Math and Logic Puzzles: Redux

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Post Post #18 (isolation #0) » Mon Apr 30, 2018 3:01 pm

Post by 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).
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Post Post #22 (isolation #1) » Tue May 01, 2018 3:46 am

Post by StrangerCoug »

brassherald is closer to the answer I get for the last plate on January 5, 2018 than Harambey180 is.
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Post Post #28 (isolation #2) » Tue May 01, 2018 7:16 am

Post by 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.
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Post Post #29 (isolation #3) » Tue May 01, 2018 9:23 am

Post by StrangerCoug »

The error in the last date was mine, so Harambey180 has corrected all his mistakes.
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Post Post #30 (isolation #4) » Wed May 02, 2018 1:01 am

Post by 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?
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Post Post #32 (isolation #5) » Wed May 02, 2018 3:48 am

Post by StrangerCoug »

That is correct :D
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Post Post #42 (isolation #6) » Mon May 21, 2018 4:16 am

Post by 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.)
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Post Post #44 (isolation #7) » Mon May 21, 2018 6:32 am

Post by StrangerCoug »

Looks good :)
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Post Post #61 (isolation #8) » Fri Jun 29, 2018 7:00 am

Post by StrangerCoug »

Never mind; I think I had
too
easy a question :P Let me think of a new one...
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Post Post #81 (isolation #9) » Sun Jul 28, 2019 5:46 am

Post by 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?
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Post Post #88 (isolation #10) » Sun Jul 28, 2019 12:21 pm

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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.
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Post Post #105 (isolation #11) » Tue Oct 29, 2019 3:20 pm

Post by StrangerCoug »

For the ones who like puzzles, solve this sudoku:
9236
651
5678
894
39
457
4593
927
3854

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...)
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Post Post #107 (isolation #12) » Thu Oct 31, 2019 6:04 am

Post by StrangerCoug »

That is correct :D
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Post Post #113 (isolation #13) » Mon Jan 20, 2020 3:12 pm

Post by 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
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Post Post #116 (isolation #14) » Mon Feb 03, 2020 8:13 am

Post by 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.
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Post Post #118 (isolation #15) » Mon Feb 03, 2020 8:47 am

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Looks good :)
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Post Post #125 (isolation #16) » Sun May 24, 2020 8:19 am

Post by StrangerCoug »

Bringing back the sudoku puzzles:
452
5137
9
547
318
652
7
2169
873
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Post Post #127 (isolation #17) » Sun May 24, 2020 8:51 am

Post by StrangerCoug »

Sudoku is good :)
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Post Post #130 (isolation #18) » Sun May 24, 2020 11:14 am

Post by StrangerCoug »

Here's a more challenging sudoku I came up with that should be visually interesting as well:
684
7913
85
6891
426
2674
74
5342
179
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Post Post #132 (isolation #19) » Sun May 24, 2020 12:21 pm

Post by StrangerCoug »

Correct :D
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Post Post #133 (isolation #20) » Mon May 25, 2020 11:07 am

Post by 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:
Image
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.
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Post Post #135 (isolation #21) » Mon May 25, 2020 5:53 pm

Post by StrangerCoug »

Very good :) Now...

Spoiler: Part two
Take the solution:
Image
...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:
Image

Prove that this is true in general for any hyper sudoku puzzle.
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Post Post #137 (isolation #22) » Tue May 26, 2020 4:03 am

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Looks good to me :)
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Post Post #141 (isolation #23) » Fri Sep 18, 2020 4:03 am

Post by 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).
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Post Post #143 (isolation #24) » Fri Sep 18, 2020 5:21 am

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I will understand you if you use j. Thanks for the heads up.
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Post Post #145 (isolation #25) » Sat Sep 26, 2020 5:45 am

Post by StrangerCoug »

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).
Spoiler: Hint
How do you convert between the rectangular form and the polar form of a complex number?
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Post Post #147 (isolation #26) » Fri Oct 30, 2020 1:11 pm

Post by StrangerCoug »

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?
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Post Post #149 (isolation #27) » Fri Oct 30, 2020 4:18 pm

Post by StrangerCoug »

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
Spoiler:
Looks good :)
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Post Post #156 (isolation #28) » Wed Dec 09, 2020 2:27 am

Post by StrangerCoug »

Still waking up, but I believe it's 1/4. Will write out a full proof later if someone hasn't by then.
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Post Post #158 (isolation #29) » Wed Dec 09, 2020 4:24 am

Post by StrangerCoug »

Oh, yeah, missed that.
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Post Post #173 (isolation #30) » Fri Dec 24, 2021 10:43 am

Post by 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.
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Post Post #175 (isolation #31) » Fri Dec 24, 2021 2:26 pm

Post by StrangerCoug »

Looks good :)
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Post Post #177 (isolation #32) » Fri Dec 24, 2021 7:22 pm

Post by StrangerCoug »

In post 176, tris wrote:
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 :)
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Post Post #181 (isolation #33) » Mon Feb 14, 2022 3:35 pm

Post by StrangerCoug »

2/3
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Post Post #205 (isolation #34) » Wed Feb 16, 2022 2:11 am

Post by StrangerCoug »

0.99999
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Post Post #209 (isolation #35) » Fri Mar 04, 2022 4:13 pm

Post by StrangerCoug »

I should probably come up with a good logic puzzle, come to think of it. Hmm...
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Post Post #211 (isolation #36) » Fri Mar 04, 2022 5:07 pm

Post by StrangerCoug »

Here's an easy sudoku puzzle I created:

6739
362
951
64
73
14
812
723
1534

Edit:
Sudoku corrected; three of the numbers were one cell too far to the right.
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A cabin?
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Post Post #213 (isolation #37) » Fri Mar 04, 2022 5:47 pm

Post by StrangerCoug »

If you prefer a little more of a challenge, I've got this sudoku, too:
185
9583
819
29
6
65
371
6419
184
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Post Post #214 (isolation #38) » Sun Mar 06, 2022 3:48 am

Post by 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.
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Post Post #216 (isolation #39) » Sat Mar 12, 2022 3:51 am

Post by StrangerCoug »

In post 215, Kcdaspot wrote:
Spoiler: This look right to you?
Image
That is correct :D

The sudoku in #213 is still available to solve.
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Post Post #217 (isolation #40) » Sat Mar 12, 2022 9:06 am

Post by 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.
Image
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Post Post #219 (isolation #41) » Mon Mar 14, 2022 10:27 am

Post by StrangerCoug »

In post 218, Charles510 wrote:
In post 213, StrangerCoug wrote:If you prefer a little more of a challenge, I've got this sudoku, too:
185
9583
819
29
6
65
371
6419
184
Spoiler:

Code: Select all

921348657
467925813
358716294
296587341
175463928
843291765
532674189
684139572
719852436
Looks good to me :)
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Post Post #220 (isolation #42) » Sun Mar 20, 2022 10:19 am

Post by StrangerCoug »

In post 217, StrangerCoug wrote: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.
Image
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:
Image
Should still be plenty hard, though ;)
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Post Post #222 (isolation #43) » Tue Nov 08, 2022 7:01 am

Post by StrangerCoug »

I think it's fair to just put up another puzzle.
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Post Post #224 (isolation #44) » Tue Nov 08, 2022 7:09 am

Post by StrangerCoug »

After work :P
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Post Post #226 (isolation #45) » Wed Nov 09, 2022 2:23 am

Post by StrangerCoug »

No, I was busy watching election results.
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Post Post #228 (isolation #46) » Wed Nov 09, 2022 5:41 am

Post by StrangerCoug »

Here in Texas it largely went as expected.
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Post Post #282 (isolation #47) » Sat Apr 29, 2023 4:26 pm

Post by StrangerCoug »

Here come the sudoku puzzles again, and this one should be easy:

896
146
5634
537
48
135
6395
314
261
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Post Post #284 (isolation #48) » Sun Apr 30, 2023 8:04 am

Post by StrangerCoug »

In post 283, biancospino wrote:
Spoiler:


842967153
317458692
596213784
268149537
479635218
135782469
681374925
723591846
954826371
That is correct :D
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Post Post #285 (isolation #49) » Sun Apr 30, 2023 10:57 am

Post by StrangerCoug »

Another one that should be easy:
175
716
3829
3251
5642
4653
835
218

Yes, the middle row, column, and region are all supposed to be empty :P
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Post Post #286 (isolation #50) » Sun Apr 30, 2023 11:44 am

Post by StrangerCoug »

Or, if you prefer a different kind of puzzle, have a cryptoquote to solve:

AQUM ITW PMTOJ JN FM AQXMK, TOK ZVYQNWQJG PVWJ FM CMBJ TAQXM. NOM PVWJ OMXMY, UNY ISTJMXMY YMTWNO, JVYO SQW FTZC NO AQUM. —MAMTONY YNNWMXMAJ
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Post Post #288 (isolation #51) » Sun May 21, 2023 11:30 am

Post by StrangerCoug »

In post 287, Aisa wrote:
Spoiler: sudoku
2175
7162
368291
3251
1725
56142
149685273
6783215
523178

I will continue to edit this until either I finish it or I get home

I got home first. Maybe I will finish this tomorrow. Maybe it will be incomplete forever.
Spoiler:
What you have so far is correct. Keep going :)
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Post Post #290 (isolation #52) » Mon Jul 17, 2023 8:16 am

Post by StrangerCoug »

In post 289, Something_Smart wrote:
In post 286, StrangerCoug wrote: Or, if you prefer a different kind of puzzle, have a cryptoquote to solve:

AQUM ITW PMTOJ JN FM AQXMK, TOK ZVYQNWQJG PVWJ FM CMBJ TAQXM. NOM PVWJ OMXMY, UNY ISTJMXMY YMTWNO, JVYO SQW FTZC NO AQUM. —MAMTONY YNNWMXMAJ
Spoiler:
Life was meant to be lived, and curiosity must be kept alive. One must never, for whatever reason, turn his back on life. -Eleanor Roosevelt

I don't know much about cryptogram techniques, but with this one I guessed that "JN FM" was "TO BE", then saw that would make the author's name "E_E..." which immediately broke the whole thing open.
That is correct :D
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