Coloured Tiles

[Lepton]

When will the world end?

Happy Birthday mathgrant!

[mathgrant]

Thanks for the happy birthday wish!

Also, the ASCII art doesn't show up well in that font. This picture should be more useful.

[Deathstream]

I just did it realy quick so i'm not too sure about my answer. It's [142,857 and 1/7 days or about 391 years]. If this is right then you need to find some harder questions. Happy Bday Mathgrant. Here's your present $1,000,000. Don't spend it all on one thing.



[This message has been edited by Deathstream (edited 05-26-2004 09:15 PM).]

[kevinatilusa]

Actually, I think the world will last a bit longer than that, perhaps 332,239 or so days.

[Deathstream]

Well from the picture [i got a 3:2 ratio of red to yellow. I think you have it the other way around.]

[This message has been edited by Deathstream (edited 05-26-2004 09:10 PM).]

[pokerfaced]

I got the same answer as kevinatilusa, namely [332239].

Deathstream, [ I don't think the puzzle says that the red rectangle needs to be bigger than the yellow for the solution, it just says that 'similar' patterns can be made. Furthermore, I think God gives the priest all the tiles in one go every day. Thus the variables should all be integers, and having a fractional part of a day doesn't make much sense. ]

I'll hold off on posting my method for now.

[This message has been edited by pokerfaced (edited 05-26-2004 09:30 PM).]

[Deathstream]

Yea the 1/7 didn't make sense to me either. [if God gave the priest 999,999 on the first day does your strategy give you 142857]

[This message has been edited by Deathstream (edited 05-26-2004 09:57 PM).]

[pokerfaced]

Deathstream, [ I don't know what result I'd get if there were just 999,999 on the first day. I used some equations and ran it through a calculator program... later I'll try it for 999,999, but don't have the time or inclination to do so right now =). ]

Happy birthday Mathgrant!

[mathgrant]

To clarify, on Day 0 (today), the priest gets 1,000,000 red tiles from God.
On Day 1 (tomorrow), God will give the priest 5 extra red tiles, and 8 yellow tiles.
On Day 2 (day after tomorrow), God will give the priest 5 extra red tiles, and 8 extra yellow tiles.
On Day 3 (three days hence), God will give the priest 5 extra red tiles, and 8 extra yellow tiles.
On Day 4 (four days hence), God will give the priest 5 extra red tiles, and 8 extra yellow tiles.
...
On some day N (N days hence), God will give the priest 5 extra red tiles, and 8 extra yellow tiles. With all of the tiles the priest has accrued over the past J+1 days, he will be able to arrange all of the red tiles into a rectangle, and the yellow tiles into a rectangle with width and height 1 tile shorter than those of the red tile rectangle. The priest will not be able to do this on any future date, although he may have been able to do so on some previous dates. The world will end. Boom.

Find N.

[This message has been edited by mathgrant (edited 05-27-2004 11:38 PM).]

[Beartalon]

Like all puzzles here, I tend to see a simplistic theory that never seems to be part of the solution by the time we're done.

Isn't this as simple as calculating the maximum N₁ from 8N₁ > 5N₁ + 1,000,000? At some point the yellow tiles outstrip the red. At a slightly earlier day, N₀ is when 8N₀ can be factored into any xy=8N₀, while there aren't enough red tiles to create (x+1)(y+1)?

[This message has been edited by Beartalon (edited 05-26-2004 11:32 PM).]

[pokerfaced]

Just went over my calculations, and found that for the answer kevinatilusa and I posted above the red rectangle is indeed larger than the yellow one. This should be the last possible day. Luckily I won't be around when it happens.

If you want to save the world, go steal a tile (would that quicken it's demise?).

[pennywise_rule]

Now, I might be way off here. But according to the question, God said that 'Suppose I give you only 21 tiles today, but still would give you 5 red tiles, and 8 yellow tiles daily, you could rearrange to make this pattern' Well, 60 tiles were needed to make the pattern.

Chances are I am way off, but the way I worked it out, you get 21 tiles on day 1, then 13 tiles each day after that. Pattern is made in 3 days, using 60 tiles.

Basically I just kept adding 13 for every day until I got to a multiple of 60.

My answer - 40 days.

Probably way, way off, but simple minded person here. haha

[Lambert]

Mathgrant please note "i before e except after c", so receive not recieve.

After 1 day there will be 1000000+5 red tiles and there will be 8 yellow tiles. The way the question is worded suggests that it is now possible to make two rectangles, the red one being (m+1) by (n+1) while the yellow one is m by n. How can this be so, even after one day? 5x200001 (= 5x3x163x409) is much wider than 4x2, etc. However if we allow 3D stacking, then 5(163) by 3(409) is one greater than 4(1) by 2(1) and thus it is possible to make a 5x3 red pattern as opposed to a 4x2 yellow pattern. Is this what God wants us to do each day until the end of the world?

[dARKhORN]

[i dont think the world will ever end. the riddle states that you will receive 5 red tiles and 8 yellow tiles on a daily basis to add to your surplus of 1000000 red tiles and that "You will be able to make similar patterns to this one as you recieve more tiles. The world will end on the last day you have opportunity to do this." if im receiving more tiles every day i will have an infinate source of tiles to work with which will allow me to build infinatly. the only real change that i see you having is going to be in about 333,334 days your surplus of tiles will containe more yellow than red.]

[mikegoo]

Lambert:
The puzzle does not state that everyday until the end of the universe you will be able to make such a pattern. It states that after a certain day, which will be the end of the universe, you would never be able to make such a pattern again.

[dnwq]

Well, either it does now, or I'm missing something.

[mikegoo]

Originally posted by from puzzle:

---

God: You will be able to make similar patterns to this one as you recieve more tiles. The world will end on the last day you have opportunity to do this.

---

The line "you will be able to make similar patterns to this one as you receive more tiles" does not state that everyday you will be able to make similar patterns. You may be interpretting the statement that way, but it is not what it says.

[dARKhORN]

ok, the riddle stats that the las t day you can make a simalar structure to the one displayed the world will end. the last structure you will be able to make that will be "two rectangles, the yellow one being one tile shorter in both dimensions" and using all of the avalable tiles as in the example will be [399999x5 (red) and 399998x4 (yellow) and will take you 199999 days or 547 year and 344 days.
399998x4=1599992 yellow tiles
399999x5=1999995 red tiles
1000000+(199999x5)=1999995 red tiles
199999x8=1599992 yellow tiles
]

[Samadhi]

I get [332244

My method: First, there needs to be more red than yellow. A minimum of (sqrt Y + 1)^2 - Y where Y is the number of yellow. The last day this can be done is on day 332246. On this day you have 1,000,000 + 332,246 * 5 Red tiles (2,661,230), and 332246 * 8 Yellow tiles (2,657,968).
Unfortunately, the number of yellow tiles for that day factors into 2^4, 271, and 613. None of these can be combined close enough to a square to allow you enough red tiles to meet the guidelines.
On the day before, the tiles factor into 2^3, 5, and 66449. Not even close to a square.
On the day before that, the tiles factor into 2^5, 3^2, 11, and 839.
This allows you to make a rectangle 1678 * 1584 out of yellow tiles and a rectangle 1679 * 1585 out of red tiles. Five tiles to spare!]

Not tremendously elegant, but Excel is my friend.

[pokerfaced]

Samadhi, [I don't think you're allowed to have tiles left over. Otherwise you could just make a 3*2 red rectangle, and 2*1 yellow rectangle everyday, despite the left-over tiles. This would mean the world would never end. God does say, "you can rearrange all your tiles..."]

[Samadhi]

You have to have left over red tiles. God gives you a million at the beginning. And it implies that you need to make a new pattern.

However, we should expect ambiguity. It is God after all.

[pokerfaced]

Samadhi, you don't have to have leftover red tiles. Those 1,000,000 red tiles are to be incorporated in the pattern. Try finding rectangles for the number of tiles you have on the day kevinatilusa and I suggest as the last day possible - you'll see that you can make the yellow and red rectangles with all the tiles.

Since God said to use all the tiles, I don't think there's too much ambiguity

[Mr. Lettuce]

God gives the priest one million red tiles to start with, but the rate at which the red tiles increase is less than the rate that the yellow tiles increase. It's gonna take 125000 days for there to even be one million yellow tiles. After that many days, there are 1625000 red tiles.
The world will definitely end before the number of yellow tiles equals the number of red tiles. After 333,333 days there are 2,666,665 red tiles and 2,666,664 yellow tiles.
This is not a guess and check problem.

[Martin.]

How about day 332239? That's 1828x1454 for yellow and 1829x1455 for red.

[edit: Invisibled the answer. -Antrax]

[This message has been edited by Antrax (edited 05-28-2004 12:28 PM).]

[pokerfaced]

Martin, [ that's what kevinatilusa and I got, if you look the invissed answers above. ]

In case you're new to the forums, you need to highlight with your cursor to see invisible text (try above).

[hellmanap]

On day 332239 you will have 1000000 + 5*332239 = 2661195 = 1455*1829 red tiles and 8*332239 = 2657912 = 1454*1828 yellow tiles

I used Excel to help me search for the answer, testing on the shorter side of the red rectangle. On day 333,334 there are more yellow tiles and on that day there are 2666670 red tiles and 2666672 yellow. So, the shorter red side must be less than or equal to SQR(2666670)= 1632. This gives us an upper bound and I worked down from there.

There are only 13 days which meet the conditions they are days 199,999 - 217,390 - 263,155 - 304,339 - 305,254 - 315,774 - 321,715 - 326,275 - 327,184 - 330,334 331,240 - 331,615 - 332,239 with the shorter red side of 5 - 6 - 11 - 29 - 30 - 49 - 75 - 125 - 144 - 305 - 456 - 581 - 1,455 rspectively



------------------
hellmanap
_{I edited your post to make the answers invisible. To see how I did it, click on the "edit post" icon at the top of this post. -Lepton}



[This message has been edited by hellmanap (edited 06-01-2004 03:29 PM).]

[smich011]

I come up with any where from 999,995 days to infinity.


Similar: not differing in shape but only in size or position so a 3x2 red and 2x1 yellow triangle would be similar. If I can keep using the tiles over and over I can make the above rectangles infinitely. If I can't keep using the same tiles then the following exists: I get 5 red and 8 new yellow tiles every day, I would use 6 red and 3 yellow per day. This gives me a net loss of 1 red tile per day. When I have less than 6 red tiles left I can no longer make such a rectangle. 1,000,000- 1(X) =5; (X)=999,995 days.

Since God said to use all the tiles, I don't think there's too much ambiguity
God does say, "you can rearrange all your tiles..."]


God does NOT say you must use all the tiles, he does however say the following: Suppose I gave you only 21 red tiles today, but still would give you 5 red and 8 yellow tiles daily. Then, in 3 days, you could rearrange all your tiles to make this. I also could rearrange them in 3x2 and 2x1 rectangles

[winterHLepsilon]

Does that mean you can use the tiles over and over agian? Then is there any end of the world, really?

------------------
*Hypotenuse Leg*
HL--^H^e^L^en
A girl from Guangzhou, China, Asia

[Jedo the Jedi]

I am not much of a puzzle solver because i suck, but I thought I'd try this one out. I used some of your methods and came up with 43 . Some how I just don't think that is right. Oh well, I won't do puzzles anymore.

[shadyforce]

Hmm, it seems it's all about finding the point where given a certain amount of red tiles, you can still make a smaller rectangle of yellow tiles, but with an increase in yellow tiles greater than red tiles, that will never be possible any more.

[The size of the yellow rectangle = L x W = 0 + 8 * N
The size of the red rectangle = (L + 1) x (W + 1) = 1000000 + 5 * N
For what value of N (>0) will the above equations not be solveable for positive values of L and W.

Solving, I get W = 8N/(1000000-3N), which becomes negative when N (integer) reaches 333334.]

=> The world will end in [333334] days???

Ok, I looked again, and it seems that all I've calculated is when does the amount of yelow tiles exceed red tiles, so that's no use. I must give it some more thought.

[Lambert]

I am pleased to see that God has learnt to spell receive but He still talks in riddles. He says "The world will end on the last day you have opportunity to do this" but does He mean "The world will end on the last day you have the opportunity to do this" ? or "The world will end on the last day you have an opportunity to do this" ?

All replies so far assume that God is two dimensional but surely that is not the case. Why not go into 3 dimensions and make similar patterns but of differing heights?

Thus after 1000000 days you will have 6000000 red and 8000000 yellow tiles and you could make a 6x5(x200000 high) red rectangle and a 5x4(x400000 high) yellow rectangle. This ingenuity will please God and we can continue to pray to Him, repeating after me "World without end, amen".

[Termital]

Assumptions:

1. All pieces are used.
2. 2 layers.
3. Yellow over Red.

R(ed pieces)=1000000+5*d(ays)
Y(ellow pieces)=8*d

R>YÛd£333,333

Cases:

1. The Red rectangle can have a side of 1.
   Then R=Y+1Þd=333,333 days till the end of the world.
   
2. The Red rectangle's sides must have a minimum length of two. Then, using this piece of (pretty readable) Delphi Pascal:
   

   code:

   ---

   
   
   
   var
   
   
   days,Redarea,Yellowarea,a:integer;
   
   
   begin
   
   
    for days:=333333 downto 200001 do begin
   
   
     Redarea:=1000000+5*days;
   
   
     Yellowarea:=8*days;
   
   
     for a:=2 to ceil(sqrt(Redarea)) do
   
   
      if (Redarea mod a=0)
   
   
      and ((a-1)*((Redarea div a) -1)=Yellowarea)
   
   
      then begin
   
   
       Memo1.Clear;
   
   
       Memo1.Lines.Add('Days: '+inttostr(days));
   
   
       Memo1.Lines.Add('Red Area: '+inttostr(Redarea));
   
   
       Memo1.Lines.Add('Yellow Area: '+inttostr(Yellowarea));
   
   
       Memo1.Lines.Add('a: '+inttostr(a));
   
   
       exit;
   
   
      end;
   
   
    end;
   
   
   end;
   
   
   

   ---

   I find the answer of 332,239 days (one side of the red rectangle being 1455)

However, this thought eats at me.

 Let a,b be the sides of the yellow rectangle. Let P=ab, S=a+b. Then a,b are the roots of
x²-Sx+P=0.
P=YÞab=8d.
(a+1)(b+1)=RÞab+a+b+1=1000000+5dÞ8d+S+1=1000000+5dÞS=999999-3d.
 So, a and b are the roots of the equation of x²-(999999-3d)x+8d=0. So d is of the form (999999-x²)/(3x+8). Which is about the point I draw a blank in my quest for a more elegant solution.


------------------
Yearn brightly



[This message has been edited by Termital (edited 05-31-2004 09:01 PM).]

[Lepton]

nip

[kevinatilusa]

A solution requiring only a calculator (or pencil, paper, and accuracy I cannot match):

After t days we have 1000000+5t red tiles and 8t yellow tiles. If our rectangle for the red tiles is x by y, we must have

x*y=1000000+5t
(x-1)(y-1)=8t.

Multipling the first equation by 8, the second by 5, and subtracting them,
8xy-5(x-1)(y-1)=8000000, so 3xy+5x+5y=8000005.
Multiplying by 3, adding 25 to both sides, and factoring,
9xy+15x+15y+25=(3x+5)(3y+5)=24000040=2*2*2*3*5*19*23*1373.

We want to maximize t, which is the same thing as maximizing x*y. Since we know that 3xy+5x+5y is a constant, this is the same as minimizing x+y or minimizing (3x+5)+(3y+5). We know the product of (3x+5) and (3y+5) is a constant, so me make the sum smallest when there are closest together, i.e. as close to sqrt(24000040)=4898.98... as possible. The 1373 has to go somewhere, and the best we can do is pair it up with two of the twos.

3x+5=1373*4 => x=1829
3y+5=4370 => y=1455. t=(1829*1455-100000)/5=332,239

[jims]

With the following assumptions:
- Only two levels of tiles allowed
- You do not have to use all the red tiles in a pattern
- You must use all the yellow tiles
- The pattern must be rectangular

I believe the answer is 250,007 days
I'll post my calculations later on, but if I'm correct the previous answers are all missing a prime element in their analysis.

Jimbo



[This message has been edited by jims (edited 06-01-2004 01:11 PM).]

[Termital]

 Solutions where all red tiles are used are a subset of solutions where all red tiles need not be used. The numbers Kevin and I suggest demonstratably work in our more specific case, so they are also applicable in your wider assumption. And since your result is smaller than ours, something must be wrong with the calculations[/showoff]

------------------
Yearn brightly

[jims]

The number I found was also under the assumption that you must make a pattern each day, which may not be the correct assumption after reading the other posts. I will try, try again...

Jimbo

[IrishJoe]

There's another apparent assumption to watch for.

For purposes of the end-of-the-world calculation, the 1000000 red tiles don't exist.

Originally posted by God:

---

Suppose I gave you only 21 red tiles today, but still would give you 5 red and 8 yellow tiles daily...

---

God goes on to describe the rectangle pattern which can be constructed three days hence, with 21+5+5+5 red tiles and 8+8+8 yellow tiles, as well as how to determine the end of the world calculating how long you can continue the hypothetical rearrangement of tiles.

Oh, and assuming it has to be two rectangles and all tiles in the hypothetical must be used, as the statement seems to imply, [the world ends one day prior to the first day there is a prime number of either red or blue tiles. The world will end in three days.]

[pokerfaced]

It's been over three days since the problem was published, and I'm still here, so that can't be right . I do believe that the 21-red-tile situation was given purely as an example of what was required, and still stand by the solution posted by kevinatilusa, myself, martin, hellmanap and termital get.

Look at reply 8 where mathgrant clarifies the problem - here he states that there are 1,000,000 red tiles.

[IrishJoe]

It's true that he states that in reply #8.

It doesn't fit the original puzzle post, but it is what he posted in reply #8.