Coloured Tiles

[jims]

Okay, I concur, the correct answer is day 332239 if you make the following assumptions:
- You must use all the tiles in the patterns when it is possible
- The tiles must be in rectangles with yellow NxM and red (N+1)x(M+1)
- Only two levels are allowed
- The priest can arrange tiles very fast

[hellmanap]

If a = a side of the YELLOW rectangle, and N = the number of days then:

N = (999,999-a)*a/(8+3a) The question is then: For what values of a is N an integer? I get the same 13 solutions as in my post above. Since there less than a couple of thousand values of a to test, you COULD given time, test all of them using only pencil & paper. In fact, using stone tablets you colud hammer out a solution within a lifetime.

[This message has been edited by hellmanap (edited 06-03-2004 10:09 AM).]

[ohtochooseaname]

The world will never end. Red is not the only set of tiles that can represent a larger rectangle....yellow, when it gets big enough, can as well. Ever single day does not have a set of tiles that can represent the proper arrangement, but a future one does, and always will. The only way for there to be an end in which this pattern cannot be completed any more is when Y=2(R+1), which is not possible because as they go to infinity, Y=8/5, which is not the greater than 2 that is required for the arrangement to be at its maximum. Essentially, Yellow will never get large enough to outstrip red far enough that the arrangement will never be made again, and the world will last forever. However, if yellow must be the inner rectangle, then the world ends on day 3, because that is the very last day that red will ever be large enough for the arrangement to last: in all of the subsequent days, red is less than 2(Y+1), and thus can never be placed in that arrangement again.

[smich011]

There are two rectangles, the yellow one being one tile shorter in both dimensions

This shows that no stacking is allowed.
It also shows that all tiles don't have to be used since only .0036% of the red tiles are used (36/1,000,015).

I checked my solution again and thought that god's statement, depending on emphasis and what the correct answer is, could lead to a trick question solution. In his last statement he says: You will be able to make similar patterns to this one as you receive more tiles. The world will end on the last day you have opportunity to do this.

The world ends on the day the priest dies.

[smich011]

There are two rectangles, the yellow one being one tile shorter in both dimensions.

This also shows yellow has to be smaller.

[S A P]

solve (1E6+5x)- 2sqrt(1E6+5x)-8x = 0 for X = days = 332246 (assume red square as solution, subtract length of sides, when yellow exceeds, then *whimper*)

[hellmanap]

Ohtochooseaname’s is interesting. If we let the yellow rectangle be the larger rectangle then there is an easy solution. We know that a smaller red rectangle cannot be something * 1 because on day 1, the red rectangle would be 1,000,005 * 1 and the yellow rectangle would be 4 * 2. Each day, the red rectangle would get 5 tiles longer, and the yellow would get only 4 longer so it would never catch up. But, if the red rectangle were 2 * (1,000,000+5N)/2, then the yellow would be 3 * [(1,000,000+5N)/2 +1] = 3 * (1,000,002+5N)/2 = (3,000,006+15N)/2

Since this number represents the number of yellow tiles, it equals 8N so

(3,000,006+15N)/2 = 8N
3,000,006+15N = 16N
N= 3,000,006 or a little over 8,219 years.

I vote for letting yellow being the larger rectangle

[Deathstream]

this isn't a democracy!

[Speeder]

Just got back from holiday to find a new puzzle that has already been solved!

I agree with kevinatilusa that the answer is (332239 days)

I must admit I wrote some code to calculate this rather than use a more elegant solution.

I would just like to add that the priest might have felt pretty worried round about day (321715 (red=75*34781)) (if he hadn't already died of old age several centuries before.)

I wonder how many times it was possible to create the pattern from day one to the last day?

Speeder.



[This message has been edited by Speeder (edited 06-09-2004 11:46 AM).]

[mathgrant]

I have a list of all §....thirteen....§ days.

[Speeder]

Excellent mathgrant, did you generate them with code or a snazzy piece of maths wizardry?

[mathgrant]

Um. . . both?

[JToomey]

I'm just chiming in to say that, as usual, I am astounded by the simplicity of kevinatilusa's solution. That is all.



------------------
The Big, Stupid Puzzle:
http://www.yark.org/puzzle

[Speeder]

What is the concensus on whether a pattern containing (x+1)*(y+1) yellow and x*y red is still a 'similar' enough pattern? (i.e when the yellows exceed the reds.)

I would imagine this pattern only occurs once. But when?

Edit: I've just calculated #336928 days# for this scenario.
Then I go and get #339183 days# so that's my 'occurs once' theory out the window.

Speeder.



[This message has been edited by Speeder (edited 06-10-2004 12:52 PM).]

[jims]

The following table shows the dimensions, the day number, and the number of tiles (red and yellow):

code:

---

Yellow Rectangle


   Dimensions


   N        M       day    N_yellow  N_red  


--------x-------- -------- -------- --------


    1454 1828     332239    2657912  2661195


     580 4574     331615    2652920  2658075


     455 5824     331240    2649920  2656200


     304 8693     330334    2642672  2651670


     143 18304    327184    2617472  2635920


     124 21050    326275    2610200  2631375


      74 34780    321715    2573720  2608575


      48 52629    315774    2526192  2578870


      29 84208    305254    2442032  2526270


      28 86954    304339    2434712  2521695


      10 210524   263155    2105240  2315775


       5 347824   217390    1739120  2086950


       4 399998   199999    1599992  1999995

---

I have a quick little script that does it all in a second or two.

Jimbo

[Speeder]

Hey Jims, I'd like to see your code.

What do you get for the 'yellow exceeding the red' version I mentioned in my previous post?

Speeder.

[dsghgf]

ummm... just to ask didnt god say in the riddle that the presit could not make that pattern until three days went by. therefore tomorrow he could not make the design therefore the world would end. may that be right? i dont know but its a thought i dont think you need to use that much math and stuff to figure that out. he cant make it on the first day.

[dsghgf]

Oh btw did anyone notice there are green tiles in the picture???

[jims]

Speeder, this is a Linux script that does it about as quickly as I know how. The line numbers are just there for reference. This ought to work on any Unix/Linux box.

code:

---

     1	#!/bin/tcsh 


     2	@ N_max = 1630


     3	@ N = $N_max


     4	printf "Yellow Rectangle\n"


     5	printf "   Dimensions\n"


     6	printf "   N        M       day    N_yellow  N_red  \n"


     7	printf "--------x-------- -------- -------- --------\n"


     8	while ( $N > 0 )


     9	   @ M = ( -8 * ( $N - 999999 ) ) / ( 3 * $N + 8 )


    10	   if ( ( 5 * $N * $M ) == ( 8 * ( ( $N + 1) * ( $M + 1 ) - 1000000 ) ) ) then


    11	      @ d = ( $N * $M ) / 8


    12	      @ Ny = $N * $M


    13	      @ Nr = ( $N + 1 ) * ( $M + 1 )


    14	      printf "%8d %-8d %-8d %8d %8d\n" $N $M $d $Ny $Nr


    15	   endif


    16	   @ N--


    17	end

---

Comments about the code:
I did a little math beforehand to find the limit (1630 on line 2) for N and the equations needed to satisfy the requirements of the puzzle. Also note that the divisions on lines 9 and 11 are integer divisions, no floating point numbers are used at all (TCSH scripts don't allow any floating point calculations).

[This message has been edited by jims (edited 06-11-2004 11:35 AM).]

[jims]

If the yellow rectangle can be the larger (which does not appear to be allowed), there are 16 days when that occurs which are given in the table below. Note that the last case where N=1 gives negative values for everything else. It is simply an impossible case because of the ratio or yellow to red tiles. The other 16 are valid and the answer would be the largest value under "day" = 3000006 days.

code:

---

    Red Rectangle


     Dimensions


    N          M       day     Nyellow   Nred  


----------x---------- -------- -------- --------


      1167 2290       334486    2675888  2672430


       917 2915       334611    2676888  2673055


       730 3663       334798    2678384  2673990


       293 9155       336483    2691864  2682415


       255 10528      336928    2695424  2684640


       155 17393      339183    2713464  2695915


        65 42107      347391    2779128  2736955


        63 43480      347848    2782784  2739240


        40 69567      356536    2852288  2782680


        27 105265     368431    2947448  2842155


        17 173915     391311    3130488  2956555


        15 200002     400006    3200048  3000030


         8 421055     473688    3789504  3368440


         5 800003     600003    4800024  4000015


         3 2000005    1000003   8000024  6000015


         2 8000015    3000006  24000048 16000030


         1 -4000005   -1000001 -8000008 -4000005

---

This is the script that generated the table (again line numbers are for reference only, they should be taken out if you try running the code).

code:

---

     1	#!/bin/tcsh 


     2	@ N_max = 1634 


     3	@ N = $N_max


     4	printf "    Red Rectangle\n"


     5	printf "     Dimensions\n"


     6	printf "    N          M       day     Nyellow   Nred  \n"


     7	printf "----------x---------- -------- -------- --------\n"


     8	while ( $N > 0 )


     9	   @ M = ( 5 * $N + 8000005 ) / ( 3 * $N - 5 )


    10	   if ( ( 8 * ( $N * $M - 1000000) ) == ( 5 * ( $N + 1) * ( $M + 1 )  ) ) then


    11	      @ d = ( ( $N + 1 ) * ( $M + 1 ) ) / 8


    12	      @ Nr = $N * $M


    13	      @ Ny = ( $N + 1 ) * ( $M + 1 )


    14	      printf "%10d %-10d %-8d %8d %8d\n" $N $M $d $Ny $Nr


    15	   endif


    16	   @ N--


    17	end

---

[Speeder]

Interesting code Jim. Much better than my convoluted method.

Speeder.

[O.T.Kaverappa]

Here is the solution to the 'Colored Tiles' Puzzle. The world will come to an end after 250007 days.

On day no. 250007,
number of yellow tiles = 2000056
number of red tiles = 2250035
The best way to arrange 2000056 yellow tiles in the form of a rectangle involves the dimensions 8 and 250007.
Thus, number of red tiles required to make the desired pattern is (8+1)*(250007+1)= 2250072, which is less than the number of red tiles the priest has.

How I found the solution:

After 'm' days,
number of yellow tiles = 8*m
number of red tiles = 5*m + 1000000

The best way to arrange the yellow tiles so that the number of red tiles is minimum, involves arranging the yellow tiles in the form of a square of side sqrt(8*m). (sqrt(8*m) is in general a real number. But let us allow real numbers at this stage.)
The number of red tiles required is, therefore (sqrt(8*m)+1)^2
When we do not have enough red tiles, the world comes to an end. i.e., we must find 'm' such that
5*m + 1000000 < (sqrt(8*m)+1)^2

Solving this we get m=332247

But we have allowed sqrt(8*m) to be a real number and hence from the above discussion, we can conclude that the actual solution is <= 332247.

Then I wrote a C program to check every number between 1 to 332246 to be a solution. This exercise provided the final correct answer 250007. Here is the code I wrote, which I hope is self-explanatory.

code:

---

for(i=1;i<=332247;i++)


{


 yellow=8*i;


 red=5*i+1000000;


 j=floor(sqrt(yellow));


 for(j;j>=2;j--)


 {


  if(yellow%j==0)


  break;


 }


 required=(j+1)*((yellow/j)+1);


 if(required>red)


 {


  printf("%lu",i);	// i is the solution.


  break;


 }


}

---

[merc: code tags]

[This message has been edited by Mercuria (edited 06-19-2004 04:47 PM).]

[lostdummy]

well, even after reading solution to "Coloured Tiles" puzzle, I still think that we may prolong our days a bit and make "end of world" few days later.

Key point is in puzzle text where you need to "make similar patterns" to one shown in picture. Well, similar pattern is also one where yellow tiles surround red tiles, ie where yellow is on the outside.

That gives ground for more solutions, for example:

End of world after 334798 days, when we will have 3664x731 yellow tiles and 3663x730 red tiles

As you can see, that would give us over 7 years more to live

[ralphmerridew]

Problem: Show, without using a calculator, that 2*2*2*3*5*19*23*1373 != 24000040

[pokerfaced]

lol

The right hand side should be divisible by 3.

2+4+4+5(0) = 10

10 is not divisible by 3, so neither is 24,000,040.

Yay, I got one

Note: Remove the factors of three, and the solution is correct.
(3x+5) = 2*2*1373
(3y+5) = 2*5*19*23

[BlackiePas]