Math question

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

airvent

.
10+ Year Member
5+ Year Member
15+ Year Member
Joined
Jul 5, 2004
Messages
323
Reaction score
0
I am not sure how to approach this one. Is there a method or formula. 😕

A wire is cut into 3 parts. Then each segment is cut into 4, 6, 8 parts respectively. If each resulting segment has an integral length, what is the minimal length of the wire?

A) 24
B) 36
C) 48
D) 54
E) 72

Thanks
 
The answer is E) 72

Unless I'm missing something, the original length of the wire (which must be present among the answer choices) has to be evenly divisible by 3; unfortunately, they all are. However, one's selection of an answer is further constrained by the qualification that each of those 3 remaining segments must be divisible by 4, 6, and 8 without remainder. So, for example, analyzing choice C, we have:


48/3 = 16 (so we now have 3 segments of length 16 after the first cut)

16/4 = 4 (fine so far)
16/8 = 2 (still ok)
16/6 = 2.66... (not an integer--BZZT, wrong answer 😛)


For choice E (72), we have:

72/3 = 24

24/4 = 6
24/6 = 4
24/8 = 3

(Note: all integers)


Ergo, we can see that only an original wire with length 72 can be divided into 3 equal segments (though it does not specifically say they are "equal length", the logic and math needed if one were to NOT assume as much would be quite complicated 😛), with the remaining 3 segments each integrally divisible by the given divisors (4, 6, and 8). Feel free to verify this by doing the above calculations with the other answer choices if you like.


Hope this helped (I also hope I'm correct). 😀 🙂
 
Also, please realize that, like I mentioned above, this only holds if they are assuming 3 segments of EQUAL length; if they aren't, though one could go through various mental and mathematical contortions so as to arrive at the answer, the easiest way I could see would be like so:

Assume that the length of the first segment was 4, the length of the second 12, and the length of the third 8 (after the first cut and before the second cut). Now, the sum total of these is 24, which is choice A. The reason why the second segment couldn't be 6 instead of 12 (using the lowest integer, 1, for the length of the final integral segments instead of 2 as I do here) is because this would add up to 18 (4+6+8), which is not among the choices given. Ditto for the other various permutations (8+6+8 = 22; 4+6+16 = 26; neither of which are given). Now, one knock on this method (and the question in general) is that it assumes that the secondary segments are cut into 4, 6, and 8 equal integral pieces, respectively, with no remainder.


However, this assumption is, in fact, implicit in the entire question; if we are to view it as such, my first answer is still the more plausible one, as if we have to "read meaning" into the question (re: equal-length segments), we must do so consistently (both prior to the first AND second round of cuts), and this would necessitate the answer I gave previously of E) 72.

In either instance, however, the question is poorly phrased now that I think about it. Though my first reaction was to solve it in the manner illustrated in my previous post, this would mean that the fact that the 3 segments were of equal length would have to be merely understood, rather than explicitly stated as a condition; this reflects poor question writing imo.


In the second condition (the alternative explanation seen above), we would still have to assume the same exact thing, as even if we are to assume that the original 3 segments are unequal in length, we are then assuming that the second cut results in equal-length segments (for instance, you can cut a strip of length 8 into either 4 pieces of length 2, or 4 pieces of lengths 1, 2, 4, and 1; as you can see, either way the notion of equality of lengths is implicit). Furthermore, the "alternative" method also exposes the piss-poor wording of the question even further, as "24" wouldn't be the "minimal length of the wire" (as they ask for), but rather would be "the minimal length of the wire among the choices given" (and there is a difference between these two). Because, clearly, we could assume that the length of each of the 3 resultant segments present after the first cut (assuming unequal lengths) was 4, 12, and 8; these would then be cut again into 4, 6, and 8 segments (now changing it up and assuming equal-length cuts), which would result in 12 segments of length 1 (4/1 + 8/1) and 6 of length 2 (12/2) for a total of 24, which would be choice A.


As you can see from either scenario, somewhere along the line, we must make the implicit decision that we are assuming equal cuts; the first scenario (previous post) is the "cleaner" one imo, and was more intuitive to me, so I stand by it. My sister, after looking at this problem, got me further into thinking about this notion of unequal cuts, because right away she stated that it didn't mention that they were 3 segments of equal length (it's funny how people's minds work differently 🙂; though I did grant this in my previous post-- see the parenthetical), and I quickly came up with the above rationale as to why, even if we assume such a thing, the first scenario is still more plausible.

In summary, the question is too ambiguous for its own good, but if I were a betting man, I'd still wager on the answer given in my previous post. I wanted to post this as sort of an exploration of alternative thought processes, since my sister got me started down that road, despite the fact that I was aware of the assumption I was making about equal length segments. I also realize that this might be hard to follow, as writing about math and logic is not my forte-- it's convoluted (and quite possibly insubstantial 😉), I know. For this, I apologize. 😛


And if this all boils down to the fact that you forgot to copy down the part question which stated that they were 3 (and eventually 4, 6, and 8) pieces of equal length, well, you've cost me 10 minutes of my life that I want back. 😀 Maybe I just think too much-- that's why I'm about to go play basketball, where I don't have to think at all. 😳
 
Holy shmuck, that was a heck of an answer! I started reading it and got hives. CJMPre-Med - if you are a pre-med student then you must truly love math to have wandered into the dental forums, devote at least a half hour of your life to formulating a response, and post it. Wow...
 
Sprgrover said:
Holy shmuck, that was a heck of an answer! I started reading it and got hives. CJMPre-Med - if you are a pre-med student then you must truly love math to have wandered into the dental forums, devote at least a half hour of your life to formulating a response, and post it. Wow...

Thanks for the kind words. 🙂 Actually, the second post took me only a minute or so to conceive of, but then about 10-15 minutes to write and express it in as clear a manner as I wanted to (I still feel it's too convoluted in writing, but whatever 😛). The original post only took a few minutes to write, and I originally solved the problem in about 20 seconds or so; this is not to brag, as I'm sure that many others here can do the same, since I don't feel it to be a terribly difficult problem as far as intuitive math goes-- neither do I mean to disparage the original poster for having problems with it, however, as we are all weaker in certain areas and have some specific problems that we can't seem to figure out for whatever reasons sometimes. 🙂 Point being, that somebody would have to be paying me for me to devote "at least a half hour" of my time in answering a question-- especially on my vacation. 😛


For the record, though I've always been very good at math, I've never progressed beyond calculus 1, as I didn't have to. I may take calc 2 before I graduate, but I'm not sure yet. Intuitive math/logic stuff like this, however, comes very easily to me (as it doesn't really rely on learned knowledge and/or proofs/rules etc., which I'm terribly out of practice with), so I figured I'd help him out. 🙂 I just happened to be scrolling down the main forum listing page, and I usually look at the most recently posted topic to the right of each forum listing; I saw that it said "math question" and curiosity got the better of me. And yes, I am a pre-med student in undergrad, though I am (recently) 26 years old. Now if I was 19 years old, you would have gotten a really good answer, as I've gotten dumberer 😉 over the years.


So though I thank you for the compliment, I'm not going to sit here and take a bow as if I just solved some daunting mathematical problem which was baffling the eminent minds in the field, because I didn't (and, quite frankly, couldn't). 😛


This question is just one of those things that you either see or you don't; Airvent may have been overthinking the problem and making it harder for himself (heh, I'm one to talk about "overthinking", I know 😉), as evidenced by his query about whether there's a "formula" or something for this. Sometimes we get so caught up in studying facts and theorems that we overlook the simpler, more obvious solutions to some intuitive problems. 🙂
 
Even your reply to my post is publishable. For such an informal forum your English is impeccable. Judging by your use of syntax and choice of words I can tell that you, my friend, have an exacting mind for numbers and precision. I, on the other hand, do not and am glad there are people like you that are out there. I wish you the best of luck in your pre-med adventures. 🙂
 
Why, thank you-- that's very kind of you. 🙂 You're making me blush, however. 😳 😀

Time to go grab some munchies before my head gets too swollen. 😉 😛
 
Well, the answer is 72. I think I was overthinking it. I was looking for a general method to solve such problems. I had originally gotten 72 as the answer, but had no way of knowing if I arrived there by design or random chance.

Oh, 3 equal parts, then 4, 6, 8.

Thank you CJMPre-Med
 
Idiopathic said:
The way the question is written, any multiple of 18 would do the trick. The key is the equal distribution, which I dont see in your post, so I go with 36. 😉

Are you serious? 🙂 Because I've always been terrible at picking up sarcasm online-- the winking smiley (and your italicizing of "equal", possibly poking fun at my own penchant for highlighting/italicizing 😀) doesn't help matters much. 😛


But anyway, assuming you were serious, no, not just any multiple of 18 will do in this case. 36 doesn't work as the original length of the wire because this is first divided into 3 segments of equal length (12 each); these are, in the next step, divided into 4, 6, and 8 equal pieces of integral length. Now, dividing 12 by 8, we see that the resultant length of the segments would not be an integer, but rather a real number (1.5). So 36 cannot be the original length of the wire. If you try this same procedure for the other choices besides 72, you'll see that the eventual 3 equal segments are not integrally divisible (i.e., no remainder) by one or another of the given divisors (4, 6, or 8).


Now, 72 works as the length of the original wire because this is first divided into 3 equal parts each of length 24. Now, 24 is evenly divisible by 4, 6 and 8; thus, we can see that 72 must be the correct response. 🙂
 
Hmm, I was just talking to my sister about why, even if we are to assume unequal lengths for the 3 segments, we still must implicitly assume equal lengths for the final 4, 6, and 8 segments derived from those 3 (as proffered in my second post). In the course of explaining my reasoning, it became clear that one could not, in fact, reason in such a manner, as if the first round of 3 cuts was unequal (say, it resulted in segments with lengths 4, 12, and 8 as I proposed), then the second round of 4, 6, and 8 cuts could possibly be unequal. Now, for the segments of lengths 4 and 8, this is not an issue, as 4 can only be divisible integrally by 4 one way (e.g., 4 segments of length 1), and the same for 8 being divisible by 8 (8 seg. of 1 each). For the segment of length 12, however, things become tricky; this is because it doesn't matter whether you divide the 12 into 6 pieces of length 2 or 6 pieces of lengths 1,1,2,2,2, and 4-- the total length of the segment is still 12, and would still count towards the total of 24, which is choice A (4+12+8).


This is why, as I mentioned above, the wording on the question is very poor-- although you later clarified and said that it was assuming 3 equal segments, which makes the correct answer 72, as I originally posted. So basically, my entire second post above is hogwash and is logically unsound. I'm quite peeved at myself for not seeing such a thing and for horribly skewing the issue and making it more complicated than it had to be. 😛 Though also, if the notion of unequal cuts was to be entertained, not only would the questioner be remiss in not explicitly specifying equal vs. unequal cuts, but they would also be at fault for phrasing the ultimate question asked in an incorrect, overly broad, and vague manner (i.e., "what is the minimal length of the wire" instead of "what is the minimal length of the wire among the choices given"). Because, if we are to take the notion of unequal cuts to its logical conclusion, clearly the answer would be 18 (4+6+8, as these segments would be the smallest which are integrally divisible by 4, 6, and 8; not 4, 12, and 8, which were only used to conform to the choices given).


Anyway, I guess the moral of the story is that we all have our moments. 😉 😛


The other moral is that next time you should post the entire question verbatim-- I lost brain cells on this. 😀
 
Top