How would you go about solving these math problems?

[astro]

I'm mainly interested in the approach and methodology-

Problem 1:

There are 999 cards numbered 1 to 999, and three boxes colored red blue and white. Find the number of ways of putting the cards into the boxes such that there does not exist a,b belonging to different boxes and c,d belonging to different boxes (the former pair of boxes being not the same as the latter pair of boxes) such that a+b=c+d.

Problem 2:

John buys a motorcycle insurance policy from an insurance company which covers partial damage or total loss of his motorcycle for one year period. This policy is subject to a deductible of $1000, and a maximum payment of $10,000. During the policy year the probability of partial damage to John's motorcycle is 0.04 and the probability of total loss of his motorcycle is 0.02. If there is partial damage to his motorcycle, the amount of damage X follows a distribution with the density function f(x) = 1/12500, x belongs to (0,12500), 0 otherwise

Answer choices: (A) 408 (B) 410 (c) 424 (d) 450 (e) 550

Problem 3:

How many perfectly spherical marbles with a diameter of 1 cm each could fit into a perfectly spherical sphere 1 meter in diameter? How many marbles into a cube 1 meter long on all sides?

Problem 4:

If a perfectly spherical vessel, a perfectly cubic vessel, a perfect toric (like a doughnut) vessel, and 3 sided equilateral pyramidal vessel each encompass 100,000 gallons of water and the walls of each vessel are 5 CM in thickness, what is the volume of paint required in liters for each vessel, if each meter of surface area requires 1/2 liter of paint and the entire job requires 3 coats of paint.

[CookingWithGas]

How many marbles into a cube 1 meter long on all sides?I'll take the easiest one. This is easily solvable if you know that Kepler (http://www.chem.ox.ac.uk/icl/heyes/structure_of_solids/Lecture1/Lec1.html) determined that the density of spheres in a closest-packed configuration is about 0.74047954. I may not be using the correct word when I say "density;" what I mean is that that proportion of the space is filled with marbles, and the other 0.25952046 cubic meter (in your case) will be left empty.

Please tell us where you got these questions. They don't sound like you just made them up staring at the clouds one day.

[astro]

I'll take the easiest one. This is easily solvable if you know that Kepler (http://www.chem.ox.ac.uk/icl/heyes/structure_of_solids/Lecture1/Lec1.html) determined that the density of spheres in a closest-packed configuration is about 0.74047954. I may not be using the correct word when I say "density;" what I mean is that that proportion of the space is filled with marbles, and the other 0.25952046 cubic meter (in your case) will be left empty.

Please tell us where you got these questions. They don't sound like you just made them up staring at the clouds one day.

The geometry problems were spurred by thinking about municipal water tanks. The probability questions were based on some actuarial test questions I didn't know how to approach.

[CookingWithGas]

Problem 4:

If a perfectly spherical vessel, a perfectly cubic vessel, a perfect toric (like a doughnut) vessel, and 3 sided equilateral pyramidal vessel each encompass 100,000 gallons of water and the walls of each vessel are 5 CM in thickness, what is the volume of paint required in liters for each vessel, if each meter of surface area requires 1/2 liter of paint and the entire job requires 3 coats of paint.You want methodology so I'm not going to look up the details for you.

The key here is to get formulas for volume and surface area of these shapes, then it's just plug & chug. Cubes are trivial; in high school we learned those for spheres (both by rote and how to derive with calculus). I would have to look up or derive how to do this for a toroid, not sure what you mean by "perfect" here. The pyramid sounds easy at first but that's another one that I would have to think about.

Start with the formulas for volume for each of these shapes. Start with 100,000 and convert gallons to cubic meters. Solve for radius for the sphere, length of a side for the cube, [something] for the toroid and the height of the pyramid. Then make adjustments to blow each one up by 5 cm of thickness (might be a little trickier with the pyramid since the length of a side or the height would not be increased by 5 cm, need to think about this one) and use surface area formulas to determine surface area. Multiply by 3 and divide out the 0.5 liters.

[panamajack]

#2 appears to be incompletely stated, but I'm guessing it's something like figuring out either expected payout or expected price of the insurance.

It's also not clear what the price of a 'total loss' is but apparently more than $10,000. I also assume only one potential loss.

If this is the question, I'd set up a 'final' distribution function of what the insurance company actually pays. Each dollar amount has a probability associated with it P(x). All values below the deductible go to 0 (including the no accident case). All values over the max payout go to the max payout.

To get the expectation, either take the sum of (discretely), or integrate, x P(x) from -inf to inf.


For this particular one, the probability of max payout = P(total loss) + sum (P(partial loss) from $11,000 to $12,500) and assign that to x = $10,000.
Assign to 0 the P(loss <= deductible) + P(no loss = 1-P(either partial/total loss))
Intervening values are assigned probability by the uniform function f(x). (P( partial loss) = P(any partial loss) * f(x) as given).

[panamajack]

By the way, I calculated it and did get one of the answers given.
I have a feeling, however, the test may have been looking for something else (perhaps a certain fixed percentage for the insurance company).

[jawdirk]

Then make adjustments to blow each one up by 5 cm of thickness (might be a little trickier with the pyramid since the length of a side or the height would not be increased by 5 cm, need to think about this one) and use surface area formulas to determine surface area. Multiply by 3 and divide out the 0.5 liters.

The "blowing each one up by 5 cm" is not necessarily easy or well-defined. Actually, the question is ambiguous. Is the outside of the containers a perfect shape, or is the inside (for cube and equilateral pyramid)? If both, then the walls aren't uniformly 5 cm in width. For the torus, what is "perfect?" You've got four values to consider: two radii for outside and inside. The only easy one is the sphere!

[Kevbo]

I'll take the easiest one. This is easily solvable if you know that Kepler (http://www.chem.ox.ac.uk/icl/heyes/structure_of_solids/Lecture1/Lec1.html) determined that the density of spheres in a closest-packed configuration is about 0.74047954.

I don't know how to solve this one, but I know enough about such packing problems to know that the e method suggested won't give the correct answer.

The Kepler approximation will allow you to figure out about how many whole and fractional marbles exist in a spherical volume of a closely packed array. That is not the question though.

The problem as stated is much more difficult. As stated ("how many") the problem requires an integer answer, and stipulates marbles and a container. The walls of the container create a problem.

Using In a real container however, there will be significant voids where there is not quite enough room for one more marble to fit in a given layer. There might also be conditions where slight lifting of a few marbles on one layer might allow one more marble to be included in that layer, such shifting may well be acomidated by the layer above, which will also have it's voids.

One approach to solving the problem is to examine the surface of the sphere and eliminate all fractional marbles of an assumed closly packed array. This would require optimization of the supposed center of the sphere in relation to the array.
(you'd need to try all points within one lattice of the "crystal")

Gut instinct tells me this answer would be slightly low, as mentioned above, I have a strong hunch that you could jam a few more marbles in by disturbing the packing of other areas of the array that are surrounded by voids.

[jawdirk]

Here's a stab at question #1:

You can have all the cards in each of the 3 boxes since there will be no a+b "belonging to different boxes." (3)

You can have any combination of 1-999 in 2 of 3 boxes, and the 3rd box empty since there will be no combinations a,b,c,d that have "former pair of boxes being not the same as the latter pair of boxes." That is something like 2^998*3?

Now, for cards in all three boxes, I believe there are only 6 possibilities. (1,999,rest) distributed in the various combinations between the boxes.

My reasoning goes like this:
1 is involved in sums 3-1000
2 is involved in sums 3,5-1001
3 is involved in sums 4-1002
...
997 is involved in sums 998-1996
998 is involved in sums 999-1995,1997
999 is involved in sums 1000-1997

The only place where there isn't multiple overlap between two numbers is 1 and 999. They only overlap at 1000. For any other choice of 2 numbers, they are involved in at least 2 common sums.

For example, if you try (2,999,rest) then you will have a=2,b=998,c=999,d=1
Similarly, if you try (1,998,rest) then, again, you will have a=1,b=999,c=998,d=2.

All the other combinations of 2 numbers can be used to form at least one a+b=c+d.

Any flaws?

So my total answer would be 2^998*3+9.

[jawdirk]

Er, I think it is 2^998*6, sorry.

[CookingWithGas]

The "blowing each one up by 5 cm" is not necessarily easy or well-defined. Actually, the question is ambiguous. Is the outside of the containers a perfect shape, or is the inside (for cube and equilateral pyramid)? If both, then the walls aren't uniformly 5 cm in width. For the torus, what is "perfect?" You've got four values to consider: two radii for outside and inside. The only easy one is the sphere!
Wasn't sure what you meant by "perfect shape" here.

Seems like the cube is straghtforward to me. Assume a cube x meters on a side with walls of zero thickness, where x is the cube root of the volume expressed as cubic meters after conversion from gallons. (I didn't realize I was assuming that the initial thickness of the walls was zero until now. To really solve this exactly you have to know the initial thickness.) Make the walls 5 cm thick. (My definition of 5 cm thick means 5 cm thick measured perpendicularly at the center of the wall--maybe that's the ambiguity you're talking about. If you measure from outside corner to inside corner you would get a different number, but the conventional meaning of "thickness of a wall" is as I described.) Now you have a cube with sides x + 0.1 meters (surface area left as an exercise :D ).

Also have to know how thick a coat of paint is, since the surface area will increase after the first coat.

[Freddy the Pig]

Why are you reading old actuarial exams? :eek: Either (a) you're taking them; or (b) you're thinking of taking them; or (c) you have waaaaaay too much time on your hands.

If (a), my condolences; if (b) Don't do it!; and if (c) find a nice girl! :)

Problem 2 is easy. Of the partial losses, 8% are below the deductible, 12% are over the policy limit, and on the remaining 80% you have a payment uniformly distributed between 0 and 10,000. This is equivalent to every payment being 5,000.

So your chance of a 10,000 payment is 2% + (4%*12%) = 2.48%.
Your chance of a 5,000 payment is (4%*80%) = 3.2%.

10,000 * 0.0248 = 248
5,000 * 0.032 = 160

Sum = 408 (A)

[jawdirk]

Also have to know how thick a coat of paint is, since the surface area will increase after the first coat.

Haha! And I guess you have to know how the paint adheres to the surface as well. Wouldn't want any error from neglecting the surface characteristics near the corners.

All I was saying is that if the interior of the container is a "perfect cube" per the OP, then you add minimal 5 cm thick walls around that interior, the exterior will not be a cube. It will have rounded corners with radius 5 cm. That is, measuring 5 cm outward, normal to the interior. That, by the way, is the only solution that results in uniformly 5cm-thick walls. If you use your solution, then the walls will be thicker at the corners. That's why I say the OP is ambiguous. And that affects the answer a lot more than the thickness of the paint!

[MaxTheVool]

All the other combinations of 2 numbers can be used to form at least one a+b=c+d.


What if you divided the numbers up between the 3 containers based on their mod 3?

So, for instance, you'd have 1, 4, 7, 10, 13... in the first, 2, 5, 8, 11, 14... in the seocnd, and 3, 6, 9, 12, 15... in the 3rd?

[CookingWithGas]

All I was saying is that if the interior of the container is a "perfect cube" per the OP, then you add minimal 5 cm thick walls around that interior, the exterior will not be a cube. It will have rounded corners with radius 5 cm. That is, measuring 5 cm outward, normal to the interior. That, by the way, is the only solution that results in uniformly 5cm-thick walls. If you use your solution, then the walls will be thicker at the corners. That's why I say the OP is ambiguous. And that affects the answer a lot more than the thickness of the paint!I agree with your statement, but I felt the only reasonable interpretation of "walls of each vessel are 5 CM in thickness" means the interior surface is a cube and the exterior surface is a cube, and the walls are 5 cm thick. If a builder asks the architect how thick he wants to make the walls, the architect isn't going to say, "Well, they're all different thicknesses. They're 5 cm thick if you measure in the middle of the wall, but they're 7.07 cm thick if you measure from an inside edge to an outside edge, and 8.66 cm thick if you measure from an inside corner to an outside corner."

[CookingWithGas]

. . .the method suggested won't give the correct answer.

The Kepler approximation will allow you to figure out about how many whole and fractional marbles exist in a spherical volume of a closely packed array. That is not the question though.Well, technically you are correct, of course. I guess there is a question of what precision is required for the answer.

This is like the old joke about the mathematician and the engineer. I am just trying to get an answer good enough for all practical purposes.

[astro]

I agree with your statement, but I felt the only reasonable interpretation of "walls of each vessel are 5 CM in thickness" means the interior surface is a cube and the exterior surface is a cube, and the walls are 5 cm thick.


Yes that was my intent.

[jawdirk]

What if you divided the numbers up between the 3 containers based on their mod 3?

So, for instance, you'd have 1, 4, 7, 10, 13... in the first, 2, 5, 8, 11, 14... in the seocnd, and 3, 6, 9, 12, 15... in the 3rd?

Brilliant! So that is at least one more combination. Any more?

[aliencase]

For Problem 1, I think this will work
A= 1-499 odd
B=1-499 even
C=500-999 odd
D= 500-999 even
I am not exactly sure this would work if it doesnt could you supply an example of where it doesnt

[aliencase]

For Problem 1, I think this will work
A= 1-499 odd
B=1-499 even
C=500-999 odd
D= 500-999 even
I am not exactly sure this would work if it doesnt could you supply an example of where it doesnt
Sorry I misunderstood the question now that I look back

[Omphaloskeptic]

I'm mainly interested in the approach and methodology-These are four very different sorts of problems (as other people have mentioned). Problem 1 is a typical math-contest problem (this is a variant of an International Math Olympiad problem, though I don't know if it originated there); these are typically designed to have a short, clever solution not requiring too much higher math. (The IMO is a high-school competition.) There are books that teach "problem-solving skills" (like Pólya's How to Solve It (http://www.amazon.com/gp/product/069111966X/)), but solving these problems is largely a matter of practice and inspiration. Problem 2 is a straightforward problem in probability, and problem 4, though incompletely specified, is a straightforward problem in geometry (well, the formulas for area and volume of a torus are not all that common, and their easiest derivations would use calculus). Problem 3 is an instance of a hard research problem; optimal finite packings are not generally known. (Unsolved Problems in Geometry (http://www.amazon.com/gp/product/0387975063/) (Croft, Falconer, and Guy) has some discussion of known optima and bounds, and is generally fun to look through.)

Problem 1:

There are 999 cards numbered 1 to 999, and three boxes colored red blue and white. Find the number of ways of putting the cards into the boxes such that there does not exist a,b belonging to different boxes and c,d belonging to different boxes (the former pair of boxes being not the same as the latter pair of boxes) such that a+b=c+d.A solution is spoilered below. As a hint (this is how I solved it), try filling the boxes starting from 1, and see if you can find any patterns to which arrangements work and which don't. For example, put card 1 in the red box, 2 in the blue box, and 3 in the white box. Where can card 4 go? Then how about 5? (For an answer, see Lemma 2.) Now put 1 in the red box, and 2 and 3 in the blue box. Where can 4 and 5 go now? (See Lemma 1.)

Apart from the arrangements with at least one empty box, there is only one arrangement (up to permuting the boxes) besides the one MaxTheVool found.

As jawdirk has said, the case in which at least one box is empty trivially satisfies the conditions, so we don't need to worry about these cases any more. Let's think about the remaining cases, in which each box contains at least one card. The two examples mentioned above lead to two lemmas that help us. (These lemmas deal with arrangements of sets of 3 or more cards in arithmetic sequences; for example, with the first k cards, or with cards 2, 4, 6, 8, ... . They are stated somewhat generally; when reading through them, think of the case a=1, n=1.)

Notation: Label the three boxes X, Y, Z. Write X+Y = {x+y: x in X, y in Y} for the set of all sums of pairs of elements from X*Y.

Lemma 1:Suppose that the k>2 values a, a+n, a+2n, ..., a+(k-1)n are all in either X or Y, with at least one such value in each; without loss of generality say that a is in X. If a+nk is in Z, then the X+Z and Y+Z sums contain all of the values 2a+kn, 2a+(k+1)n, ..., 2a+(2k-1)n; so these values are not allowed in the X+Y sum. Now Y, by assumption, contains at least one of the values a+n, ..., a+(k-1)n; say a+yn is in Y, with 0<y<k. If X also contains one of these values, a+xn with 0<x<k, then since a+(k-1)n is in either X or Y, either (a+(k-1)n)+(a+xn)=2a+(k+x-1)n or (a+(k-1)n)+(a+yn)=2a+(k+y-1)n is in X+Y. But k-1<k+x-1<2k-1 (and similarly k-1<k+y-1<2k-1), so this is equal to an element of either X+Z or Y+Z, violating the requirement. So if a+kn is in Z, then X must contain none of a+n, ..., a+(k-1)n, and we must have X={a}, Y={a+n, ..., a+(k-1)n}.Lemma 2:Suppose that a is in X, a+n in Y, and a+2n in Z. If a+3n were in Y, we would have a+(a+3n) = 2a+3n = (a+n)+(a+2n) in both X+Y and Y+Z; similarly if a+3n were in Z, the sum 2a+3n would be in both X+Z and Y+Z. So we must have a+3n in X. Similarly a+4n must be in Y, etc. By induction, then, the values a+3kn (for all k) must be in X, a+(3k+1)n in Y, and a+(3k+2)n in Z. (This is MaxTheVool's example.) Note that the induction works symmetrically in both directions: n may be positive or negative.Putting it all together:Now if 1 and 2 are in the same box (call it X), then we can see by the above that at least one box must be empty. (For suppose 1, ..., k are in boxes X and Y only; Lemma 1 (with a=1, n=1) shows that k+1 cannot be in Z, since both 1 and 2 are in X.) So suppose 1 and 2 are in different boxes (1 in X, 2 in Y). Now 3 can be in Z, in which case by Lemma 2 we must have X={1,4,7,...}, Y={2,5,8,...}, Z={3,6,9,...}; or in X, in which case Z must by Lemma 1 remain empty; or in Y. Now further consider this last case: Suppose that 1 is in X, and 2 and 3 in Y. As before, if 4 is in X then (by Lemma 1) Z must be empty; 4 may also be in Y. But if 4 is in Z, then where can 5 go? 5 cannot go in X (by Lemma 2), nor in Y nor Z (in each case, violating the form of Lemma 1 with a=5, n=-1). So this case does not work if we need to add another card to the boxes.

Then by inductively adding cards, we get one more possible arrangement of cards, using the final case considered above: X={1}, Y={2,3,4,...,998}, Z={999}.

So, now let's count all the arrangements: The mod-3 filling is unique up to permutations of boxes, so it gives 3!=6 possible arrangements. The arrangement above is also unique up to permutation, so this gives us 6 more. All that is left is counting the cases with at least one box empty. There are 3 cases with all the cards in one box. With two boxes nonempty, there are 6 ways of choosing the two boxes, and 2998-1 ways of partitioning 999 cards into two nonempty subsets, giving 6(2998-1) more ways. Collecting these all together gives us

6(2998-1)+6+6+3 = 3*2999+9.