Wikipedia:Reference desk/Archives/Mathematics/2006 December 8

= December 8 =

Confusing little puzzle
A friend of mine is trying to solve a puzzle, and we're both stumped. Here it is, word for word - anybody have a clue? We know for sure that there's an answer to be found.

"If 'A' is two, and 'B' is three, and 'F' is eight...how high must you count to get to Heaven?"

--134.173.92.139 04:08, 8 December 2006 (UTC)


 * Infinity. Counting to any finite number is conceivably possible, but it will not get you to Heaven, so it is necessary to count to infinity. Counting to infinity is impossible, barring cheating, so it is sufficient to achieve any goal, up to and including transfiguration.
 * The A, B, F stuff is just there to distract you. Melchoir 05:18, 8 December 2006 (UTC)

Are you sure F isn't 7? yandman 08:01, 8 December 2006 (UTC)
 * yes it should rhyme. 87.102.37.17 10:50, 8 December 2006 (UTC)


 * Mabye it's just a poem? :-) —Bromskloss 12:45, 8 December 2006 (UTC)
 * "If 'A' is two, and 'B' is three, and 'F' is eight...how high must you count to get to Heaven's gate?"87.102.37.17 13:53, 8 December 2006 (UTC)

Assuming F should be 7, and not 8, we get the following:

a=2 b=3  c=4  d=5  e=6  f=7  g=8  h=9  i=10 j=11 k=12 l=13 m=14 n=15 o=16 p=17 q=18 r=19 s=20 t=21 u=22 v=23 w=24 x=25 y=26 z=27

Thus, H-E-A-V-E-N is 9-6-2-23-6-15. Now, perhaps they want to run the numbers all together to get 96223615 ? StuRat 13:01, 8 December 2006 (UTC)


 * I was thinking of 9+6+2+23+6+15=61. I was hoping it would give 42, but never mind... yandman  13:38, 8 December 2006 (UTC)


 * If I add one to the count at every vowel this gives H=10,E=7,A=2,V=27,E=7,N=17. Curious it's 1's 2's and 7's but doesn't seem to make any extra sense. The sum is 70 ("three score years and ten"? isn't that a quote from somewhere?)87.102.37.17 14:07, 8 December 2006 (UTC)


 * Think base 27. —The preceding unsigned comment was added by 203.27.159.30 (talk) 00:54, 9 December 2006 (UTC).

unsolved problem
I cannot solve this small problem

x^x+ y^y =31 -eqn (1)

x^y+ y^x =17 -eqn (2)

The answer I have found by trial and error method that x=2,y=3 or x=3,y=2. but what is the procedure to solve the problem. —The preceding unsigned comment was added by 203.145.188.131 (talk) 07:50, 8 December 2006 (UTC).


 * These equations look like they cannot be solved in the general case (that is, there is no standard function which can express the solution). It can be solved numerically in various ways, and if you know in advance that you are looking for integer solutions, you can try x=0,1,2,3 and find which one allows an integer solution for y (sort of an educated version of trial and error). -- Meni Rosenfeld (talk) 10:55, 8 December 2006 (UTC)


 * The equations are transcendental so you have to solve them numerically. However, it is possible to solve by inspection. From (1), you can see that the highest value that x or y can take is 3, as $$3^3=27<31$$ and $$4^4=64>31$$. Obviously one of x or y cannot be zero. If one was 1, then from (1) it would require that the other was greater than 3, which is not allowed. This leaves 2 and 3 as allowable values, which cannot coincide as (1) would not be satisfied. So $$(x,y)=(2,3)$$ or $$(3,2)$$ are the only solutions. Readro 11:47, 8 December 2006 (UTC)


 * Also, a graph of the two equations is quite interesting, because (according to GrafEq, at least) the first equation has a split that's quite hard to see, but occurs somewhere around (3.1, 0) (and (0, 3.1) but I'm going to focus on the 1st and 2nd quadrants and just assume that the reflection about y=x is given). The result of this split is that besides the (3,2) solution, there are two more apparent solutions quite close to each other, at about (-2.516, 3.067) and (-2.518, 3.064). Of course, proving the actual existence of these is tricky because of the nasty behaviour of things like x^x and x^y when x is negative. Confusing Manifestation 13:18, 9 December 2006 (UTC)

Check out Lambert W function for related problems. Robinh 15:00, 15 December 2006 (UTC)

Equation for determing Occupancy for an office building
I need an equation to help me determine the Occupancy of an office building or meeting room, conference room, etc —The preceding unsigned comment was added by 24.208.156.155 (talk) 19:42, 8 December 2006 (UTC).


 * You have to compute the volume of the room by using the formula height*width*length, and then divide it by the volume of a human.(Igny)


 * I think it has more to do with area and the number of tables, doors, etc. --Daniel Olsen 05:47, 9 December 2006 (UTC)


 * If you mean the maxmimum legal occupancy for fire codes, for example, that's a legal question, not a math question. See fire safety. Friday (talk) 21:24, 8 December 2006 (UTC)


 * I'd say it's both. There will be some legal rule for the amount of space to be allocated for each person, based on your location/jurisdiction, I suspect in the form of square footage per person.  Then, to determine the occupancy, you will need to do some basic math. StuRat 07:15, 9 December 2006 (UTC)

I'd interpret occupancy differently, but as there are various meanings the questioner should be clearer. Anyway:

Take just one room. If it is fully occupied all the time, it has an occupancy of 100%; if half-occupied all the time, of 50%; if fully occupied for half the time, of 50%; etc. So a formula would be:

Occupancy (%) = 100 X Σ(No. of people X Length of meeting)/(Size X Total time)

So for example, a meeting room holds 20 people and is available for 8 hours in a day. There are 2 meetings, one with 8 people lasting 3 hours and one with 12 people lasting 2 hours. Then the occupancy will be given by 100 X (8 X 3 +12 X 2)/(20 X 8) = 30%.

This can be extended to more than one room and to more than one day in what should be an obvious way.Semiable 13:27, 9 December 2006 (UTC)


 * Yea, that might be how they meant the question. StuRat 12:28, 10 December 2006 (UTC)