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Bringing back the sudoku puzzles:

4				5	2
	5		1			3		7
		9
5					4		7
	3			1			8
	6		5					2
						7
2		1			6		9
			8	7				3

Edit:

Sudoku is good

In post 126, Mitillos wrote:

Spoiler: @Scigatt

I don't think your proof works. You appear to be proving some kind of converse to the statement? You end up with the assertion that the lines drawn are of equal length, but that's given by the problem statement.

Edit: Thinking about it some more.

Edit2: Yeah, I missed a couple of critical sentences there. It works.

Here's a more challenging sudoku I came up with that should be visually interesting as well:

6				8				4
	7		9		1		3
		8				5
	6		8		9		1
4				2				6
	2		6		7		4
		7				4
	5		3		4		2
1				7				9

Correct

Here's another fun sudoku one, and this one is part one of a two-parter:

The below is a hyper sudoku puzzle, also known as a "windoku" puzzle. The standard rules apply except with one additional restraint: four "window panes", partly overlapping each of the nine boxes, are shaded gray, and each pane must also have the digits from 1 to 9 exactly once. Here is the puzzle:

Part 1, as in my last two questions, is to solve the windoku puzzle. Part 2 will be posted when the correct solution is posted.

Very good Now...

Looks good to me

hmm
Ill try to keep alive this section

I guess a lot of different ppl come to this thread, so here is one constructive in nature that can (supposedly) be enjoyed by ppl with different degrees of knowledge; it also serves as basis to expand ur knowledge of math while we r at it (if u didnt know about it already)!
To solve this u will need basic algebra (think about working with the x and decomposing a number in prime factors) and know about sum and multiplication.

So first lets begin with an operation: .
We will write the module operation between 2 numbers x and y with the symbol % (x%y) and read it as "x module y".
The intuition behind module is the remainder of an integer division; 9%2=1, for example, because 9/2= 8/2+1/2; 8/2=4, but 1 is left as a remainder, as it can not be divided by 2 into an integer number.
Formally, we say that p%q=r if and only if there exists an integer n such as p=n*q+r, with r>=0 and r<q.

U can prove it if u want, but it is not necessary, that the operation is well defined and whatever formality nitpicks u have.

Modules generalize a lot of the concepts we work with, and we use that operation constantly, even if we dnt realize it. Want an example? Lets say it is 5:30 Pm; wich minute well have in 92 minutes? Those quick at calculating stuff will now it will be 7:02; it is 60+30+2 +the initial 30, rendering 60+60+2. The answer to wich minute is it is the answer to (30+92)%60; with hours, it works the same way. if it is 23:00, in 48 hours it will be 23:00, and in 5 hours it will be 4:00; that is a module by 24.

so, lets go to the challenges:


1. Prove that % is distributed over addition, that is:
(x+y)%z=(x%z+y%z)%z

Examples:
(5%3+11%3)%3=(2+2)%3=1=(5+11)%3

Hint:



2. Prove that the sum of 2 numbers is even if and only if both numers are even or both numbers are odd

Hint:



3. Prove that the same thing above occurs with multiplication:
(x*y)%z=(x%z*y%z)%z

Example:
(7%3)*(11%3)%3=(1*2)%3=2=(7*11)%3=77%3 (75=3*25)

Hint:



Now comes the FUN stuff! Well, if u r stilll with me it must mean u r having fun, right? Right? Well, if u r:
Remember those tricks we used when we were learning the multiplicaction table and what nots? now well prove some of them:

4. Prove that a number is a multiple of 2 if the last digit is a multiple of 2, and that a number is multiple of 5 if the last digit is 0 or 5.

Hint:



5. Prove that a number is multiple of 3 if the sum of its digits is a multiple of 3

Hint:



6. What is the last digit of 7^49?

Hint:



So there it is! The idea was to be doable by a lot of ppl (hopefully), and at the same time answering some of the mysteries we might have had on the school. Remember to spoiler ur answers, and leave more puzzles/exercises/challenges!

In post 138, word321 wrote:1. Prove that % is distributed over addition, that is:
(x+y)%z=(x%z+y%z)%z

I will note that #2 is much easier to do by directly applying the definitions of even and odd versus using the module operation.

the solution is correct!
and u r right, it is "easy" to prove 2 by hand; but I blv it is more of an issue with notation. The core principle of doing it directly and using the % notation is the same, calling the same thing with different names; but I left it there just to illustrate the logic to be used in further demostrations (so ppl can "adapt" to "adding things module something").

Let z₁ and z₂ be two complex numbers whose magnitudes are r₁ and r₂ respectively and whose phases are φ₁ and φ₂ respectively. Prove algebraically that z₁z₂ = r₁r₂ cis(φ₁ + φ₂), where cis(φ) = cos(φ) + i sin(φ).

I’d love to see if I remember how to do this, but as an electrical engineer I’m pretty sure I’d get forced out of this thread for my use of j.

I will understand you if you use j. Thanks for the heads up.

I mean if anyone else wants it go ahead, I’m only about a year and a half removed from my complex variables class...

In post 141, StrangerCoug wrote:Let z₁ and z₂ be two complex numbers whose magnitudes are r₁ and r₂ respectively and whose phases are φ₁ and φ₂ respectively. Prove algebraically that z₁z₂ = r₁r₂ cis(φ₁ + φ₂), where cis(φ) = cos(φ) + i sin(φ).

In post 141, StrangerCoug wrote:Let z₁ and z₂ be two complex numbers whose magnitudes are r₁ and r₂ respectively and whose phases are φ₁ and φ₂ respectively. Prove algebraically that z₁z₂ = r₁r₂ cis(φ₁ + φ₂), where cis(φ) = cos(φ) + i sin(φ).

In post 147, StrangerCoug wrote:

Spoiler: Response to Ircher's attempt

In post 146, Ircher wrote:= r1 * r2 * (cos(phi1) * cos(phi2) + i * cos(phi1) * sin(phi2) + i * cos(phi2) * sin(phi1) - sin(phi1) * sin(phi2)

You have an unmatched parenthesis on this line. Where should the other closing parenthesis go?

In post 148, Ircher wrote:

In post 147, StrangerCoug wrote:

Spoiler: Response to Ircher's attempt

In post 146, Ircher wrote:= r1 * r2 * (cos(phi1) * cos(phi2) + i * cos(phi1) * sin(phi2) + i * cos(phi2) * sin(phi1) - sin(phi1) * sin(phi2)

You have an unmatched parenthesis on this line. Where should the other closing parenthesis go?

Spoiler:

Ah, it should go at the end of the line, so revised proof:

Let z1 and z2 be complex numbers with magnitudes r1 and r2 and phases phi1 and phi2 respectively. Then (in polar form), z1 = r1(cos(phi1) + i * sin(phi1)) and z2 = r2(cos(phi2) + i * sin(phi2)). Then:

z1z2 = (r1*cos(phi1) + i * r1 * sin(phi1)) * (r2 * cos(phi2) + i * r2 * sin(phi2))
= r1 * r2 * cos(phi1) * cos(phi2) + i * r1 * r2 * cos(phi1) * sin(phi2) + i * r1 * r2 * sin(phi1) * cos(phi1) + i * i * r1 * r2 * sin(phi1) * sin(phi2)
= r1 * r2 * (cos(phi1) * cos(phi2) + i * cos(phi1) * sin(phi2) + i * cos(phi2) * sin(phi1) - sin(phi1) * sin(phi2))
= r1 * r2 * (cos(phi1 + phi2) + i * sin(phi1 + phi2))
= r1 * r2 * cis(phi1 + phi2).\box