/puzzle/ - Puzzles and recreational math

You have 6 bags and 31 doo-dads. Arrange it so that each bag has an odd number of doo-dads in it.

Testing math tags.

[math]F(\omega )\; = \; \frac{1}{\sqrt{2\pi }}\int_{-\infty }^{\infty}\;f(t)\, e^{-i\omega t}\;dt[/math]

$$F(\omega )\; = \; \frac{1}{\sqrt{2\pi }}\int_{-\infty }^{\infty}\;f(t)\, e^{-i\omega t}\;dt$$

NEW OWNER EMAIL IS puzzleOwn@8chan.co

Do you know the exact alloy you are using? We need to know the density of the alloy to get going.

>>143

7850 kg/m^3

>>144

For a solid cylindrical 20kg weight, the thickness is 16.11 mm. If you punch out a circular hole 159mm in diameter off of the centre hole, the weight is 2.5 kg lighter. If you add a 1.5 thickness outer flange down to a diameter of 319mm, the disk is 5 kg heavier.

>>135

I give up. Answer please.

I can tell you that any two cow's weight must be congruent mod 4, but that's as far as I can get.

Okay, I finally got the answer. This was really hard, where did you get it?

Also you don't need an equal number of cows in each herd.

Zero

>>66

This sounds like homework. Read the rules. (If not it's still too easy for a puzzle.)

>>124

Notation: Given an arbitrary projectivity, its name can be its expression in half-perspectivities(e.g. A<b>C<d) or a string of letters enclosed in vertical bars (e.g. |A|, |def|). Given compatible projectivities |a| and |b|, their composition is |ab|. Additionally, the domain and co-domain forms of a projectivity may be explicitly noted by combining the two notations (e.g. A>|Cb|>d). Given a perspectivity |D|, |/D| is the name of its inverse and can be incorporated into previous notation(e.g. |F/G|, A<|/hj|>B, |C/(ab)D|). Given a line a in the domain of a projectivity |B|, |B|(a) is its image. The identity perspectivity which leaves all members of a form fixed is named |1|.

Also, ends of proofs will be noted by hashtags(#).

——————————–

In the last few posts, we introduced a prospective new axiom, but left open questions about the consistency of this Axiom P with our previous axioms. The goal of this instalment is to answer this question, but we first need a few more theorems in our belt.

One question we can tackle right now is how Axioms P and D fit together. If we look at their corresponding perspectivity theorems, the UPT(and Theorem 4, >>117) implies that a particular point has a special property with respect to an applicable projectivity, while the SPT(and Theorem 3, >>117) only say that some point within a particular range has the corresponding property. Intuitively, one might suggest that the UPT could be 'stronger' than the SPT. This is, in fact, the case.

>>128

>>129

The corollaries below immediately follow from the proof:

Corollary 1.1(Theorem of Hessenberg):Axiom P implies Axiom D

Corollary 1.2(Perspectivity Theorem(PeT))(P):Given a projectivity a>|E|<c with a and c distinct, if ac is self-corresponding under |E|, then there exists a point E such that a>|E|<c = a>E<c.

Proof: Starting with a>|E|<c, use the 2-Perspectivity Theorem to reduce to an identical projectivity a>P<b>Q<c. Since ac is self-corresponding under the projectivity, Theorem 2, >>117 implies that either the SPT or the UPT can be applied for a final reduction to an identical perspectivity a>E<c.#

Corollary 1.3:Given the plane axioms alone, the Perspectivity Theorem implies Axiom P.

Proof: The UPT, equivalent to Axiom P, is a special case of the Perspectivity Theorem.#

Other important theorems follow relatively easily as well.

Theorem 2(Projectivity Theorem(PrT))(P): Given two projectivities b>|D|<c and b>|E|<c, if there are three distinct points G, H, and I on b whose images are the same under |D| and |E|, |D| and |E| are identical.

Proof: Call the images of G, H, and I under |D| and |E| G', H', and I' respectively. Let {A, B, C, D} be a set of distinct points such that no three are collinear.

By Theorem 2, >>99, there exist projectivities |C| and |F| such a(ac, C, D)>|C|<b(G, H, I) and c(G', H', I')>|F|<d(ac, A, B). Composing, we have a(ac, C, D)>|CDF|<d(ac, A, B) and a(ac, C, D)>|CEF|&Post too long. Click here to view the full text.

>>130

Theorem 4: Given the plane axioms alone, the Three-Fix Theorem implies Axiom P

Proof: Let a>|M|<b and a>|N|<b be two projectivities to which the PrT would apply and let P, Q, and R be the required distinct points on a with P', Q', and R' their respective images.

Assume that |M| and |N| are not identical. Therefore, there exists a point X such that |M|(X) = Y and |N|(X) = Y' distinct from Y. Now consider the projectivity |/MN|. The points P', Q', and R' are self-corresponding under this composition, so by the Three-Fix Theorem, |/MN| = |1|. However, |/MN|(Y) = Y' in contradiction to previous statement. Therefore, |M| and |N| are identical, proving the PrT, which was already proven equivalent to Axiom P.#

We are are now finally ready to tackle the issue of the consistency of Axiom P. In particular, we will exhibit a particular plane in which the Three-Fix Theorem must hold.

Theorem 5: Given a plane with exactly 4 distinct points in each range, the Three-Fix Theorem holds.

Proof: Say n>|A|<n is a projectivity on a line n in the given plane such that the distinct points P, Q, R, are self-corresponding. What we require is that any other distinct point on n is self corresponding as well. However, there is only one other distinct point S on n. Since |A| is a bijection and all the other points are accounted for, S is forced to be self-corresponding, as required.#

Corollary 5.1: Given a plane with exactly 4 distinct points in each range, Axiom P holds.

Similarly,

Theorem 6: Given a plane with exactly three distinct points in each range, the Three-Fix Theorem holds.

The existence of the required planes is guaranteed by the PDSs {0, 1, 3}(mod 7) and {0, 1, 3, 9}(mod 13). Assuming the consistency of the natural numbers, we have now shown Axiom P is consistent with thePost too long. Click here to view the full text.

>>85

>>131

Rewrite of the posts is occurring at >>>/sci/3170

>>46

link pls

>>47

when I was into it, I managed to get my 3x3x3 averages down to lower 30s with roux method. I still play around with it but mostly for FMC stuff.

>>20

sort of same for me, it was fucking up my wrists pretty badly also

>>48

I should mention that this video is rather old. At the time, it was believed that it was best practice to always start your cross on the same side. Nowadays I believe that being able construct the cross an any side is considered better practice.

Not so much a website as a nice river crossing puzzle game.

http://www.digyourowngrave.com/cross-the-river-puzzle/

When one stands facing North, West is on the left and East is on the right.

But, when one stands facing South, the East is on the left and the West is on the right.

Left and right are directions that are oriented by one's body.

>>69

On the Earth, how does one define north vs. south anyways? Of the two poles, how did we decide to call one the north?

>>70

I mean, east and west can be defined be where the sun/stars rise and set, but what about north and south?

You could use the fact that you share a language to explain direction. If, for example, they understand written english, you could say them that they're reading from left to right.

>>83

>excepting that related to handedness.

That includes standard direction of reading, etc.

That's the answer I got when I was setting out the problem.