions, you add a positive integer to each of x,y,z, which translates to all the solutions being 301+111*k, for k a non-negative integer.
That's okay because the question implies multiple possible answers.
Since "some" implies more than zero, much easier to start with 1 box of 10 clips instead of starting with the number of leftover clips. That gives 1 box of 10, 4 boxes of 100 and 2 left over or 412 clips as a valid answer.
About 10 seconds. 523. 634, 745, etc. Why is this “baffling” people? Actually, sadly, I know why.
This problem given to second graders is designed to help them think about place value and words like “more” and “fewer”. It has also been crafted to allow for multiple solutions, probably to generate a group discussion. I’m sure second graders would handle this just fine.
Any controversy is the result of adults reading too much into the problem or thinking there has to be just 1 correct answer.
I honestly think that *the WORDING* of the problem automatically precludes the possibility of the possible answers where he has *0 **_OR_** 1* boxes of 10 clips.
Notices that it says "he has *SOME boxes* "... This translates to me like "he has *AT LEAST 2 (TWO)* ...
Otherwise the wording would have specified that he *COULD* have 0 or 1 box of that particular type... :P
.
Where are exactly the adults that get stumped at these questions?
@@ekatvakushvaha1814 the ones who have been out of school in a while lol. Glad my kids are grown 😅😅😅
Good question. If anyone finds them, please let me know so I can avoid them
By the fact that they start on the wrong foot assuming "some" = "one" or hell, even the most idiotic of all "some" = "none" as in their first answer, 301, that has no 10 paper clips boxes, even though the start says Brian has some of those.
"Some" means at least "2".
This video creator is one of those adults that got stumped.
523 is the minimum number
Where are these adults? Sadly, they're all around you.
@@ekatvakushvaha1814 XD I can find so many in this comment section getting confused about the leftovers
I dont agree with Presh on this one. My vote is for 412
"some" generally means more than one, so 301 shouldn't be a valid answer
Definitely valid. Mathematically, "some" can mean 0. Also, this is 2nd grade math where place values is most likely being taught, so this is clearly about finding a 3-digit number whose hundreds digit is 3 more than its tens digit and 2 more than its ones digit.
When teaching mathematics, I train my students to use as few variables as possible, especially when they are related by the context of the problem.
If you let the number of boxes with 100 clips be n then from the wording of the problem the total number is 100n + 10(n-3) + (n-2). I know that the last set of brackets is not needed but I think it aids understanding.
Weak students often start a problem by introducing too many variables and then lose the plot.
I started out by recognizing that H and T both had to be odd and that H and L had to be both even or both odd and worked from there.
This problem is terribly worded. Why couldn't he have say 65 clips left over from a box of 100? There is not a parameter specifying if there are available 10 clip boxes, so any number like 11 for example will make this equation look silly. 65 + 67 + 64 so 67 h 64 t 65 l to reduce down to 7605 should be a viable solution number (I really hope I did that right) given there is no presumption that 10 clips will be put into a box that can contain 10 clips. I should be able to presume that there is not an available box of 10 clips to put any overflow from a bigger box since the parameter is not specified.
I do not like these questions that have implied, though never stated, parameters outside of the actual written problem. I think there needs to be some level of clarification that there can not be clips left over that will fit into a box, small or large. By the way this is written, the answers are basically all numbers algorithmically that do not specifically have 10 or 100 clips left over and loose and a box is available to hold them.
The first assumption is wrong.
Nothing says he have to put all clips in a box... He could have more than 10 left over clips if he doesn't have a box to put them in.
There is nothing in this problem to limit leftover clips to a maximum 9. He could have 5,000 leftover clips and simply not have any boxes for them. If he had ten leftover clips, nothing implies that he would be able to manifest a new box to hold them.
Furthermore, the phrase 'some boxes' implies multiple to me. I don't understand why 0 & 1 are acceptable values for 'some boxes' while 10+ is an unacceptable amount for 'leftover clips' (save for the fact that it would ruin the handy place value trick). The issue is really that this is a poorly worded question.
My answer: 523+111c for all non-negative integer values of c.
5:57 - no, the answer *_can't_* be 301 because we know he has *_some_* boxes of 10 and 0 isn't "some".
If some can be 0 boxes of ten, than some can also be 0 left over clips. It doesn't make much sense that way, as you'd be left with a box of -10 clips for a mathematically sound answer of 190, which is pretty silly given the question.
I find it odd to point out in one part of the video that some lose clips must be more than zero, but then to say that zero is fine for some boxes later.
This entire chat doesn’t understand mathematical problems.
The term "some boxes" excludes the possibility of there being zero boxes.
I think that this is actually a digit and place value math problem, just with a twist.
And also it was stated "some" boxes hold 10 paper clips suggesting not only that there is at least one 10 paper clip box but at least two so the first possible number is technically 523 and the formula for the consecutive possible numbers would be like
f(n)=412 + 111n assuming n starts at 1 and goes to positive infinity
Why does “some” mean at least 2?…. That seems ridiculous
Not fair for 2nd graders. Many people find word problems unclear, and confusing for younger kids.
Even 4th graders would be stuck with this.
Gonna say 523 in order to assume the lowest/simplest possible values for common sense
Let
x = # of boxes with 10 paper clips
y = # of boxes with 100 paper clips
and z = # of left over paper clips
They’ve established in the problem that we can base the other variables in terms of x as they say that y is three more than x and z is two less than y which also means that z is one more than x so y is just x+3 and z is just x+1
Now there’s no viable way to get one correct answer here as the variables aren’t and cannot be defined in this situation so I’m assuming the teacher was aware of this as the problem says “could he have” rather than “does he have” so
Problem says “some” boxes hold 10 paper clips and that “some” is our x value. “Some” assumes more than one, and for simplicity’s sake I’m therefore gonna give x a value of 2 as its a value that satisfies the function without being too large a variable for a second grader so therefore
Let
x = 2
y = 2 + 3 = 5
z = 2 + 1 = 3
5(100)+ 2(10) + 3 = 523 total paperclips (but you can technically use any real number greater than 1 in place of x to satisfy the equation as long as you keep y and z in the same terms of x that are used in the problem)
"he has some paper clips left over" -- left over from what? Maybe they're left over from when he went on a box-shredding binge, having shredded a bunch of paperclip boxes and now having several hundred paperclips with no boxes for them. The problem presents zero reason to infer that the "some clips" must be less than the smallest box size.
i swear this question was asked in a 2nd year of middle school math tournament in greece but with candy. my teacher used it as an example
Well, while the number of leftovers must be less than 10, it doesn't necessarily mean that "it's the first sequence up to 9". Since L=10 results in 1300 clips and "some leftovers" means that there is at least 1 leftover (which contradicts the 301 argument), then we could jump to 1411 clips (with 1 leftover or L_2=1) and we could go up indefinitely as long as we keep the pattern and skip all L_n=10 cases. In fact, we can generalize this pattern as:
C = C(k) = 301 + 111 . ( k - 1 ) , k = k(j , n) = j - 10 . ( n - 1 ) , n = n(j) = ceiling( j / 10 )
where "C" is the total number of clips, "k" is the number of leftovers (goes from 1 to 9), "n" is the current "k" cycle and "j" is the current case considered
In this video we only consider n=1 (the first 1~9 cycle)
(Sorry for the confusing functions. I've always been bad at modeling problems and there's probably a better way of describing this, even though this took me quite a while to figure out xD)
"Some paper clips left over":
1. There's no reason to assume the highest number could be 9. Unless one feels the need to assume because one is used to being expected to assume things.
2. There's even less reason to assume the lowest number could be 1. "Some paperclipS" paperclips left over is clear enough phrasing to how that it can't be a single one.
2:23 - strictly, he must have at least 2 paper clips left over since 1 isn't "some". You actually get L >= 3 by applying the same interpretation of the word "some" to T and H.
Any adult that is baffled by this question is a complete simpleton
As soon as I saw this, I thought that it was analogous to forming a 3-digit number when the hundreds digit is 3 more than the tens digit, and the ones digit is 2 less than the hundreds digit:
301, 412, 523, 634, 745, 856, 967
Yeah, we could also have 10 boxes of 100, 7 boxes of 10, 8 leftover = 1078 or 11 boxes of 100, 8 boxes of 10, 9 leftover = 1189, but forming a 3-digit number seems to be more in keeping with 2nd grade math, since we're dealing with groups of 100, 10, and 1. Also, figuring out place values of digits sounds like something you would do in 2nd grade.
Am I to conclude that the box holding 10 clips is smaller than the one holding 100 ? If I add a paperclip to either box does it not now hold 11 or 101 clips? Also , where are you getting these (possibly smaller sized) empty boxes from to justify that L
We'll presume non-negative integral numbers for:
t=number of boxes holding 10 paper clips
h=number of boxes holding 100 paper clips
p=number of leftover paper clips
t+3=h
p+2=h
T=total number of paper clips
T=100h+10t+p
"some" ... "left over", so, not none, and we presume not enough to fill
box of 10, so
1
Some (adjective) "being at least one - used to indicate that a logical proposition is asserted only of a subclass or certain members of the class denoted by the term which it modifies" - Miriam Webster Dictionary
Which would remove 301 paperclips as a possible solution, as there would be at least 1 box of 10 paperclips. Removing 412 paperclips is not supported by the above definition of "some"
There being no more that 8 boxes of 10 paperclips, or not more that 9 unboxed paperclips, requires that Brion is only unboxing the fewest number of paperclips required for his task. This is sensible, but not stated.
Aside from the ambiguous wording, I would have rewritten the tens equation to use L:
Since H = L + 2 and T = H - 3, T = (L + 2) - 3 = L - 1.
Or you could write a single equation to do the lot...
C = 100(L + 2) + 10(L - 1) + L = 100L + 10L + L + 200 - 10 = 111L + 190
It’s not a 2nd grade answer, but I think the only complete answer is:
Total = 111T+301, for T>0 (>1;>2)
Depending on what you think T can be based on question wording.
I really did not like the wording of this question after seeing the answer. Firstly, "He has some paper clips left over", does not automatically imply that the number is limited to 9 clips. Nowhere in the question is it explained that Brian is filling these boxes and therefore has access to empty boxes.He merely possesses said boxes and lose paperclips. To me this means that the leftover paper clips is a positive whole integer of 2 or higher, since some implies a plural. This would also disqualify both 301 and 412 as answers, since Brian was supposed to "some" boxes with 10 paper clips and "some" leftover paperclips. Finally, I dislike the phrasing of the last sentence "What number of paper clips could he have?". This could also be construed as how many leftover paper clips does he have, not the total including the boxes.
Let's name the piles:
C(lips)=100B(ig boxes)+10S(mall boxes)+L(oose clips)
B=3+S
L=B-2
With substitution, we realize that C=111B+32. Now we have to find minimal values. We know that he has "some" of each, so we can safely assume that all our variables are positive integers. Specifically, S≥1 which implies that S+3=B≥4, so we know that C=(444+32)+111n for some natural number n. Thus the minimum viable amount of clips is 412, with all other solutions 111n above that (for natural number n)
I would approach this like a second grader. You have some boxes with 10, so I would say some boxes is 2. There are 3 more boxes with 100, so that’s 2+3 is 5. That’s 520 so far, but I know there are 2 less leftovers, so that’s 3.
So he could have 523 paper clips and that just takes reading comprehension and arithmetic.
I agree with another commenter who noted that it's not stated that he has more empty boxes to use if he has extra clips. My assumption is he has what's stated in the problem. So restricting this to 9 loose clips isn't warranted by the statement of the question.
The problem says that some boxes hold 10, some hold 100. This is not a guarantee he has any of these boxes, however. We simply know he has three more 100 boxes than he has 10 boxes, and two more 100 boxes than he has loose paper clips. This guarantees he has some boxes, and he has at least three 100 boxes. It also guarantees he has at least one loose paper clip. From that, we must assume that 0 boxes of 10 is acceptable.
Finally, yes this is a decimal place question. It would be unlikely a school kid would answer 1300 paper clips, but it is 12*100 + 9*10 + 10, and this satisfies the wording of the problem.
The best answer for this, to me, is 190 + 111k where k is a positive integer.
The reasoning about left over clips is a huge stretch. It is never mentioned that left over clips that could fit perfectly either boxes must be stored in such a box.
Some boxes is not none, and it’s not one. If the question were rephrased as having “an amount” of something then that amount, acting as a variable, could be zero.
Can you tell me the logic behind partial fractions in general?
What if you put a box inside another box?
Both have clips, just the same clips 🤷♂️
That still results in the same answer(s).
EDIT: Oh, wait, that's not true. It leads to the same solutions, plus a bunch more solutions. The minimum solution would be 101 .
Allowing the number of leftover clips = L with L greater than or equal to one, then the total number of clips = N = 111*L + 190
At a bare minimum, this problem teaches kids to make assumptions that may not be true. I understand what the intention is, but I could be phrased way better you saying that he started with boxes of hundreds which are ten boxes of ten, and doesn’t open a box unless he needs to
The formula describing sets of solutions: 100*x+10*(x-3)+(x-2), x>3. 111*x-32.
Nothing said in condition regarding principles of storing clips aka if more than 10 leftover put to box 10, if more than 10 box 10 put to box 100 and etc. (if, then only, x
6:35 - I'm sorry, but saying you don't see anything in the problem that precludes having zero boxes of 10 means that (a) you have just redefined the word "some" to mean "some or none", (b) L can be zero, and (c) there is now no reason to preclude negative numbers of boxes either.
Zero isn't some, zero means none.
If this problem was truly given in school by a teacher, the teacher should be ashamed of himself. Impossible to answer without defining 'some'. And if 9 is meant to be the max free clips, it should be stated somewhere. There are infinitely many answers, arguably starting from 523. Kids should get extra credit for calling out extremely poorly worded questions.
total = 100H + 10T + L
total = 100(T+3) + 10T + T+1
total = 100T + 300 + 10T + T + 1
total = 111T + 301 (I think this shows the pattern better)
if we say L must be less than 10 and we are not allowed to have 0 boxes, then T must be 1,2,...7,8
so we have a total number of 412, 523, 634, 745, 856, 967, 1078 or 1189 paper clips
What if to start Brian had all full boxes and some leftovers from Tommy's box of 100 after Tommy had used one? This would also perfectly fit the grammar of the question, giving us 99 leftovers, 98 boxes of 10, and 101 boxes of 100 = 10100 + 980 + 99, or 11179.
I have a nephew who in the second grade would have said there's no reason the leftovers couldn't have come from a box of 1000 or a case of 144 boxes of 1000, but the question does not ask for all possibilities, any one viable solution would suffice.
Given this is for 2nd grade students, I think the question is modelling the idea of 100s, 10s and units and maybe it is the last question in a series of questions where Brian's boxes of paperclips represent the 100s, 10s and units. So I think it's asking you to present a 3 digit number where the number in the 100s place is 3 more than the number in the tens place and the number in the units place is two less than the number in the 100s place.
So:
301
412
523
634
745
856
967
Obviously this is presuming that he must have less than 10 of each for each type, but perhaps that would be more obvious in context.
It is difficult for second grade but why the hell an adult may have a problem with this problem
because differing definitions of "some" dont auto-resolve into one once you turn 18
Some adults are stumped by the poor articulation of this problem, some adults are trying to work with the incomplete data and some couldn't care less.
06:28
it also doesn't say anywhere in the problem that the leftover clips have to be less than 10...
it also doesn't say anywhere in the problem that the "number of paperclips could he have" are not the number of left over paperclips but the number of total paperclips inside the boxes and outside...
412+111(n-1) where n is number of 10 paper clip boxes. I also assume the boxes are fixed and he could easily have enough spare paperclips to fill another box that he just doesn't have. The word SOME is actually plural neutral so it means at least 1 in my interpretation and not 2 or more. 1189 would be the most if we had to stop but it is VERY poorly worded.
He said _"some boxes hold ten clips,"_ which, by definition, means that there *are* boxes (plural), which he "has," holding 10 clips each.
"Some boxes (plural) hold (not _"might/could hold")_ 10 clips."
So at least 2 boxes *DO* hold 10 clips each. That's what was said.
He *HAS* some (plural) amount of _boxes_ with the specified value of ten clips.
You wouldn't say that you _have_ cars (plural) in your garage if you did not *have* any.
You cannot say that you "have" a thing when you do not "have" that thing.
Saying you _"might"_ have that thing leaves open other possibilities. And since this type of qualifier is clearly used at the end of the equation, the only reasonable inference is that "maybe/might/could" qualifiers will be used when they apply and will not be used when they don't apply.
I'm still trying to figure out what kind of box only holds 10 paper clips. :P
I guess in an office store, you can get plain, standard paperclips that come in a box of 100, and larger, fancy colored paperclips that are sold per 10. (The latter would also cost more per paperclip.)
This is my take. The only thing in the text that limits the number of leftover clips is the number of boxes of 100 clips. Lets call this boxes Y and the leftover clips Z. So Z is always Y-2. The only thing that limits the number of boxes Y is the number of boxes of 10 clips. Lets call this boxes X. So Y is always 3 more than X. Now we have, Y=3+X and Z=Y-2. There is anything that limits the number of boxes of 10 clips. He may have a full warehouse for what we know. The minimun number of boxes he can have is 1*10+4*100+2 leftovers=412 clips. That is the minimum amount of clips, from there, he can have any number of clips that follows the next equation: 111A+301 being A a positive integer.
Looking through the comments I definitely don’t think 301 is an answer. Some can’t mean zero. Then if you look at 412 - you can’t say “1 boxes of 10 paperclips”
Let...
A = # boxes that can hold 10 clips
B = # boxes that can hold 100 clips
L = # leftover clips
___
We know...
B = A + 3
L + 2 = B
Total = 10A + 100B + L
___
This can be simplified in terms of A, since
B = A + 3
and
L = B - 2 = A + 1
So...
Total = 111A + 301
If real world common sense is assumed, then A >= 0, where A is a nonnegative integer.
301 would be one possible answer, with other possible answers varying depending on the number of boxes of A
(which depends on how the likely 8-to-9 year-old 2nd grader interprets the question).
Some implies more than zero, boxes implies more than one… 523 is the least it could be.
Total (T) = 100a + 10b + c.
From the rest of the equation we get a = b + 3 and a = c + 2. We can deduce that b = a - 3 and c = a - 2
Now we can get everything in terms of a.
T = 100a + 10(a-3)+(a-2).
T = 100a + 10a - 30 + a - 2
T = 111a - 32.
But since a, b and c all have to be 0 or more we can figure out that a ≥ 3 since b ≥ 0 and a = b + 3 ⇒ a ≥ (0 + 3).
So any solution for the equation 111a - 32 ∀ a ≥ 3 is a solution to the problem.
I'm not sure how it's a question for 2nd Grade/Year 3/7-8 years old pupils.
'Some' implies two or more (never 0 or 1). So, 5x100, plus (5-3)x10, plus 5-2=3, 523 clips. Or: 10x100, (10-3)x10, plus 10-2=8, 1078 clips. Since nothing says that he has to put left over clips in boxes, so 9999x100, (9999-3)x10 plus 9999-2=9997 is also fine.