Math and Logic Puzzles: Redux

[StrangerCoug]

Here's what should be easy ones, as was traditional for me last time:

You've been hired at the state prison and your job is to oversee the factory that makes the state's license plates. A new license plate series started on Monday, January 1, 2018. In the series, license plates have to be three letters followed by three digits. License plates cannot use the letters I or O, nor can the digits be all zeroes. The license plates are made in sequential order, with the very first plate made on the first day being AAA-001, the plate after that being AAA-002, and so on through AAA-999, with the plate following AAA-999 being AAB-001. Assume your prison makes exactly 25,000 plates per day Monday through Friday with no holiday breaks and that there are no other restrictions to the allowed plate numbers.

1. What was the number of the last plate made on January 5, 2018? What will be the number of the last plate made on December 31, 2018?
   
2. On what day will you make plate number ZZZ-999? Which plate that day will that be?*
   
3. Suppose a law is passed that, after all the available numbers as above are exhausted, the alphabetic and numeric portions change places so that after ZZZ-999 comes 001-AAA. The numeric portion is still the portion that is incremented for the next plate, with the rollover from 999 to 001 incrementing the alphabetic portion, so the sequence goes ..., ZZZ-999, 001-AAA, 002-AAA, ..., 999-AAA, 001-AAB, ... If ZZZ-999 and 001-AAA are made the same day and you still make 25,000 plates total that day, when will you make plate number 999-ZZZ?

* For clarity, I'm looking for an ordinal number to the second part of the question, such as "the 12,345th plate" (which is incorrect, by the way).

[StrangerCoug]

brassherald is closer to the answer I get for the last plate on January 5, 2018 than Harambey180 is.

[StrangerCoug]

I have the same answers as Harambey's new ones except for the date the prison makes 999-ZZZ. I'm at work on a short break, so I'm going to have a closer look at whether I'm the one who messed up after work.

[StrangerCoug]

The error in the last date was mine, so Harambey180 has corrected all his mistakes.

[StrangerCoug]

Here's a logic puzzle:

You're a detective that has caught a serial thief. The thief had stolen a different item (a car, a phone, a ring, a TV, a wallet, and a watch) from a different person (Bradley, Eric, Jessica, Michelle, Richard, and Vivienne) each day from Monday through Saturday before you finally caught him on Sunday, before he could steal again. Your job now is to determine not only what items were stolen from each person, but also on what day each item was stolen. Your notes on the case so far are as follows:

1. Bradley was targeted on Thursday, Jessica was targeted on Friday, and Richard was not targeted on Saturday.
2. Michelle had her wallet stolen at some point after the thief stole from Bradley.
3. The watch was stolen on Wednesday, which is after the phone was stolen.
4. The thief stole from Vivienne and Eric in some order on consecutive days.
5. The ring was stolen from a woman three days before the TV was stolen from a man.

Can you solve the case?

[StrangerCoug]

That is correct

[StrangerCoug]

Let's breathe some new life into this:

Prove that no refactorable number is perfect. (A refactorable number is a number that is divisible by the number of divisors it has, including 1 and itself.)

[StrangerCoug]

Looks good

[StrangerCoug]

Never mind; I think I had

too

easy a question Let me think of a new one...

[StrangerCoug]

Here's an interesting one in addition to #76 above for those who like proofs: Prove that there are only finitely many triangular numbers that are also tetrahedral numbers. How many are there, and what are they?

[StrangerCoug]

It only matters that both sides of the equation come out to the same number. You can use one variable on one side and another on the other.

[StrangerCoug]

For the ones who like puzzles, solve this sudoku:

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

Bonus points if you can prove you've done it manually without the help of a solver, since it's apparently really hard >=) (I created it, so I know the solution, but still...)

[StrangerCoug]

That is correct

[StrangerCoug]

(0!+0!+0!)! = 6
(1+1+1)! = 6
2+2+2 = 6
3*3-3 = 6
sqrt(4)+sqrt(4)+sqrt(4) = 6
5+5/5 = 6
6+6-6 = 6
7-7/7 = 6
cbrt(8)+cbrt(8)+cbrt(8) = 6
sqrt(9)*sqrt(9)-sqrt(9) = 6

[StrangerCoug]

Here's one I just found on YouTube:

ABC = A! + B! + C!, where ABC represents a number with A as the hundreds digit, B as the tens digit, and C as the ones digit. The digits may or may not repeat, but none of the digits are 0. Find the three-digit number ABC.

[StrangerCoug]

Looks good

[StrangerCoug]

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

[StrangerCoug]

Sudoku is good

[StrangerCoug]

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

[StrangerCoug]

Correct

[StrangerCoug]

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.

[StrangerCoug]

Very good Now...

Spoiler: Part two

Take the solution:

...and note that that not only do the four window panes contain every number from 1 to 9, but the cells colored pink, yellow, green, and blue in the solution grid as well as the cells left white below also contain every number from 1 to 9:


Prove that this is true in general for any hyper sudoku puzzle.

[StrangerCoug]

Looks good to me

[StrangerCoug]

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).

[StrangerCoug]

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

[StrangerCoug]

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).

Spoiler: Hint

How do you convert between the rectangular form and the polar form of a complex number?

[StrangerCoug]

Spoiler: Response to Ircher's attempt

= 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?

[StrangerCoug]

Spoiler: Response to Ircher's attempt

= 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

Spoiler:

Looks good

[StrangerCoug]

Still waking up, but I believe it's 1/4. Will write out a full proof later if someone hasn't by then.

[StrangerCoug]

Oh, yeah, missed that.

[StrangerCoug]

Here's one to breathe some life in this:

Prove that the cube of any integer ending in 5 is always an odd multiple of 125.

[StrangerCoug]

Looks good

[StrangerCoug]

Spoiler: alternatively

any number ending in 5 is an odd multiple of 5 and can be expressed as 5n where n is an odd integer. cubed, this becomes 125n³. as the product of odd numbers, this is also odd.

That also works

[StrangerCoug]

2/3

[StrangerCoug]

0.99999

[StrangerCoug]

I should probably come up with a good logic puzzle, come to think of it. Hmm...

[StrangerCoug]

Here's an easy sudoku puzzle I created:

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

Edit:

Sudoku corrected; three of the numbers were one cell too far to the right.

What lives in the woods, is brown, and is made of concrete?

A cabin?

[StrangerCoug]

If you prefer a little more of a challenge, I've got this sudoku, too:

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

[StrangerCoug]

Any takers for the Sudoku puzzles? I had a solution to the first one DM'd to me on Discord a couple nights ago, but I can't officially accept it unless and until it's posted here.

[StrangerCoug]

Spoiler: This look right to you?

That is correct

The sudoku in #213 is still available to solve.

[StrangerCoug]

Or, if you want a harder puzzle than the one in #213, here is a Sudoku X puzzle. On top of the normal rules, each corner-to-corner diagonal must also contain each number from 1 to 9 exactly once.

[StrangerCoug]

If you prefer a little more of a challenge, I've got this sudoku, too:

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

Spoiler:

Code: Select all

921348657
467925813
358716294
296587341
175463928
843291765
532674189
684139572
719852436

Looks good to me

[StrangerCoug]

Or, if you want a harder puzzle than the one in #213, here is a Sudoku X puzzle. On top of the normal rules, each corner-to-corner diagonal must also contain each number from 1 to 9 exactly once.

This puzzle turns not to have a unique solution (not sure why SudokuWiki didn't catch this and gave it a grade anyway), so I'll give you two more clues from the intended solution:

Should still be plenty hard, though

[StrangerCoug]

I think it's fair to just put up another puzzle.

[StrangerCoug]

After work

[StrangerCoug]

No, I was busy watching election results.

[StrangerCoug]

Here in Texas it largely went as expected.