:

Err. Show working

Where are our esteemed mathematicians? Cc richiez, fynestboi, karmanaut, geniusdavid, kolooyinbo, mathschic.

Lol. I guess this place is a no-go area.
Oya, who ah go mention?
Cc. Seun, sirwere, all4naija, teempakguy, darkhorizon( you watch xmen evo season 4?), sukkot(maybe this question might just esoteric)

timonski:
Lol. I guess this place is a no-go area.
Oya, who ah go mention?
Cc. Seun, sirwere, all4naija, teempakguy, darkhorizon( you watch xmen evo season 4?), sukkot(maybe this question might just esoteric)

lol bros, this one pass my power. i dont even understand the question sef.

sukkot:
lol bros, this one pass my power. i dont even understand the question sef.

lol
what they are asking is if the red ball and other 2 blues balls are placed to touch each other, how many such distinct arrangement can be made?

HINT: this is a question on PERMUTATION and COMBINATION.
Cc. Durentt, agentofchange1

Where is our household mathematics ninja, MATHSCHIC

Cc. Proffemi

timonski:
lol
what they are asking is if the red ball and other 2 blues balls are placed to touch each other, how many such distinct arrangement can be made?

thats a very good question to which i have no answer

the 3m cube can be placed in the big cube in one way. every other way is a rotated or mirrored perspective.
the 2m cube can be placed on besides the 3m cube in 4x3 ways.

for each placement of the 1m cubes can be positioned beside the 3m cube and the 2m cube in 5x4 ways.

so the final combination should be 4x3x5x4 = 240 different combinations.

cc timonski

Teempakguy:
the 3m cube can be placed in the big cube in one way. every other way is a rotated or mirrored perspective.
the 2m cube can be placed on besides the 3m cube in 4x3 ways.

for each placement of the 1m cubes can be positioned beside the 3m cube and the 2m cube in 5x4 ways.

so the final combination should be 4x3x5x4 = 240 different combinations.

cc timonski

wrong sir. Look again
What happens when the 3m cube is placed at the centre of a face, at the edges and at the vertices? Take it slowly this time.

timonski:

wrong sir. Look again
What happens when the 3m cube is placed at the centre of a face, at the edges and at the vertices? Take it slowly this time.

then the 3m cube can be placed three ways. at a corner, at the middle of one side, or at the center of a face. any other way can be dismissed as a rotation or a mirror. otherwise, it will be outside the cube.
that would add a factor of two. making a total 720 ways. when the cube is placed at the middle of one edge, the 3m cube can then be placed in 4x4 ways . . . .

I'm going to have to reevaluate my entire working.

I'll be back to edit.

Teempakguy:
then the 3m cube can be placed three ways. at a corner, at the middle of one side, or at the center of a face. any other way can be dismissed as a rotation or a mirror. otherwise, it will be outside the cube.
that would add a factor of two. making a total 720 ways. when the cube is placed at the middle of one edge, the 3m cube can then be placed in 4x4 ways . . . .

I'm going to have to reevaluate my entire working.

I'll be back to edit.

continue... But try to be a little bit gentle when explaining. U know u could tell us why it should be So X So ways.

timonski:

continue... But try to be a little bit gentle when explaining. U know u could tell us why it should be So X So ways.

well, you see, the problem is 3 dimensional one. It's sort of hard to explain . . . anyway.

let me go at this again.

we can divide the big box into grids. each face of the box would have 25 small squares. where the length of one square is 1m.

then, we simply get rid of everything except one face. put it on the ground horizontally. and put the 3m cube on it. we can find that it occupies nine possible places.(because if we label the face of the large box with numbers, the 3m box can stand on squares 1-3, 6-8 11-13, at first. then we can move it around)

out of this nine possible places, only three possible places can allow the smaller boxes to be stacked on the 3m box. the earlier configuration I described, moving to the squares 2-4, 7-9, 12-13, and moving to the squares 7-9, 12-14, 17-19. every other configuration would be considered a rotation or mirroring.

I think my earlier postulations for the arrangements of the smaller cubes can hold here. so the final config should then be 3x4x3x5x4 = 720.

how about that?

Teempakguy:
well, you see, the problem is 3 dimensional one. It's sort of hard to explain . . . anyway.

let me go at this again.

we can divide the big box into grids. each face of the box would have 25 small squares. where the length of one square is 1m.

then, we simply get rid of everything except one face. put it on the ground horizontally. and put the 3m cube on it. we can find that it occupies nine possible places.(because if we label the face of the large box with numbers, the 3m box can stand on squares 1-3, 6-8 11-13, at first. then we can move it around)

out of this nine possible places, only three possible places can allow the smaller boxes to be stacked on the 3m box. the earlier configuration I described, moving to the squares 2-4, 7-9, 12-13, and moving to the squares 7-9, 12-14, 17-19. every other configuration would be considered a rotation or mirroring.

I think my earlier postulations for the arrangements of the smaller cubes can hold here. so the final config should then be 3x4x3x5x4 = 720.

how about that?

The 2m and 1m cubes still have to be permuted about the free surfaces of the 3m cube. So you should analyse the 3 cases!
Case1. The 3m cube at the corners
case2. At the sides and at face centres.
You still havn't done all these.

The answer is actually in thousands.

http://www.matrix67.com/iqtest/

Bros this question was gotten from the test above... Read about the test so you'll see what's special about it

timonski:

The 2m and 1m cubes still have to be about the free surfaces of the 3m cube. So you should analyse the 3 cases!
Case1. The 3m cube at the corners
case2. At the sides and at face centres.
You still havn't done all these.

case 1, the 3m cube at the corners.
I can simply do the math for a single corner. every other corner, I will dismiss as a rotation or mirroring as directed by the question.

case 2, the cube at the sides. the same logic will apply here. at the side, every other side, can be considered as a rotation. the important leverage here is that we're dealing with squares. squares are symmetrical across all the spatial dimensions. so whatever works for one sides, can be duplicated for the others by simply rotating or reflecting. which is just rotating 180 degrees.

now, I will admit. I only stacked the cubes. they still have a lot more room.

just thinking about the implications of that makes me cringe.

**Strolls in**

** Sees Teempakguy handling the question **


**Strolls out without attempting the question**


Cc Durentt (I think).

yorex2011:
http://www.matrix67.com/iqtest/

Bros this question was gotten from the test above... Read about the test so you'll see what's special about it

Ah, I see. a set of difficult questions meant to measure IQ.

Hey . . . It says here that the average professional mathematician or physicist would barely be able to answer half of them correctly. This IS a cringe worthy test.

Also, the use of calculators and computers are permitted. what . . .

Teempakguy:
Ah, I see. a set of difficult questions meant to measure IQ.

Hey . . . It says here that the average professional mathematician or physicist would barely be able to answer half of them correctly. This IS a cringe worthy test.

Also, the use of calculators and computers are permitted. what . . .

lol. The haselbauer test is a fun test. But this particular question is more math than brains.
Anyway how many have you attempted?

okay, timonski, I give up. tell us the answer.

timonski:

lol. The haselbauer test is a fun test. But this particular question is more math than brains.
Anyway how many have you attempted?

three, as of now.
plus this question. making four.
but it's kind of depressing because I can't find any expo sheets.

EDIT

Teempakguy:
okay, timonski, I give up. tell us the answer.

case1: the 3m cube at a corner.-
At the corner, the 3m cube is bound by 3 surfaces. So it has only 3 free faces each of area 9m^2. The 2m cube(with surface area 4m^2) can move about each 9m^2 surface in 4ways.
How? On the 9m^2 surface(nine 1m squares) squares 1245, 2356, 4578, and 5689 are the 4 possible squares. (note this!)
since we have 3 free 9m^2 surfaces(total of 27m^2 area), the 2m cube can arrange along the 3m cube in 4*3 ways.
Now, for each 4m^2 of area occupied by the 2m cube, there is (27-4)m^2 area left for the 1m cube. So it can occupy in 23 ways. Total number of permutation for case 1 is 23*4*3 ways= 276ways...(1)

case 2: 3m cube on edge.
We have 2 bounded surfaces, and thus 4 free faces of the 3m cube. Wait o! Kasala don burst! The 2m actually occupy only 2 faces. Sorry o!
So the number of ways the 2m cube can arrange on the 4 faces is........4*2 ways.
For each 4m^2 area occupied by the 2m cube, there (4*9 - 4)m^2 i.e 32 1m^2 area left for the 1m cube. So it can occupy in 32 ways.
Thus, total permutations for case2
is 32*4*2=256 ways.
Note that for each case the other 89 red balls arrange themselves in only 1 DISTINCT way.
Case3. 3m cube at a face centre.
Here we have 5 free faces.
WTF! No! There is actually only one space for the 2m cube to roam. Hence total of 5*9m^2 i.e 45m^2 area.
The 2m cube occupies 4m^2 square in 1*4 different ways.
For every 4m^2 occupied we are left with (45 - 4)m^2 area for the 1m cube. Therefore. Number of distinct arrangements is 41*1*4=164 ways
Finally, we add the different combinations in the 3 cases to get
276+256+164= 696 distinct arrangements. Okay. I am a joke.

Teempakguy:
three, as of now. plus this question. making four. but it's kind of depressing because I can't find any expo sheets.

lol. The test was stopped since 2001. If you get 8 then you are a real genius.

yorex2011:
http://www.matrix67.com/iqtest/

Bros this question was gotten from the test above... Read about the test so you'll see what's special about it

yea, but that particular question is more mathematical.

Oya make una clap for me na.
Where my mathschic?

Teempakguy:
three, as of now. plus this question. making four. but it's kind of depressing because I can't find any expo sheets.

What did you get in that "circle and area" question?

timonski:

case1: the 3m cube at a corner.-
At the corner, the 3m cube is bound by 3 surfaces. So it has only 3 free faces each of area 9m^2. The 2m cube(with surface area 4m^2) can move about each 9m^2 surface in 4ways.
How? On the 9m^2 surface(nine 1m squares) squares 1245, 2356, 4578, and 5689 are the 4 possible squares. (note this!)
since we have 3 free 9m^2 surfaces(total of 27m^2 area), the 2m cube can arrange along the 3m cube in 4*3 ways.
Now, for each 4m^2 of area occupied by the 2m cube, there is (27-4)m^2 area left for the 1m cube. So it can occupy in 23 ways. Total number of permutation for case 1 is 23*4*3 ways= 276ways...(1)

case 2: 3m cube on edge.
We have 2 bounded surfaces, and thus 4 free faces of the 3m cube.
So the number of ways the 2m cube can arrange on the 4 faces is........4*4 ways.
For each 4m^2 area occupied by the 2m cube, there (4*9 - 4)m^2 i.e 32 1m^2 area left for the 1m cube. So it can occupy in 32 ways.
Thus, total permutations for case2
is 32*4*4=512 ways.
Note that for each case the other 89 red balls arrange themselves in only 1 DISTINCT way.
Case3. 3m cube at a face centre.
Here we have 5 free faces. Hence toal of 5*9m^2 i.e 45m^2 area.
The 2m cube occupies 4m^2 square in 5*4 different ways.
For every 4m^2 occupied we are left with (45 - 4)m^2 area for the 1m cube. Therefore. Number of distinct arrangements is 41*5*4=820ways.
Finally, we add the different combinations in the 3 cases to get
820+512+276= 1608 distinct arrangements.

Okay. well, I think the bolded is problematic.
the 3m cube cannot arrange on two of the sides because then it would be outside the 5m cube. it can only arrange on the top, and the side directly opposite the side bounded by the larger cube. the two sides adjacent to that have only 1m of free space left. meaning that if you try to put a 2m cube there, half of it will be outside the cube. which is prohibited by the question.

don't you think?

timonski:

What did you get in that "circle and area" question?

would that be number three?
64.

I simply assumed it would progress uniformly.

Teempakguy:
Okay. well, I think the bolded is problematic.
the 3m cube cannot arrange on two of the sides because then it would be outside the 5m cube. it can only arrange on the top, and the side directly opposite the side bounded by the larger cube. the two sides adjacent to that have only 1m of free space left. meaning that if you try to put a 2m cube there, half of it will be outside the cube. which is prohibited by the question.

don't you think?

whoa did I really miss that? I beem think that thing before.