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/puzzle/ - Puzzles and recreational math

For any and all fun with math and logic

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File: 1434270453993.jpg (130.48 KB, 630x230, 63:23, committee_flash.jpg)

92c781 No.68[Reply]

This text is modified from the original, but the idea is the same.

A particular company organizes its employees into committees. Each employee can be a member of multiple committees. Several of the committees can be in turn organized by a directory, A committee may be listed in several directories.

The employees, committees, and directories follow these rules.

1. For any pair of employees, they are both in exactly one committee.

2. For any employee and directory, the employee is in exactly one committee listed by the directory.

3.All committees have an odd number of employees, and a committee has 2k+1 employees iff it is listed in k directories.

The company has 2401 employees. How many directories are there?


File: 1432280878916.png (33.62 KB, 1280x1280, 1:1, 1420009998419.png)

3436ad No.60[Reply]

can someone explain to me what is going on in the sticky picture, I've been really curious and can't seem to find anything, is it some sort of logic chart or something?

27348e No.62

It's a cleaned-up version of one of my notebook diagrams. The bottom left is supposed to be the DFT of the top inputs.


27348e No.63

>>62

Also, sorry for not noticing this post until now. It did use the same pic as a sticky.



File: 1431678324431.jpg (281.7 KB, 1248x1063, 1248:1063, solids.jpg)

f128d9 No.57[Reply]

Of the 5 platonic solids, the cube, the the octahedron, and the tetrahedron alone have regular analogues in all positive dimensions. (Respectively, the n-(hyper)cube, the n-orthoplex, and the n-simplex.) Also, in their respective space, it is possible to circumscribe these regular polytopes with the boundary of some n-ball. What is the ratio of the edge length over the radius of the ball for these three polytopes in terms of n?


File: 1426544845227.png (479.88 KB, 600x600, 1:1, 1-art-of-π-600x600.png)

e8f61b No.30[Reply]

What is the probability that the (10^100)^100th digit of pi is 7?

bd4b01 No.37

1 in 10

5a1bda No.38

>>30
The digits of pi are effectively determined by its mathematical properties and the laws of logic. They are thus 'fixed', and the fact of the specified digit being 7 is either logically necessary or logically contradictory. Thus the probability is either 0 or 1.

7e350c No.49

Most likely, yes. http://en.wikipedia.org/wiki/Normal_number

Pi has not been proven to be a normal number, but statistics show that it is significantly likely.



File: 1428735097069.png (36.01 KB, 828x707, 828:707, root_example.png)

ecaaa5 No.43[Reply]

Show that this procedure for finding the real roots of a real quadratic equation finds all and only the real roots.

1.On the x-y plane, given the equation ax^2 + bx + c = 0, a != 0, start with the points S = (0, -1), P = (-b, a-c-1) and the line y=(a-1).
2. Construct the midpoint M of S and P and the circle centered at M passing through P.
3. Mark any points where the circle touches y = (a-1). Draw lines from these points to S. Where the line(s) cross the x-axis, the root is given by the x co-ordinate.

(Picture: Example construction using 4x^2 - 8x + 3 = 0)

084c27 No.41[Reply]

A group of 7 pirates find a treasure chest. Opening it up, they find only 2 coins within. The pirates divide the treasure in the traditional pirate way: The captain proposes a distribution of coins and there is a vote on whether to accept it. A proposal is rejected if and only if a majority vote against it. If the proposal is rejected, the captain is killed and the highest ranking pirate still alive becomes captain. Each pirate's preferences are as follows, with strongest factors listed earlier.

-Staying alive preferred to being killed
-Being captain preferred to not being captain
-Receiving two coins preferred to receiving one coin and receiving one coin preferred to receiving none
-Killing someone preferred to not killing someone.

Assuming all pirates vote logically, which pirate(s), in terms of relative rank in the group at the start, get the treasure?

084c27 No.42

>>41
Also, if the proposal is rejected, the process starts again with the new captain.


File: 1423304742441-0.png (7.08 KB, 194x397, 194:397, agg-ch1.png)

File: 1423304742441-1.png (7.76 KB, 192x387, 64:129, agg-ch2.png)

0f9c50 No.18[Reply]

I found a web app that lets you make compass and straightedge constructions. It also gives you some specific challenges to aim for, but they're not mandatory. The pics show how well I've done the challenges.

http://sciencevsmagic.net/geo/
(Note, the origin circle is the circle centered on the left starting point passing through the right starting point.)

0f9c50 No.40

My records so far(OC:origin circle)

Triangle:5 (OC:6)
Hexagon:9 (=OC)
Circle Pack 2:5 (OC:7)
Square:8 (=OC)
Circle Pack 3:9 (OC:11)
Circle Pack 7:13 (OC:15)
Octagon:15 (=OC)
Dodecagon:20 (=OC)
Circle Pack 4:12 (OC:14)
Pentagon: 13 on my own, 11 possible(=OC)


File: 1426324912385.png (1.8 KB, 341x231, 31:21, points.png)

1b914a No.28[Reply]

Given two points A and B, and a ratio q > 0, !=1, Construct the set of points S such that for all P in S, PA/PB = q.

292213 No.31

Testing latex, will post soon.
$$line~1. \\ line~2. $$
$$line~1. \\* line~2. $$

292213 No.32

Set A to be the origin, with B on the x-axis. Let AB = a.
AP = x^2 + y^2. BP = (x-a)^2 + y^2
PB/PA (sorry I changed it) = q, so PB = PA*q
Let's see if this latex mess manages to work…
$$(x-a)^2 + y^2 = qx^2 + qy^2 = x^2 -2ax + a^2 + y^2
\\
(q-1)x^2 + 2ax + (q-1)y^2 = a^2
\\
x^2+\frac{2a}{q-1}x + y^2 = \frac{a^2}{q-1}
\\
x^2 + \frac{2a}{q-1}x + \frac{a^2}{(q-1)^2} + y^2 = \left(x + \frac{a}{q-1}\right)^2 + y^2 = \frac{a^2}{q-1} + \frac{a^2}{(q-1)^2} = \frac{(a^2)((q - 1) + 1)}{(q - 1)^2} = \frac{qa^2}{(q-1)^2}$$


This is an equation for a circle

292213 No.33

>>32
well it posted…don't know why it got scrunched up
Posting it math again.
$$(x-a)^2 + y^2 = qx^2 + qy^2 \\
=x^2 -2ax + a^2 + y^2
\\
(q-1)x^2 + 2ax + (q-1)y^2 = a^2
\\
x^2+\frac{2a}{q-1}x + y^2 = \frac{a^2}{q-1}
\\
x^2 + \frac{2a}{q-1}x + \frac{a^2}{(q-1)^2} + y^2\\
= \left(x + \frac{a}{q-1}\right)^2 + y^2 \\
= \frac{a^2}{q-1} + \frac{a^2}{(q-1)^2} \\
= \frac{(a^2)((q - 1) + 1)}{(q - 1)^2} \\
= \frac{qa^2}{(q-1)^2}$$
33 get

292213 No.34

>>33
fuck it… use ant vision or something

1b914a No.35

>>33
Now do it with a compass and straightedge.(Assume line segments of length 1 and q in the distance)

Try breaking it up into separate double-$ tags, anon
Post last edited at


File: 1425725511259.jpg (66.17 KB, 600x671, 600:671, 05321_FootballSHOT-600x.jpg)

cc863f No.26[Reply]

A rotating square table contains four pockets, one for each corner. Within each pocket, a shot glass lies either upright or upside down. The pockets are such that the shot glass orientation can only be determined by touch. The goal is have all the shot glasses in the same orientation.

One is allowed to flip the shot glasses in the following manner: First the table is rotated so that it is impossible to keep track of any particular pocket and stopped in a random orientation. Then, any two pockets are chosen for examination. These pockets are then examined by hand and shot glasses flipped as desired. After this is complete and the hands are out of the pockets, a bell is rung if and only if the goal is met. This entire procedure constitutes a 'move'.

Is there a procedure that guarantees that the goal is met in a finite number of moves?

40595c No.29

Turn 1:On the first spin, flip the opposite corners to orientation A (A or B for short).
Turn 2:Assuming the bell has not rung, you know that at least on glass is in B. If you choose two adjacent pockets and one glasses are not in A, by flipping it to A you know that at least 3 glasses are in A. If they are both in A, then there is one glass in B. If both are in B, flipping both means you won, since all glasses would be in A. You can't have two adjacent glasses in B, since one of the glasses we choose must be one of the glasses we flipped in the previous turn.
Turn 3: Look at two opposite corners. If one is in orientation B, you can make them all in A. If both are in A, both are adjacent to the glass in B. Choose one to be in B. This makes sure that you have to adjacent glasses in A and two adjacent glasses in B.
Turn 4: Choose two adjacent glasses.
If both are of the same orientaion, you know the other two glasses are both of the other orientation, so if you flip them you have won. If they are different, flip both glasses, so you have opposite corners having a different orientation, with adjacent ends having different corners.
Turn 5 (last one): Choose two opposite corners and flip both of them. You know the other corners are in the other orientation, so you have won.

You could probably do this in a shorter number of steps, but this is what I came up with. Nice puzzle


5f9a66 No.22[Reply]

This is a topic that always interested me. For instance, Mersenne primes, happy numbers, and so on. Post any unique number properties and give examples, I guess.

>Lucas-Carmichael Numbers
Take a number a and split it into its prime factors. An L-C Number is one in which the prime factors, when added to by 1, will multiply to give a+1

The smallest Lucas-Carmichael Number is 399.

399 = 3 /times 7 /times 19
400 = 4 /times 8 /times 20

5f9a66 No.23

>tfw can't use [tex] right

5f9a66 No.24

Liar Numbers are another I rather like.

This involves a test for prime numbers, which is as follows:
\[{a}^{p}-a = n, \, \textup{where} \, 1 < a \leq p\] will give a prime number if n is divisible by p for every value of a.

So, say, for instance, 5:

2^5 - 2 = 30, 30 is divisible by 5
3^5 - 3 = 240, 240 is divisible by 5
4^5 - 4 = 1020, 1020 is divisible by 5
5^5 - 5 = 3120, 3120 is divisible by 5

Thus, we know that 5 is definitely prime. Ideally, this test will prove and disprove any potential prime, but there are numbers that shouldn't pass this test, yet do, and these are liar numbers. The smallest number that satisfies this situation is 341, which is 11 * 31.


File: 1424520474795.png (16 KB, 360x480, 3:4, Tschirnhausen_cubic.svg.png)

6b659e No.21[Reply]

On the x-y plane, for a particular point (a,b), there exists a plane curve such that if a line is incident with (a, b) and a distinct point on the curve (p, q), the slope of the line = q. Find f(x,y) so that f(x,y) = 0 iff (x,y) is on the plane curve.

File: 1420013024869.png (4.72 KB, 797x91, 797:91, F_seq_def.png)

342359 No.5[Reply]

Given any the ordered integer pair(a,b), define the doubly infinite sequence F_{a,b} as shown in the picture. It may be the case that two distinct pairs (a,b) and (c,d) generate the same sequence with merely shifted indices(e.g. (1,1) and (3,5)). Find an expression containing a, b, c, and d that evaluates to 0 if and only if this happens.

Solution:
=:>W9\\m_XD:?WA:YW=?W9Z2Z3YA9:X\=?W9Z4Z5YA9:XX^=?WA9:XXjA9:lW`ZDBCEWdXX^a
1 post omitted. Click reply to view.

53ec2f No.7

>>6
Set theory, and maybe a healthy dose of either LSD or ska maria pastora.

342359 No.9

>>6
You don't need anything higher that pre-calc to get this.

342359 No.10

>>6
>>7
I should probably mention that everything before the first comma on the second line can be translated as 'For all integers'.

f9f366 No.12

>>9

Calc 3 + Intro Linear Algebra & PDEs. Didn't ever have to touch on any of this.

342359 No.13

>>12
The gap between you and the solution will not be filled with more formal math education. Take a look at the recurrence relation in the picture. Does it remind you of anything?


File: 1420218770674.jpg (14.35 KB, 300x173, 300:173, alien23.jpg)

6377fd No.8[Reply]

This looks like a fun board, so please don't be too offended by my drunken incursion.
However, I would argue that "vedic mathz" and other such shortcuts are directly relevant to the subject matter here.
>https://www.youtube.com/watch?v=kvLjpQ0XGao
>https://www.youtube.com/watch?v=grkWGeqW99c
As examples. Nothing amazing, just some kewl shortcuts for people who's calculator ran out of batteries.
Also, I'm not saying study of mathematics makes one insane, but eh, mathematicians tend to be batshit…
>http://www.ams.org/journals/tran/1969-137-00/S0002-9947-1969-0236393-5/S0002-9947-1969-0236393-5.pdf
So, via these premises, how do we be mathematicians without going completely batshit insane?

d92022 No.11

>>8
I'm partial to Russian Peasant Multiplication myself.
https://www.youtube.com/watch?v=qHXsKyVSPOU
Also, remember that this board does math just for fun.


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