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How to explain this?

  1. Jul 31, 2015 #1

    julian

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    I buy a packet of 28 biscuits with intention of eating two biscuits every lunch time (including weekends). However, sometimes I get greedy and eat an extra biscuit at lunch time (I always eat 2 or 3 biscuits each day). If I DONT eat any extra I would end up eating the last two biscuits on say a Thursday lunch time, and as such would have no biscuits left to eat on the Friday lunch time. Now say I did eat extra - question: how many extra biscuits have I eaten in order to have no biscuits left to eat Wednesday (in the second week) lunch time?

    The answer is...if I get to Wednesday and find I dont have the two biscuits I was going to eat that lunch time and I dont have the two biscuits I was going to eat Thursday lunch time then I must have eaten 4 more than I should have, so the answer is 4 extra.

    A person I know is saying I ate 6 extra biscuits (errh?). I try telling him that if I get to Wednesday and I'm missing 4 biscuits (the two for Wednesday lunch time and the two for Thursday lunch time) then I must have eaten 4 extra.

    He cant work out why this is correct. How would you explain it to a person?

    Here are a couple of arguments you can give:

    Here is an argument that directly equates number of extra eaten to number left come Wednesday when sticking to two biscuits per day:

    On a certain number of lunch times I ate an extra biscuit so as to end up with none left to eat on the Wednesday (of the second week).

    Here is something you can do...Do the whole over again but this time instead of eating the extra biscuit that you remove from the packet, place it to one side. You do this enough times in order to have none left in the packet come Wednesday. The number you have put to one side is then exactly equal to the number of extra biscuits I ate!

    All we have to do now is work out how many biscuits I would put to one side, but that is easy...because I'm not eating the extra biscuits I take out of the packet, it means that I am eating two biscuits per day. But we know what happens when I stick to two biscuits per day, I end up with 4 biscuits left come Wednesday (two for Wednesday lunch time and two for Thursday lunch time). So 4 is the number I will have put to one side because that is how many I'm supposed to have left over. So 4 is the number of extra biscuits eaten.

    Here is something you would physically do:

    Get 14 boxes. Label each box by a day of the week. Get your 28 biscuits and put two biscuits in each of these boxes - the last box being labelled the "Thursday of the 2nd week". When you want to eat an extra biscuit on one of the lunch times you draw from one of the biscuits in the "Thursday of the 2nd week" box. On another day when you want to eat an extra biscuit you draw from the one biscuit left in the "Thursday of the 2nd week" box. If you want to eat a 3rd extra you draw from one of the biscuits in the "Wednesday of the 2nd week" box. When you want to eat a 4th extra you draw from the one biscuit left in the "Wednesday of the 2nd week" box.

    And then you stop eating any more extra biscuits because that would involve eating one of the biscuits in the"Tuesday 2nd week" box.

    See what arguments you would give...
     
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  3. Jul 31, 2015 #2

    collinsmark

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    You didn't specify which day you started eating the biscuits. Usually these types of riddles mention that.

    We can work backwards though based on your statement, "If I DONT eat any extra I would end up eating the last two biscuits on say a Thursday lunch time." That means, given that you had 28 biscuits to a batch, that you started eating the first biscuits in the batch on a Friday.

    So from there, every two "extra" (above and beyond the standard two that you would normally eat) biscuits eaten reduces the final day by one. 0 extra biscuits corresponds to finishing the last one on Thursday (meaning you have none for Friday). 2 Extra biscuits corresponds to finishing the last one on Wednesday (meaning you have none for Thursday), and so on. [Edit: which gives that if your first day with no biscuits was a Wednesday of the second week, you must have ate either 4 or 5 biscuits (less than 6 but greater than or equal to 4).]
     
    Last edited: Jul 31, 2015
  4. Jul 31, 2015 #3

    julian

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    I did start on a Friday, you can work that out from the information given. Yes your "spoiler" answer is what I get - but not what this person gets...
     
  5. Jul 31, 2015 #4

    collinsmark

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    I suggest guiding the other person to reach the correct conclusion himself/herself by taking it one biscuit at a time. Have the other person map extra biscuit to days, starting with 0 extra biscuits and moving up from there.
     
  6. Jul 31, 2015 #5

    collinsmark

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    I should point out that usually these riddles specify the day of the week that the biscuit batch was acquired, and specify that they started eating the biscuits that same day. The conundrum that most people have is that if no extra biscuits are eaten they incorrectly conclude that they will finish their last biscuit on the same day of the week they started eating them. The resolution is that that if no extra biscuits are eaten, then the day that they started eating is the same day of the week that they first have no biscuits -- not the day that they finish the last biscuit.

    If using continuous time, rather than discrete time, it might be useful to consider/specify whether the biscuits were acquired before or after lunch of that day.
     
  7. Jul 31, 2015 #6

    julian

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    The biscuits were acquired before Friday lunch time and then ate the first biscuits that day, the Friday. There is no riddle to it.

    People sometimes make the mistake of thinking if you eat the first biscuits on a Friday, and dont eat extra, then you end up eating the last two also on a Friday - which is wrong because that would correspond to 15 lunch times.
     
    Last edited: Jul 31, 2015
  8. Jul 31, 2015 #7

    Borek

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    Deja vu. Wasn't this (or very similar riddle) posted just a few weeks ago?
     
  9. Jul 31, 2015 #8

    collinsmark

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    I remember a thread/riddle nearly identical, but several months ago. Or was it over a year ago? Then again, maybe it was just last week. I'm pretty bad when it comes to continuous time.

    This sort of riddle is a good exercise at getting students into the mindset of thinking about discrete-time systems. The purpose of such exercises is to show that sloppily applying continuous-time equations to their corresponding discrete-time systems can lead to incorrect conclusions. The math works out fine if applied correctly, but sloppy application is all too easy with many people -- at least until they get into the right mindset.
     
  10. Jul 31, 2015 #9

    julian

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    I'm still having trouble getting the guy to understand and I'm trying to come up with other ways of explaining it to him. Here in particular I'm giving more rigorous arguments than before that are foolproof (this is the point of my two spoiler answers). His mind set is something I'm trying to figure out exactly so I can get him to correct it. I think the guy is being very sloppy with his arguments - (but then he is not exactly a student of maths). Sorry I'm off to bed now though. Tomorrow maybe we could try to going into more specifics.
     
  11. Jul 31, 2015 #10

    collinsmark

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    This sort of mindset is not limited to discrete-time systems, but can be generalized to discrete systems altogether.

    Consider an example of mapping temperature to an 8 bit, unsigned number. Suppose the temperature range you are interested in representing is 0o C to 100o C. [Edit: Also suppose temperatures outside of this range, i.e., < 0o C and > 100o C, are of no interest and can be neglected.]

    Method 1:

    The simplest method is to realize that range of temperature is 100 degrees, and the cardinality of an 8 bit number is 256 (the number of states that can be represented by an 8 bit number). So you have a resolution of 100/256 = 0.390625 degrees per count.

    So you can convert that number back to degrees by reading the 8 bit (unsigned) number and multiplying by 0.390625 degrees per count, and you have your temperature in degrees.

    And this is a good method! It's usually the best type of approach. But it's imperative to realize that each number actually corresponds to its own small range of temperatures. Each number represents a range of temperature 0.390625 degrees wide.

    Method 2:

    This is usually the method that causes problems.

    It should be noted that 100o C cannot be represented, specifically, by the above method. The maximum count of an 8 bit number is 255, not 256. Thus the maximum temperature represented by the above method is (100o C)*(255/256) = 99.609375 degrees. This isn't necessarily a bad thing depending on how the system was designed, and what its purpose is. For many applications, this is just fine.

    But what a good engineer must realize is that each number represented by the 8 bits is actually representing a small range of temperatures, where 100o C falls at the upper end of 255. [Edit: well technically, the upper end of 255 approaches 100o C, getting infinitesimally close, without ever quite reaching it -- but infinitesimally close all the same.]

    But an engineer not realizing the original design intent might suddenly redefine the resolution of the 8 bit number to be (100o C)/255 = 0.392156863 degrees/count (instead of dividing by 256) in the middle of the project causing all sorts of problems down the road.

    ---

    If it's absolutely necessary to detect both transitions of 0o C to 100o C [Edit: such as if the system triggers alarms if the temperature goes out of range, rather than neglecting those temperatures], then the temperature range represented by the 8 bit number needs to be greater than 100o, not equal to it -- and in this example [Method 2] the maximum temperature should have been chosen to be something > 100o C from the beginning of the project.

    Whatever the case, it is important to realize how discrete numbers relate to their corresponding continuous counterparts such that consistency is maintained, for whatever system you are working with.
     
    Last edited: Jul 31, 2015
  12. Jul 31, 2015 #11

    collinsmark

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    Again, my best advice is to let the other person figure it out. And you can guide the other person one biscuit at a time.

    Start with asking the other person:
    "Which day do you finish the last biscuit if you ate 0 extra biscuits?"
    Confirm that the other person got the correct answer. Then continue,
    "Which day do you finish the last biscuit if you ate 1 extra biscuit?"
    "Which day do you finish the last biscuit if you ate 2 extra biscuits?"

    Continue until you reach the first Wednesday, then continue further until you reach the first Tuesday for good measure. That way the other person can determine for himself/herself the actual full range of biscuits that fit into that Wednesday.
     
    Last edited: Jul 31, 2015
  13. Jul 31, 2015 #12

    Borek

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    Perhaps put pennies (matches?) in boxes (or just put them on the checkerboard or something like that). Once it become less abstract he may have to accept he is wrong, which is often a good first step in the right direction.
     
  14. Jul 31, 2015 #13

    collinsmark

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    If you'd like to form a continuous-time, mathematical equation to get you answer, you might start with simple algebra, finding the equation for a line that intercepts two points. (This is a y = mx + b type of equation).
    • Find the slope of the line. You could graph it out and calculate the rise/run, if you wish. How many extra biscuits eaten per days lost? Or if you'd rather switch your axes, how many days lost per extra biscuit?
    • Find the offset (y-intercept). What day of the week corresponds to 0 extra biscuits?
    The tricky part is that since the final answer will ultimately be in discrete time [Edit: or discrete biscuits], a floor function (or ceiling function, depending on which way you go) may/will be involved. It's important that you apply that floor function (or ceiling function) correctly.

    It's probably easier to explain it heuristically with coins though as @Borek suggests.
     
    Last edited: Jul 31, 2015
  15. Aug 1, 2015 #14

    Gaz

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    you said you finished your last two on Thursday then had none for Friday. If you had none for Wednesday then you would have finished your last on Tuesday leaving Wed,Thurs and Friday. Last I checked 3 times 2 is 6 so your wrong your friend is right.
     
    Last edited by a moderator: Aug 1, 2015
  16. Aug 1, 2015 #15

    Borek

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    In the basic case the last two are eaten on Thursday, leaving none for Friday. Why do you count the Friday biscuits, if it was said there would be no biscuits for Friday even if he didn't eat the extras?
     
  17. Aug 1, 2015 #16

    Gaz

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    Haha I didn't read that even though it was in caps. Oh well I am wrong and your mate is wrong' =)
     
  18. Aug 1, 2015 #17

    julian

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    Hello collinsmark. In your spoiler answer (given below)

    So from there, every two "extra" (above and beyond the standard two that you would normally eat) biscuits eaten reduces the final day by one. 0 extra biscuits corresponds to finishing the last one on Thursday (meaning you have none for Friday). 2 Extra biscuits corresponds to finishing the last one on Wednesday (meaning you have none for Thursday), and so on. [Edit: which gives that if your first day with no biscuits was a Wednesday of the second week, you must have ate either 4 or 5 biscuits (less than 6 but greater than or equal to 4).]

    you say 4 or 5....so when you say 5 you mean by eating 5 extra I would still have one biscuit left to eat on the Tuesday. But note in my question I say "I always eat 2 or 3 biscuits each day", this requirement rules out 5 extra as an answer. So 4 is the only answer then.
     
    Last edited: Aug 1, 2015
  19. Aug 1, 2015 #18

    julian

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    That's what you meant by 5 extra wasn't it?

    Thanks for people's suggestions. I may have to actually demonstrate to him by drawing 14 boxes on a piece of paper, labeling them by days of the week, placing two coins on each square, and then ask/show him what happens when you start taking extra, taking it step by step. If he cant understand then...oh dear!
     
  20. Aug 2, 2015 #19

    julian

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    It isn't just this one guy who doesn't understand, it is a group of people - they are deluded, because collinsmark didn't get back me they wont listen to how 6 is wrong even though collinsmark stated it has to be less than 6....oh dear.
     
  21. Aug 2, 2015 #20

    Gaz

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    Ok I get it now it's basically just a trick question. for instance :

    If you don't eat any extra and eat the last two on Thursday. Then that's your 14 days of biscuits ate and the reference to having none left for Friday is irrelative as the Friday has no connection to your two weeks quota of biscuits.
    Thus it is only put in to create confusion.

    If you remove the : and as such would have no biscuits left to eat on the Friday lunch time. From you original question and give that to your friend i think he will understand and come to the right answer.
     
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