The following question (number 15 of this test) has become infamous as a poor "Common Core" question. What is the correct answer?

Juanita wants to give bags of stickers to her friends. She wants to give the same number of stickers to each friend. She's not sure if she needs 4 bags or 6 bags of stickers. How many stickers could she buy so there are no stickers left over?

  • $\begingroup$ This is perhaps a bit old by now, but reading online discussions reveals a fair bit of confusion, so I thought it was worth it to make this question here. $\endgroup$ – Eric M. Schmidt Dec 19 '17 at 5:52
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    $\begingroup$ It's a terribly written question, but that's hardly exclusive to common core. It is a risk in writing any word problem. $\endgroup$ – Thomas Andrews Dec 19 '17 at 6:04
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Daniel Fischer Dec 19 '17 at 20:02
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    $\begingroup$ Better phrasing (assuming multiple choice): Juanita doesn't know whether she has 4 friends or 6 friends. She wants to buy them some stickers and give the same number of stickers to each friend. How many stickers could she buy? $\endgroup$ – Reinstate Monica Dec 20 '17 at 21:55
  • $\begingroup$ @Solomonoff'sSecret Or for a unique answer without multiple choice: Juanita doesn't know whether she has 4 friends or 6 friends. She wants to buy them some stickers and give the same positive number of stickers to each friend. What is the smallest number of stickers she could buy? $\endgroup$ – Solomon Ucko Apr 3 '19 at 11:37

This "infamous" question is so poorly worded, but any mathematical answer that can be given would reinforce the perception that "the question is OK, see, it has a valid answer!"

The question is not OK.

Word problems should not be about mulling over "what did the author want to say?". Even the "trick" questions are normally about spotting some well-defined linguistic (or mathematical) trick, rather than based on complete lack of clarity. Plus, note it was not meant to be a trick question, but some standard question meant to assess how well students of certain age comprehend divisibility and common multipliers.

I can personally think about a dozen different ways to write a word problem which would boil down to the same mathematical problem and will be better worded.

As for this question, I hope if it was on some exam, that it did not affect anyone's passing or failure. I can relate to a poor methodical soul who got stuck for two hours on that question, trying to get their head around it and failing to comprehend it (and also wasting precious time for the following questions) not realising that it is not their fault.

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    $\begingroup$ @Dunk As mentioned in the comments on the question, it reads like there's information missing in the question. I've actually taken tests in highschool where the teacher forgot something and had to announce it at the beginning of the test, so it really isn't that far-fetched. $\endgroup$ – Izkata Dec 19 '17 at 19:32
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    $\begingroup$ What is missing is: we don't know anything about the "bags of stickers" and how they are related to either the stickers or to the friends. Exercise: try rewriting the question without mentioning bags, and not only you will succeed easily, but the resulting question will be completely clear. It's anyone's guess why the author didn't do that. $\endgroup$ – user491874 Dec 19 '17 at 19:45
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    $\begingroup$ @user8734617 The friends are irrelevant, because Juanita has already figured out that she needs either 4 or 6 bags in order to distribute the stickers. For all we know, she might only have two or three friends who will each get two bags - it doesn't matter, because we're only concerned with filling four or six bags. $\endgroup$ – Iszi Dec 19 '17 at 22:42
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    $\begingroup$ @Joshua 0 is also multiple of 12, and it's clear that no matter how she distributes the stickers in the bags, if she buys 0 she'll have 0 left over. Hence: 0 > 120 (wrt to the partial ordering by divisibility). $\endgroup$ – Kimball Dec 20 '17 at 3:35
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    $\begingroup$ @Iszi, that only works if you've already realized that Juanita is buying individual stickers and then putting them in bags and giving one bag to each friend. The way the question is written implies that she is buying bags of stickers and that the correct answer is either "4 bags" or "6 bags" and you're supposed to somehow figure out which. $\endgroup$ – Harry Johnston Dec 20 '17 at 4:06

The difficulty arises from the confusing wording and strange premise. Juanita is buying "stickers", not "bags of stickers". Each friend is to receive a single bag of stickers, with the number of stickers in each bag the same. For reasons unexplained, Juanita does not know exactly how many friends she is giving stickers to, but she does know that it is either $4$ or $6$. Thus, the question is how many stickers to buy so that she will be able to divide them evenly among the friends, whether there turn out to be $4$ or $6$ of them. The answer, then, is any multiple of both $4$ and $6$. Equivalently, the number of stickers can be any multiple of $12$, which is the least common multiple of $4$ and $6$.

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    $\begingroup$ I would not have been able to answer that question. It seemed like there was missing info, and I did not know how to interpret the "does not know if she needs 4 or 6 bags" info. $\endgroup$ – Michael Dec 19 '17 at 8:40
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    $\begingroup$ How can we assume that she either has 4 or 6 friends? $\endgroup$ – stanri Dec 19 '17 at 9:59
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    $\begingroup$ "She is not sure if she needs 4 or 6 bags of stickers" to me means that either 4 or 6 are the acceptable answers, and nothing else. Who ever wrote that question should be tarred and feathered. Its furthermore absolutely unimaginable that she would not know how many friends she has. if its either 4 or 6, what if she has 6, but one doesn't show? Then she has stickers left over. But if left overs are allowed, the question is nonsensical, because 6 bags would also work for 4 friends if leftovers are acceptable. This does not test maths at all, it tests wether you can read minds... $\endgroup$ – Polygnome Dec 19 '17 at 11:27
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    $\begingroup$ Maybe this is better? Juanita invited her friends to a party. Three children said they were definitely coming, but Bob and Sue said they might not be able to come. Since Bob and Sue are siblings, either they both come or they both do not. So (counting Juanita herself) there will either be 4 attendees or 6, but Juanita is not sure which. She wants to cut the cake into pieces so that, independent of whether 4 or 6 people are present, each person could get an equal number of pieces. How many pieces could she cut the cake into? $\endgroup$ – Steven Gubkin Dec 19 '17 at 16:34
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    $\begingroup$ @StevenGubkin Much better but the English might be too hard for some age appropriate children. I am not referring to ESL but the subtleties of the word independent in this context or the word attendees. That stipulation does make writing these much harder. $\endgroup$ – kaine Dec 19 '17 at 18:31

Juanita buys zero bags. Then she gives zero to each friend. After that, there are zero stickers left over.

(Note that, consistent with other answers, zero is a multiple of twelve.)

Beyond that, I keep wondering how many bags of stickers Juanita intends to keep for herself after supplying her friends with some. It may well be that there are no circumstances under which there would be any "left over".


Short Version

The correct answer is "any multiple of 12", but they really probably just want "12".

Long Version

I agree that the question is poorly-worded, but after some reflection upon existing commentary and answers, I also agree it is possible to work out the proper mathematical problem and its possible solutions without requiring many leaps of logic or unfounded assumptions. However, I also believe that the actual answer to the question is not the answer that test reviewers would consider correct.

First, about how to solve the problem:

The first thing a student needs to do is eliminate what they don't need from the problem.

This is a common assessment objective of word problems - finding whether the student can determine what is or is not relevant to the question.

The scenario talks a lot about bags, stickers, and friends. We know that we need to fill bags with stickers, and the bags are being given to friends. We don't know how many friends Juanita has and, at first glance, this would seem to be part of the problem. However, the number of friends is irrelevant because Juanita has already narrowed the quantity of bags to two possibilities - 4 or 6.

Now the student can determine the mathematical problem, and its possible solutions.

We have either 4 or 6 bags to fill, and we want to buy an amount of stickers that divides evenly among the bags no matter which number is actually true. The easiest way to do this is to multiply the two numbers, for which you get 24.

It is important to note here that 24 is actually a valid answer to the question that has been presented to the student.

Now the student could conceivably extrapolate the ideal result, and determine that solution.

Considering that Juanita is buying these stickers, she probably doesn't want to pay more than necessary. And, if they understand the concept of the Least Common Multiple, the student could realize that there might be lower quantities of stickers that could satisfy Juanita's needs.

At this point, they would then do the appropriate work and arrive at the answer of 12 - which is probably the answer the instructors want to see on the standardized test.

The question as written leaves open the possibility of an infinite number of correct answers being considered "wrong".

That last part is where this question, in my opinion, fails to serve its purpose horribly. There's an important difference between this (from the original question):

How many stickers could she buy so there are no stickers left over?

And this (change in italics):

What's the fewest number of stickers she could buy so there are no stickers left over?

The former leaves all multiples of 12 as possible valid answers. In fact, the most correct answer would be for the student to write (as it's a free-form answer field anyway) "Any multiple of 12".

However, the latter form narrows the field of possibilities down to only the answer that costs the least amount of money. For that question, the only correct answer (assuming the stickers are individually priced, and there's no buy-one-get-one-free sales on) is 12.

A Better Question

Personally, I'd probably suggest a rewrite similar to this:

Juanita is going to the store to buy some stickers for her friends. To distribute them equally, she's going to put the stickers into bags. However, she left her bags at home and can't remember whether she has 4 or 6 bags to fill. She wants to be able to give the same number of stickers to each friend. How many stickers should she buy, to avoid spending any more money than she needs to while still equally dividing the stickers?

That puts the question in simpler terms, and links it to a practical real-life need (saving money), while keeping all of the elements from the original question.

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    $\begingroup$ You've made a mistake here: you've silently replaced the constraint "She wants to give the same number of stickers to each friend" with "we want to buy an amount of stickers that divides evenly among the bags". That sleight-of-hand allows you to argue that "the number of friends is irrelevant" (because suddenly all you care about is bags), but nothing in the problem statement entitles the student to replace the one constraint with the other. $\endgroup$ – ruakh Dec 20 '17 at 2:27
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    $\begingroup$ The info "each friend gets one bag" is missing in the question. Students might find out that they need some link between friends and bags, and learn that they should just assume something that looks reasonable. Later in life, they start a sticker-buying company, getting Juanita as customer. Juanita is still not clear on what she wants, so the student buys 12 stickers. Juanita becomes angry, because she 'obviously' knows that she has four friends, she was doubting about the number of bags because two of her friends think it is easier to carry two bags. ... $\endgroup$ – user193810 Dec 20 '17 at 8:07
  • $\begingroup$ ... So Juanita 'obviously' wanted to have 8 stickers to give to 4 friends, she just was not sure if she wanted them in 2+2+2+2 or 2+2+1+1+1+1. So the business fails, all because the student was taught in school to just assume something if there is not enough info, in stead of asking for clarification. $\endgroup$ – user193810 Dec 20 '17 at 8:10
  • $\begingroup$ @Pakk Yes, so when you can get clarification you do And when you don't, you presume bags=friends, determine if the risk of error is worth the cost of double checking, and usually just get on with it. $\endgroup$ – Yakk Dec 20 '17 at 18:34
  • $\begingroup$ @Yakk: With questions as bad as in the common core test, students will not learn that there is a risk associated with such assumptions. The worst thing is that it is so easy to make better questions, as Iszi also demonstrated... $\endgroup$ – user193810 Dec 20 '17 at 20:45

The question is terribly worded. Here is another solution which is consistent with the wording:

The only way Juanita can be unsure whether she needs 4 or 6 bags is if she has exactly 1 or 2 friends; any other number of friends does not divide evenly into both 4 and 6 bags and she would know in those cases that at least one choice was inappropriate. The question clearly states that Juanita has more than one friend, so she must have two. To ensure that no stickers are left over, she should buy the smaller of the two allowed choices, i.e. four bags.


The answer is 5.

Juanita cannot decide whether to buy "4" or "6" bags because she has 5 friends. In common core lingo "5" is between "4" and "6." And "4" and "6" are the clue numbers given in the problem--pay attention! LOL This is how we know Juanita has 5 friends. Since Juanita doesn't seem to know how many stickers are in each bags, she simply buys 5 stickers.

  • $\begingroup$ Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, use MathJax. $\endgroup$ – dantopa May 28 '19 at 0:58
  • $\begingroup$ or she buys their lcm lol $\endgroup$ – user645636 Dec 11 '19 at 13:27

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