Insane Math Questions. Those With Weak Hearts Beware.

[SIZE="6"]Before you come in and say "Yo mother ****er there's already a math help thread,"[/SIZE]
these levels of questions are on a different league. They are not specifically integrating or differentiating or something I learned specifically in class. No my friends, these are hard core logic questions, and to be honest I don't understand what half of them are asking for. And the other half I don't understand how to do.

I have fourteen questions, like these, that I want to do. You see, my whole school can work together on these questions and submits our best answers, it's a math league competition.

But I'm calling for outside help. I've scanned the questions from my real life sheet and uploaded them to photobucket. You are not complied to help. Do this for your own math enjoyment and to prove how good you are. But hey, a little incentive never killed anyone. Most of us play maplestory, if you get a question right and can show me the work and logic so that I can understand it too then I'll give you a certain amount of nx in game via MTS,I will cover the tax. The amount is listed with the specific problem. *cough* I doubt anyone can get these anyway.


Editted posts will be disqualified from the nx prize. Post twice if you need too.

Edit: New nx posted. I will cover MTS tax for all of them. Yeah I know I'm cheap F3.

 Spoiler

Question Seven

 Spoiler

A. 800 nx
B. 800 nx
C. 800 nx
All. 3k nx

Question Ten and Eleven

 Spoiler

10. A. Done
10. B. Done
10. C. 500 nx
10. D. 500 nx
10. E. 500 nx
10. F. 500 nx
10 All. 2.5k nx

11. As of now there is no reward for 11.

Question Twelve, Thirteen, and Fourteen.

 Spoiler

12. A. Done
12. B. 250 nx
12. C. 250 nx
12. D. Done
12. E. 250 nx
12. All. 1000 nx This one is kinda easy F3 no need to super pay.

13. A. 500 nx
13. B. 500 nx
13. C. 500 nx
13. All. 2k nx

14. A. 250 nx
14. B. 250 nx
14. C. 250 nx
14. D. 250 nx
14. E. 250 nx
14. F. 250 nx
14. G. 250 nx
14. All 2k nx

For question thirteen the word "but" is cut off a little. Nothing else is cut off though.

Holding this slot in case I post more questions.

I just looked at the first question and blinked hardly.

Is the first question guess and check?

Question 7 is easily solved with by an algorithm, but it'd be a lot faster to use a computer.

13 and 14 seem pretty easy to solve. I'll give them a try later when I have some time if no one's solved them.

lol was expecting some challenging questions...

it's just brute force, get someone who can code and they can solve these questions easily. (though you'll need to type in all the numbers)

I will definitely ask my in game buddies or buddies in real life for help, or post these on another forum. If they answer before you give me the answer, I'm sorry the nx prize is null (for you).

But don't worry if you are first I won't scam you, getting help on these will be nice and it deserves a prize.

So basically you're asking us to help you win a contest?

I don't win anything from the contest though. Except pride.

I want to be one of the people to get at least one of the questions in my school. I think I got 11 right but if I can get more, then it's worth any monetary nx I put up. Well not any. But the prices I'm putting up, yeah.

For 12 you can basically make it a large set of linear equations and then solve them.

Since each letter appears once in the solution, make 2 variables for all the letters of the alphabet, one with value equal to position, the other indicating whether it's present (A = value, a = present)
1) Alpha != Beta for all non zero (alpha, beta) pairs.
2) Alpha + Beta + Gamma + Delta = Xi, where Xi is one of the clues
3) alpha + beta + gamma + delta = xi, where xi is one of the clues
4) the increasing sequence spells a word with some mathematical meaning
5) Alpha <= 10 for all Alpha in the alphabet.
6) sum(lowercase) = 10

eg. for a), you have
S+I+D+E=17
s+i+d+e=3
F+A+C+E = 12
f+a+c+e = 2
D+I+S+K = 10
d+i+s+k = 2
C+O+N+E = 23
c+o+n+e = 3
R+E+C+T = 17
r+e+c+t = 2
R+A+Y+S = 9
r+a+y+s = 2
L+U+M+P = 13
l+u+m+p = 3
M+A+T+H = 11
m+a+t+h = 2

One assumption that may prove useful is that all 10 letters in the eventual solution are present in these clues (not stated in the question, though). This would reduce the full alphabet to
acdefhiklmnoprstuy
Only 18 letters.

We have 8 words with a total of 19 letters present in the final solution, meaning that there are 9 "repeats" of letters.
Letter frequencies are:
a3 c3 d2 e4 f1 h1 i2 k1 l1 m2 n1 o1 p1 r2 s3 t2 u1 y1

With the 8 word equations and the sum(alpha) = 10, we only have 9 equations and 18 unknowns so a unique solution is not guaranteed. But since they're binary variables it's easy to construct a tree of possibilities, starting with the most common letters and eliminating possibilities.

A simple contradiction can be found if we assume 'e' is not present:
s+i+d+e = 3, d+i+s+k = 2, e = 0 => k = -1
ie.
e = 1 -> s+i+d=2, f+a+c=1, c+o+n=2, r+c+t = 1, k = 0
a = 1 -> f = 0, c = 0, r+y+s = 1, m+t+h = 1, o = 1, n = 1, r+t = 1
s = 1 -> i+d = 1, r = 0, y = 0, t = 1, m = 0, h = 0, l = 1, u = 1, p = 1
This gives 2 possibilities
(e,a,o,n,s,t,l,u,p,i)
(e,a,o,n,s,t,l,u,p,d)
Neither of these find quick anagrams so replace one assumption

s = 0 -> i = 1, d = 1, r+y = 1
r = 1 -> y = 0, t = 0, m + h = 1, l + u + m + p = 3
We have 6 so far, so the only possibility that gives 10 letters is h, l, u, p = 1
(e,a,o,n,i,d,r,h,l,u,p) Edit: This line is proof that Stereo can't count to 10. x_x
Again, no quick anagrams, so step back to the assumption a = 1.

a = 0 -> f+c = 1, r+y+s = 2, m+t+h = 2
c = 1 -> f = 0, o+n = 1, r = 0, t = 0, y = 1, s = 1, m = 1, h = 1, i+d = 1, l+u+p = 2
This gives a whole bunch of valid possibilities:
(e, c, y, s, m, h) + o/n, i/d, 2 of (l, u, p)
Ex. ecysmhoilu, ecysmhodlu, etc. Unfortunately, none of these seem to be anagrams of actual words.

c = 0 -> f = 1, o = 1, n = 1, r+t = 1
This always results in 11 letters or more in the solution so it's invalid.

Take one of those 'potential solutions' and find positions
ECYSMHODLU
S+D+E = 20
C+E = 12
D+S = 10
C+O+E = 23
contradiction - O = 11

And so forth. Obviously doing all of these is kinda a waste of time - so try starting with just the "known" formulas and see what introduces contradictions.
ECYSMH:
E+C = 12 (FACE = 12)
E+C = 17 (RECT = 17)

EAONSTLUP:
A+E = 12
O+N+E = 23
E+T = 17
A+S = 9
L+U+P = 13
A+T = 11
E = 10 (as before), A = 2, thus contradiction: T = 7 and 9.

That leaves
EAONIDRHLUP
I+D+E = 20
A+E = 12
D+I = 10
O+N+E = 23
R+E = 17
R+A = 9
L+U+P = 13
A+H = 11
E = 10, A = 2, R = 7, H = 9
I+D = 10, O+N = 13, L+U+P = 13
These need to fit into (1, 3, 4, 5, 6, 8). Obviously 1 is L, U, or P, leaving the other 2 equal 12. So 10, 12, 13 -> 3 doesn't fit into any of these.


Since I've kinda eliminated all possibilities I guess I made a mistake.

On the other hand, I've got a better handle on the question.
E = 10, is in the solution. K is not in the solution.
so clues are reduced to
SID, 2, 10
FAC, 1, 2
CON, 2, 13
RCT, 1, 7
This leads to elimination of C (can't be both 2 and 7), so
O and N are in the solution, and their positions total to 13
FA, 1, 2
RT, 1, 7
RAYS, 2, 9
By inspection, if A is in RAYS, then another letter has 7. If R is in rays, then another letter has 2. This is only true if
a) RA are both in the solution, with R = 7, A = 2. FT are not in the solution.
b) FT are both in the solution, with F = 2, T = 7. AR are not in the solution.

a)
_A____R__E, plus O and N in (4, 9) or (5, 8)
_A_(ON)__R_(NO)E
_A__(ON)_R(NO)_E

b)
This implies MH, 1, 4. -> one or the other is in the solution.
Which in turn implies that O and N are in (5, 8)
_F_(MH)(ON)_T(NO)_E
At this point, SID, 2, 10 must be (1, 9) because (3,6) don't add to 10. SF, DF don't make sense.
IF_(MH)(ON)_T(NO)(SD)E
LUMP can only have 3 letters if M is present, and requires that the other 2 letters add to 9 and are in (LUP)
IF(LUP)M(ON)(UPL)T(NO)(SD)E
Assuming this word is constructed with vowels in typical positions
IFUMN(PL)TO(SD)E
Doesn't seem to have any good possibilities. So eliminate option b).

Going to take a second look at a)
_A_(ON)__R_(NO)E
_A__(ON)_R(NO)_E
Plus remaining useful clues:
SID, 2, 10
LUMP, 3, 13
MTH, 1, 9 - this eliminates one of the options, again.
_A__(ON)_R(NO)(MTH)E
1 3 4 6 are left. Needs to fit 2 of SID and 3 of LUMP into those 4 spots. So it's kinda gotta be an M.
_A__(ON)_R(NO)ME
4+6 for SID, 1+3 for LUP
(LUP)A(UPL)(SID)N(SID)ROME
I would say the answer is:
PALINDROME


This is just 12.a)



12. d)
HALF 3 22
ONES 3 11
TRIO 3 18
DUOS 3 15
TWOS 2 7
FOUR 2 11
FIVE 2 11
PLUG 2 12

O must be in the solution, otherwise it has 11 letters.
F must not be in the solution, otherwise FIVE / FOUR share 2 letters in the same spots (due to being 2,11)
HALO is in the solution.
NES 2
TRI 2
DUS 2
TWS 1
UR 1
IVE 2
PLUG 2
Assume S is in the solution (random guess)
-> TW are not in the solution
-> RI are in the solution
-> U is not in the solution
-> D is in the solution
NE 1
VE 1
PG 1
Since we've got HALOSRID already, N/V cannot be in the solution (too many letters)
HALOSRIDEP
HALOSRIDEG

Interestingly, an anagram of HALOSRIDEP is SPHEROIDAL
Checking the equations this works.
12. d) SPHEROIDAL.

I don't win anything from the contest though. Except pride.

I want to be one of the people to get at least one of the questions in my school. I think I got 11 right but if I can get more, then it's worth any monetary nx I put up. Well not any. But the prices I'm putting up, yeah.

NX first Terry. =]

this might get you started on doing the first problem:

Code:

class bitboard:
    def __init__(self):
        self.bitboard = [[2,2,1,5,2,6,8,0,3,2,9,4,8,5,1,4,6,3,9,1],
                         [6,0,3,9,3,6,3,5,9,5,5,7,7,9,8,2,1,8,7,5],
                         [8,6,8,5,9,4,8,1,1,1,7,4,4,3,1,0,4,3,4,3],
                         [6,3,1,1,0,4,1,8,4,2,6,7,0,6,0,5,4,3,9,0],
                         [9,5,9,4,0,1,9,9,8,5,9,8,2,2,0,2,6,3,5,3],
                         [0,9,3,8,8,5,8,3,8,7,7,9,6,3,2,8,9,7,3,4],
                         [1,3,3,7,2,6,6,0,6,2,8,0,3,4,4,5,3,9,0,8]]
        self.width = len(self.bitboard[0])
        self.height = len(self.bitboard)

        self.ShapeAssignValue = 0x80000000
        self.ShapeDescriptions = []

    def RegisterShape(self, Description):
        self.ShapeDescriptions.append(Description)
        self.ShapeAssignValue >>= 1
        if self.ShapeAssignValue < 16:
            raise Exception("Cannot assign any new shapes")
        return self.ShapeAssignValue

    def GetShapeAtOffset(self, xcoord, ycoord):
        if self.ShapeAssignValue == 0x80000000:
            return "NO SHAPES ASSIGNED"
        else:
            bitboardvalue = self.bitboard[xcoord][ycoord]
            shapeassignvalue = 0x80000000
            iterator = 0
            while shapeassignvalue > 16:
                if shapeassignvalue & bitboardvalue == 1:
                    shapeassignvalue >>= 1
                    return self.ShapeDescriptions[iterator]
                else:
                    iterator += 1
                

    def GetShapeValues(self, xcoord, ycoord, xybitfield, value):
        return sum([self.bitboard[xcoord + xbit][ycoord + ybit] for xbit, ybit in xybitfield]) * (value + 1)

class shape:
    def __init__(self, description, height, width, offsets, value):
        self.description = description
        self.height = height
        self.width = width
        self.offsets = offsets
        self.value = value

Genius
12. d) SPHEROIDAL.

I'll pay the nx when I come home. PM me your in game name and set up some noob etc item up for 250 nx in MTS. I'll buy.

Oh. Put 500 nx in MTS then sorry. I just noticed A too.

Some computer magic

Righhhhhhhht. I'm not really good with computers F3.

14.

a. This problem is simple because it's one-dimensional. It helps you think about how to do the harder problems later on, though.

You have a 3x4 grid, and you need to figure out how many ways there are to fill it with 1x2 blocks. That means there are at most 6 blocks to place. Now we just need to figure out how to place those 6 blocks.

So ask yourself, is there any situation that makes you unable to place a block? Let's start with situations that make blocks really hard to place. If you'll notice, this setup leads to only one possible solution:



In fact, as soon as you place that blue block in that spot, you've locked yourself into one solution. Similarly, if you put the red block in the center, you've locked yourself into that one solution. The only other possibility is if the blue block is below the red block, which is the same result, since you can just flip the box 180 degrees. So now let's add one to our total solutions and pretend this one doesn't exist anymore. If we can't place the blue block or the red block, that means you aren't allowed to place any blocks spanning the middle two rows. This is really important. Why? Because now you can cut the box in half down its center, and you won't have to deal with any solutions where a block sticks out halfway from the other side. This makes the problem "how many ways are there to fit 3 1x2 blocks into a 2x3 box?" This is extremely easy to solve. Is there any way you can place a block so that another block can't be placed? No. No matter how you place the first two blocks, the third block will have a place to go. So since there's only three blocks, let's just look for unique situations. Once you place the first block, there are only two ways to place the other two.



We can:
1. place all three horizontally (symmetrical) or
2. place one horizontally and have the other two pointing at it (2 facings).

Now we just need to fit the two 2x3 sections together. (1) fits with itself one way, (1) fits with (2) two ways, and (2) fits with itself two ways. That means there are 5 unique solutions.

Add the original solution, and we have a total of 6 for our answer.

I solved this problem extremely methodically, even though it could've been done much faster by guess and check, because it allows you to see how to attempt the other problems, which can't be so easily guess-and-checked. Don't have time to get around to the other parts right now, but hopefully this will help others that do

Edit: Actually, © is ridiculously easy.

You have a 3x3 cube and you have to fill it with 3x1 blocks. These blocks will always fit across the entire cube. What's more, once you place the block that passes through the very center of the cube, you lock yourself into a small set of solutions. We can think of the cube either as a set of three 3x3 squares stacked on top of each other, a three 3x3 squares stacked one behind another. If you think about it, there are no solutions where you can't rotate the cube so that this is true.

I'll choose to start with the cube as 3x3 squares stacked on top of each other. There's only two ways to fit three 3x1 bars into a 3x3 square. Horizontally (H), or vertically (V). Every solution will be based on the facing of the stacked squares. Since the problem states that the top of the box is open, HVV is different than VVH. However, since rotation doesn't matter, HVV is the same as VHH.

There are only four unique solutions.
HHH
HHV
HVH
HVV

Now let's deal with the squares stacked vertically. This time, it's backward. HVV is different than VHH, while HVV is the same as VVH. Also, HHH is an invalid solution, because it's exactly the same as the HHH from the previous part.

Unique solutions:
VVV
VVH
VHH

4 + 3 = 7 solutions.

yoooo STEREO IS ON POINT.

I didn't read the questions but I saw steroe's post and had to /applaud him


Here's a fun riddle for you guys. A guy is carrying 3 golden balls with him as he's running away from some thieves who want to take his balls. He comes across a bridge that says "Weigh limit 200 lbs". The man weighs 180 lbs, and each of the balls weigh 10 lbs each. He does not want to throw one ball across the bridge, dig a hole and come back to get it, or roll it across the bridge. How does he get across? Yes he does go across.

He could assume that the engineers have followed standard practices and 200lb is more a guideline to keep you well inside safe ranges. Or if you want the clever but physically unlikely example, he could juggle them. Man, have you ever juggled bowling balls?

yoooo STEREO IS ON POINT.

I didn't read the questions but I saw steroe's post and had to /applaud him


Here's a fun riddle for you guys. A guy is carrying 3 golden balls with him as he's running away from some thieves who want to take his balls. He comes across a bridge that says "Weigh limit 200 lbs". The man weighs 180 lbs, and each of the balls weigh 10 lbs each. He does not want to throw one ball across the bridge, dig a hole and come back to get it, or roll it across the bridge. How does he get across? Yes he does go across.

It's probably something stupid like "Two balls are his own because he's a man, and that counts toward his 180 pounds. The extra real ball is 10 so his total is 190."

...these questions look so painful. D: I might take a shot at these through C++. Or pure thinking depending...

I decided to finish up 14 while I'm at it.

14.

b.

Once again, there are 6 blocks to place. The shaded 2x2 area must be filled with two complete blocks. If you were to lay any block along the x-axis instead, you could just rotate the box 180 degrees, and the other side would have to be filled with 2 blocks. There are 2 ways to fill a 2x2 area with 2 blocks.

Now let's look at the rest of the box. We have a 2x2x2 cube, and 4 blocks to fill it with. This is simple enough to start listing possiblilities.
1. We can stack them all vertically.
2. We can stack them all horizontally, facing the x-axis.
3. We can have the ones in front stand vertically, and the ones in the back stand horizontally.
4. We can have the ones in front stand horizontally, and the ones in the back stand vertically.
That's 8 possibilities.

We have to be careful with setups where no block lies along the x-axis, because some of the possibilities are duplicated when you flip the box around. The middle row of the box is unaffected by rotation, so there are two possibilities there. There are three ways to set up the left and right ends of the box.
1. Left vertical, right horizontal
2. Both vertical
3. Both horizontal
That's 6 possibilities.

Answer is 8 + 6 = 14 solutions.

Will edit when I solve the rest.

d.

This problem is similar to part (a). From part (a), you can see that if you stand a block up so that it takes up a 1x2 area on the 3x4 grid, there are 6 possibilities. Sweet, halfway done. All our other solutions must contain at least one block that lies on its side.

I'm going to proceed in the same way as I did in part (a). Figure out which solutions involve the middle of the box, eliminate them, and cut the box in half. You can see that you are not allowed to completely fill up the middle of the bin with 3 blocks, because you can't divide 3x3 areas into 3x2 sections. Say we place a block floating in the center of the bin. There are now only two ways to fill the rest of the bin.



The other is the same thing, upside down. There are no other ways to fill the bin with blocks spanning the center. So now that we have our two solutions, we can cut the box in half.

Ways to fill a 3x3x2 box with 3x2x1 blocks not counting standing all blocks up:
1. 3 stacked on top of each other.
2. One flat on the bottom, other two vertical
3. Two vertical, one flat on the top.

3 choose 2 = 6 permutations.

Answer is 6 + 2 + 6 = 14 solutions.

e. There are 9 blocks in this one. Luckily, it is impossible to fit all the blocks into the given area. The answer is 0.

f. Not too hard to see that once you place the yellow block, there is only one solution.


All other solutions involve the blocks all lying flat or all standing up and facing the same direction. There are 4 other solutions, for a total of 5.

g. I only found 2 solutions via trial and error. I'm pretty sure there aren't any more, though.

For convenience, solutions to all parts of problem 14 here:
a. 6
b. 14
c. 7
d. 14
e. 0
f. 5
g. 2

I will compare these to the answers of my teammates and if these are the best answers, either them being right or the best answers possible, set up something in the MTS and PM me which char to look up.

I disagree on F, Dusk.

There is a simplification to make: Cut it into 2 big "L" shaped sections, 2x2 + 2x1
[]
[][] from above.
There are 3 ways to assemble 2 blocks into this shape: both directly on each other, or the "interlocked" way your picture has it (which is asymmetric, and thus can be reflected).

There are 4 distinct rotations of the 2 big Ls that are not the same under the rotation rules. (left/right, top/bottom, and mirrored versions)
For each of these, you can place the 2 small Ls in 3 ways, for 4*3*3 = 36 unique combinations.


Add to that the "lying flat" ones, where you make 2 3x2s that do not overlap into big Ls and I think there are at least 40 solutions.



Also, justification on E -
the 3x3 dimension cannot have 2x2x1 blocks fit into it. A 2x2x1 block has either 0, 2 or 4 blocks in a plane. Adding up to 9 blocks is impossible.