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June 17[edit]

A simple riddle.[edit]

Heres how it goes: three men go to a hotel room and they split the cost of their room- the cost is 30 dollars so they each pay ten. Later the manager realizes that he charged them too much- the room cost only 25 dollars, so he gives a bellboy 5 dollars to give to the men. The bellboy steals 2 dollars out of the 5 so he gives the men 3 dollars- they each paid 9 dollars in effect. But 9 x 3 =27+2=29. Why doesn't it add to thirty. It makes sense if you look it from 30-3-2 is 25 but why doesn't it work it you flat add the amount of money each man paid (27) and the 2 dollars the bellboy stole? 24.171.145.63 (talk) 06:56, 17 June 2009 (UTC)[reply]

See Missing dollar riddle. AndrewWTaylor (talk) 07:52, 17 June 2009 (UTC)[reply]

The answer is that you're adding it up incorrectly. The $2 that the bellboy stole is part of the $9 apiece ($27 total) that the men paid. It's the $3 that the men got back that you want to add to get $30, or you can subtract the $2 that the bellboy stole to get the $25 that the hotel charged.

This is, of course, all included in the article that AndrewWTaylor linked to. --COVIZAPIBETEFOKY (talk) 14:17, 17 June 2009 (UTC)[reply]

--Another way of looking at it which may make more sense is as follows:

The manager originally charges the men $30, (10 x 3 = 30)

The manager returns 5 leaving him $25, therefore the men pay (25/3 = 8.33 per individual) When the bellboy returns 1 to each man, they would be paying (8.33 + 1 =9.33 per) or (9.33 x 3) or $28

Upon adding the $2 from the bellboy, it is clear to see how multiplication can and will work in this situation. --f5centvillian143--

name of math concept[edit]

What is the name of the math concept that says that all even integers are the sum of 2 primes? How could you possibly prove that since we are always coming up with larger primes, and one could select an arbitrarily large even number? 65.121.141.34 (talk) 14:50, 17 June 2009 (UTC)[reply]

Goldbach's conjecture. Proofs in mathematics rarely work by giving examples for every possibility, you come up with a logical argument that proves it for every possibility, even ones we haven't thought of yet. This could, for example, be done using mathematical induction, where we prove that if the conjecture holds for one possibility, then it must hold for the next so (if it holds for the first one) it must hold for all of them. --Tango (talk) 14:58, 17 June 2009 (UTC)[reply]

In case it wasn't clear, Goldbach's conjecture has not been proved. So, in fact, no one knows for sure whether it's true or not (although there is great statistical evidence in favor of its truth). That's not to say that it can't be proved, but that no one has ever constructed a proof. It is an open problem. --COVIZAPIBETEFOKY (talk) 15:17, 17 June 2009 (UTC)[reply]

No one knows "for sure" even those things that have been proved. --Trovatore (talk) 23:51, 23 June 2009 (UTC)[reply]

volume of a sack[edit]

How to find the maximum capacity of a sack made by gluing together at edges two pieces of plastics or paper or something else. —Preceding unsigned comment added by 113.199.161.144 (talk) 18:40, 17 June 2009 (UTC)[reply]

Can we assume that the material of the sack does not stretch? 65.121.141.34 (talk) 18:53, 17 June 2009 (UTC)[reply]

That would be a good starting point. If the material stretches, that seems like something you would add later, as a complication, after working out the non-stretch case. That's kind of like how we describe projectile motion in a frictionless vacuum, and then learn to correct for air resistance later. On the other hand, the material had better be flexible, because gluing together two flat pieces of sheet metal along edges does not a sack make. -GTBacchus^(talk) 20:01, 17 June 2009 (UTC)[reply]

So, we're determining how much stuffing a pillow can hold? Black Carrot (talk) 05:28, 18 June 2009 (UTC)[reply]

Yes. This sounds messy to do rigorously. See isoperimetric inequality for a less constrained case. 208.70.31.186 (talk) 05:59, 18 June 2009 (UTC)[reply]

If there is not restriction on size of the pieces of plastic or paper, then there will be no maximum capacity of the sack. The volume of the sack is proportional to the cube of the linear size of the pieces, so the remaining problem depends on shape only. Choosing shapes that makes the sack approximately spherical makes the maximum volume for given surface area. Bo Jacoby (talk) 08:29, 18 June 2009 (UTC).[reply]

If the sack absolutely cannot stretch but can bend and it is glued all round the edge then it is a limiting form of a Flexible polyhedron. And by the Bellows conjecture (now proved) the volume will be zero. :) Dmcq (talk) 09:49, 18 June 2009 (UTC)[reply]

Actually that's only true if you glue the edges before you make the volume. You really need to fold the two sides first so the edges will join up and then glue them together. You might be interested in the problem of folding a shopping bag mentoned in rigid origami, there the top is open. Dmcq (talk) 09:58, 18 June 2009 (UTC)[reply]

Following on from Bo Jacoby, if the two pieces are circular and of diameter D then the maximum possible volume would be that of the sphere of diameter d where d =

{\frac {2D}{\pi }}

i.e.

{\frac {4D^{3}}{3\pi ^{2}}}

, but it is not possible to stretch two circular pieces of most materials to meet on the circumference of the sphere, or to join them with zero overlap, so the maximum volume is "a bit less than this". How much less depends on the materials and how they are joined, but this idealised optimum is about 13.5% of

D^{3}

, so the best to expect in practice would be 10% of

D^{3}

, and most situations would give considerably less. Dbfirs 12:10, 18 June 2009 (UTC)[reply]

If you're going to use shaped pieces then how about just one long piece of ticker tape? Just go round and round putting in small pleats and you'll soon have a very good approximation to a sphere. You can do this with a balloon and wallpaper paste :) Dmcq (talk) 15:11, 18 June 2009 (UTC)[reply]

... or take that to the limit as the width of the ticker tape tends to zero, and start with a pre-formed sphere.

... but if you insist on starting with right-angles, start with two squares of side D, fold D/4 along each edge, cuting out the corners or using them for strengthening overlap. This produces a cube of side D/2, with volume

D^{3}/8

, giving a lower limit of 12.5% of

D^{3}

. This can be imporoved (perhaps to 15% of

D^{3}

or more) by packing tightly and allowing the faces to bulge. The reason that this appears to be a better solution than the two circles is that you are starting with more material (though if you have to cut circles out of squares in the first place, then don't bother). I doubt whether many materials are manufactured in circular form, except possibly round tea bags? Dbfirs 16:42, 18 June 2009 (UTC)[reply]

One case is the paper bag problem. (I remembered reading some work on it but would never have found it if Dbfirs had not mentioned tea bags.) —Tamfang (talk) 21:36, 19 June 2009 (UTC)[reply]

Ah, thank you! So I under-estimated how much tea can be packed into a (very strong) square teabag. You might achieve 20% of

D^{3}

with really tight and careful packing. Dbfirs 00:35, 21 June 2009 (UTC)[reply]

Enigma puzzle from New Scientist magazine[edit]

I solved this puzzle using brute force; my question is whether there is a more elegant solution that you could get to without a computer program. Essentially the problem is to find three three-digit numbers such that:

between them, the three numbers use each of the digits 1-9 exactly once
the product of the three numbers is a nine-digit number that uses each of the digits 1-9 exactly once
the smallest of the three numbers is 7 times a prime
the middle of the three numbers is 8 times a prime
the largest of the three numbers is 9 times a prime

The answer is:

413 = 7 x 59
568 = 8 x 71
927 = 9 x 103
413 x 568 x 927 = 217,459,368

But how can you arrive at this solution, short of trying different permutations of three primes until you get it? At first I thought you might be able to narrow it down using divisibility rules, particularly for divisibility by 7, but I didn’t get anywhere with that. --Mathew5000 (talk) 23:11, 17 June 2009 (UTC)[reply]

This is a problem in constraint satisfaction. There are well known algorithms like unification for solving these problems, and logic programming languages like Prolog that have the algorithms built in. For an example of how to do it by hand, see the article SEND MORE MONEY. 208.70.31.186 (talk) 00:43, 18 June 2009 (UTC)[reply]

I wonder what's the less elegant solution. Checking all 9! permutations by prime factorization? --pma (talk) 10:44, 18 June 2009 (UTC)[reply]

It's not quite 9!, the order of the numbers doesn't matter, so you can divide by 6. --Tango (talk) 13:15, 18 June 2009 (UTC)[reply]

I meant all numbers with 9 different digits, form 123456789 to 987654321... ;) --131.114.73.61 (talk) 16:21, 18 June 2009 (UTC)[reply]

Surely the less elegant solution is to cycle through all 10 digits for each place giving 10⁹ possibilities? However one could do it on paper seeing there aren't all that many primes to check, 100<7p₁<800, 200<8p₂<900, 300<9p₃<1000, the numbers can't have repeats in them, each number is greater than the one before, the sum of the first two must be a multiple of 9. Dmcq (talk) 15:57, 18 June 2009 (UTC)[reply]

Let's see if we can reduce the size of the search space even further. There are 9x8x7=504 three digits numbers with (1) no repeated digits and (2) no 0 digit. But only approximately 1 in 9 of these are divisible by 9 - and testing for divisibility by 9 is simple. So we have about 56 choices for our first factor. For each choice of first factor we have approximately 6x5x4/8 = 15 possible second factors which are divisible by 8 - and, again, quite easy to eliminate non-qualifiers by inspection. That leaves 6 candidates for the final factor - let's say on average 1 of these is divisible by 7, but it takes a bit more work to filter out non-qualifiers here. So the size of our search space is now about 56x15 = 840 sets of candidate factors. Multiply each candidate set together and eliminate it if the product contains a repeated digit or a 0. But that's still a lot of work if you want to solve it "by hand" or just with a calculator. Gandalf61 (talk) 15:57, 18 June 2009 (UTC)[reply]

Well I just tried it and it took half an hour without a calculator just biro and paper. Now I must really go and do something slightly more useful. Dmcq (talk) 16:43, 18 June 2009 (UTC)[reply]

Very good; then maybe to make it more feasible, one thing is that the larger of the three factors X is larger then the cubic root of 123456789≈125*10⁶, so larger than 500-3, and certainly not larger than 987. Its quotient by 9 is a prime between 497/9≈55 and 987/9≈109. Removing from these well known primes the one that multiplied by 9 give a result with some repeated digits, [or 0], one is left with 61 67 71 73 89 97 ~~103~~ 107 109. These give 9 6 choices for X: 549,~~603~~,..,981. Then one lists similarly the possible choices for the medium factor Y and for the least factor Z, that are numbers with distinct digits respectively of the form Y=8p and Z=7p, with p a prime between 29 and 113. I guess some 15 possibilities for Y and around the same for Z. Then one goes on like in Gandalf's method, trying products X*Y*Z, but only with X>Y>Z and all digits different in X,Y,Z. So actually very few multiplications are to be effectively computed. I guess it's something one could do by hands, but since the OP already gave the answer, he spoiled the small satisfaction to try it... ;) It could be an exciting project in a class of kids that are learnig multiplication by hand, if they still learn it. --pma (talk) 17:50, 18 June 2009 (UTC)Oh I forgot that zero is not there, so it's even better[reply]

I did 46 multiplies in total of 7 and 8 with prime numbers and cut out the ones with repeats and zeroes. Then I got the digit sums of the multiples of 7 and 8 so I could match up pairs that gave a sum that was a multiple of 9 and had no repeat digits and with space for a multiple of 9 after them. This gave only four possibilities. I then did 12 multiples of 9 with primes in the possible range. This gave 5 possible triples. I then had to try all five of these possibilities which involved 10 big multiplies. So about 70 arithmetical operations in total, 10 of them big, plus lots of straightforward checking digits and comparisons. Two sheets of paper. Don't know why I took half an hour put like that. Dmcq (talk) 21:46, 18 June 2009 (UTC)[reply]

Thanks very much! --Mathew5000 (talk) 23:09, 19 June 2009 (UTC)[reply]