Math and Logic Puzzles

StrangerCoug · Post Post #0 (isolation #0) » Sat Jul 09, 2011 6:21 am

The rules of this game are very simple. The puzzle maker (for the first puzzle, I) will present a question that can be solved by math or logic.

Each player gets one chance and one chance only to solve the puzzle. If you get it right, you become the next puzzle maker. If three people miss, the puzzle maker wins and posts a new puzzle.

Puzzle makers: Try to make each puzzle have a unique solution. Chess puzzles in which making a different move has no real effect on the outcome (e.g. it doesn't matter if a king on g1 moves to f1 or h1) are fine, as are "fill in the grid" puzzles where a different orientation is still correct, but you shouldn't be making puzzles where making major differences from what you have still results in a correct answer.

Do not post a puzzle if it is not your turn unless the person who won says it is OK. (I'll allow exceptions for if it takes an unreasonably long time to put up a puzzle, but "an unreasonably long time" has yet to be defined.)

(

Edit 4/21/2014:

The rules stricken out above have not been enforced in a long time. It's more laid back these days, though I'd still like only one person asking a question at a time.)

I'll start with a math puzzle.

Timmy, who is a gambling addict, enters a casino with $50 at 7:00 PM. He loses $20 in his first five minutes, but in the next five minutes he is able to make $15 of it back. He loses $20 more in his third five minutes, wins $15 back in his fourth five minutes, and so on in that cycle. At what time will he run out of money?

StrangerCoug · Post Post #2 (isolation #1) » Sat Jul 09, 2011 6:27 am

Kcdaspot wrote:5 dollars gone in 10 minutes

15 in 30

25 in 50

50 in 100.

100 = 8:40 pm.

Wrong.

StrangerCoug · Post Post #4 (isolation #2) » Sat Jul 09, 2011 6:32 am

You only get one shot a puzzle, Kcdaspot.

StrangerCoug · Post Post #8 (isolation #3) » Sat Jul 09, 2011 6:34 am

DonJosh wrote:DAMMIT kcda ninja'd me

No he did not—he made a second guess and he's only allowed one, so I cannot judge the second one. 8:05 PM is correct and you get to make the next puzzle!

StrangerCoug · Post Post #32 (isolation #4) » Sat Jul 09, 2011 9:33 am

Which way will somebody from the other city say leads to the City of Truth?

StrangerCoug · Post Post #34 (isolation #5) » Sat Jul 09, 2011 9:39 am

And go the other way as before, yes?

StrangerCoug · Post Post #36 (isolation #6) » Sat Jul 09, 2011 10:14 am

Give me a moment here; I'm trying to make an original one.

StrangerCoug · Post Post #38 (isolation #7) » Sat Jul 09, 2011 10:29 am

It's taking awhile, so another Knights and Knaves type problem.

In addition to the City of Truth and City of Lies, there is also the City of Confusion, where whether people tell the truth or lies is up to its residents. You are talking to Alice, Bob, and Charlie, who is from a different city than the other two.

Alice tells you, "I'm from the City of Confusion."
Bob responds in anger, "Are you out of your mind!? It's Charlie that's from the City of Confusion!"
Charlie nods and states, "Alice, you're from the City of Truth, remember?"

Who's from where?

StrangerCoug · Post Post #41 (isolation #8) » Sat Jul 09, 2011 10:41 am

Tragedy wrote:Alice is from the City of Lies.
Bob is from the City of Truth.
Charlie is from the City of Confusion.

Yes.

StrangerCoug · Post Post #46 (isolation #9) » Sat Jul 09, 2011 11:01 am

A last name?

StrangerCoug · Post Post #52 (isolation #10) » Sat Jul 09, 2011 11:07 am

It did take awhile, but the length of the last names in the last two clues was a big hint.

Give me a moment...

StrangerCoug · Post Post #54 (isolation #11) » Sat Jul 09, 2011 11:20 am

Solve this Sudoku puzzle.

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

StrangerCoug · Post Post #63 (isolation #12) » Sat Jul 09, 2011 1:24 pm

If you need help solving it, the PDF file I got this from suggests doing these ten squares first in order:

The center left square of the top right cell
The bottom center square of the top left cell
The center square of the center cell
The center left square of the center left cell
The bottom left square of the center cell
The top left square of the center cell
The bottom center square of the bottom center cell
The center square of the bottom left cell
The top center square of the top right cell
The bottom center square of the center left cell

StrangerCoug · Post Post #68 (isolation #13) » Sat Jul 09, 2011 1:37 pm

yabbaguy, it'd be nice if you let the previous puzzle maker judge your answer before you post a new question. You're right, though.

StrangerCoug · Post Post #71 (isolation #14) » Sat Jul 09, 2011 2:20 pm

Alfred is the insane cop, Bob is the sane cop, Chris is the Mafia goon, David is the paranoid cop, and Eliot is the naïve cop.

StrangerCoug · Post Post #73 (isolation #15) » Sat Jul 09, 2011 2:41 pm

If I draw two cards from a standard deck of 52 without replacing them, what are the chances that they'll both be aces?

StrangerCoug · Post Post #75 (isolation #16) » Sat Jul 09, 2011 2:46 pm

Wrong.

StrangerCoug · Post Post #78 (isolation #17) » Sat Jul 09, 2011 3:15 pm

That's right!

StrangerCoug · Post Post #81 (isolation #18) » Sat Jul 09, 2011 3:39 pm

Five-card poker hands, I presume?

StrangerCoug · Post Post #83 (isolation #19) » Sun Jul 10, 2011 4:47 am

I've got the number of different poker hands

TOTAL

, but I'm stuck working much anything else out without cheating.

StrangerCoug · Post Post #86 (isolation #20) » Sun Jul 10, 2011 5:47 am

I can tell that's wide of the mark since you have an eight-digit answer and drawing five from the full 52 only allows for a little under 2,600,000 hands...

StrangerCoug · Post Post #88 (isolation #21) » Sun Jul 10, 2011 5:55 am

Doing it that way makes the order of the cards in the hand matter, which they don't in poker. 10S JS QS KS AS is the same as AS KS QS JS 10S, which is the same as QS AS JS 9S KS. You want combinations, not permutations.

StrangerCoug · Post Post #90 (isolation #22) » Sun Jul 10, 2011 6:05 am

29,988,000 / 5 = 5,997,600 > 2,600,000, so 5,997,600 can't be the answer either.

StrangerCoug · Post Post #91 (isolation #23) » Sun Jul 10, 2011 6:19 am

Just for clarification, PersonalIdiot4: Threes and fours of a kind and full houses count as multiple pairs, yes?

StrangerCoug · Post Post #108 (isolation #24) » Sun Jul 10, 2011 8:22 am

20 pieces of gold apiece?

StrangerCoug · Post Post #109 (isolation #25) » Sun Jul 10, 2011 8:23 am

Damn it, I just realized I outsmarted myself ><

StrangerCoug · Post Post #111 (isolation #26) » Sun Jul 10, 2011 8:26 am

34 to A, 33 to B, 33 to C, 0 to D, and 0 to E?

StrangerCoug · Post Post #113 (isolation #27) » Sun Jul 10, 2011 8:27 am

That's actually

FOUR

incorrect answers if I'm counting right, so you've beaten us. What was it?

StrangerCoug · Post Post #117 (isolation #28) » Sun Jul 10, 2011 8:40 am

StrangerCoug wrote:Each player gets one chance and one chance only to solve the puzzle. If you get it right, you become the next puzzle maker. If three people miss, the puzzle maker wins and posts a new puzzle.

KageLord wrote:
Empking wrote:A,B,C get 33
D gets 0
E gets one?

Nope.

KageLord wrote:
Kcdaspot wrote:A gets 0

BCDE get 25 a peice.

No.

Tragedy wrote:
I think it's 45?

10 for A,
9 for B
8 for C
7 for D
6 for E

Well... there are 100 coins total... but that scheme is wrong anyway.

Three misses, so KageLord couldn't even accept my answer.

StrangerCoug · Post Post #119 (isolation #29) » Sun Jul 10, 2011 8:45 am

Yeah.

StrangerCoug · Post Post #124 (isolation #30) » Sun Jul 10, 2011 9:06 am

You should switch; door #2 has two chances in three of having the car as opposed to door #1, which has only one chance in three.

StrangerCoug · Post Post #128 (isolation #31) » Sun Jul 10, 2011 9:13 am

Swap it, I believe.

As for the official puzzle, this is my first attempt at writing my own logic grid puzzle and I had a computer program to help me. Tell me how easy/hard you find this, OK?

It looks like the economy is turning back around: I just got word that six of my friends (Alexandria, Elizabeth, Jack, Millie, Nathaniel, and Oliver—I can't remember who has what last name, but the last name of one of them is Stocker) got hired yesterday! They are each of a different age from 18 to 23 and got hired at a different location. One of the places that hired one of my friends is The Great American Pizza Parlor, which is my favorite place to eat. They have such good food... now I'm hungry! Oh—you want to go? Well, I'll make you a deal: I'll take you if you can tell me my friend's full names, how old they are, and where they work. OK?

CLUES

Del Río, who is 23, works at a restaurant.
Elizabeth, who is older than the person who works at Discount Supermarkets, works at Burnie's Burgers.
Jack, who is 22, works at The Five O'Clock Bar.
Copperfield works at Jerry's Clothing Store.
The person who works at Vinnie's Videos, who is not Oliver, is 18.
The Five O'Clock Bar requires employees to be 21 or older, so White can't work there.
Millie, who does not work at Discount Supermarkets, is two years younger than Jefferson, who in turn is two years younger than Alexandria.
Harrington is younger than White.

StrangerCoug · Post Post #136 (isolation #32) » Sun Jul 10, 2011 10:42 am

KageLord is indeed correct with my puzzle.

StrangerCoug · Post Post #137 (isolation #33) » Sun Jul 10, 2011 11:11 am

I just realized that the first six clues in my logic puzzle go in alphabetical order by first name as to whom they describe. I should randomize the order of my clues for next time.

StrangerCoug · Post Post #154 (isolation #34) » Sun Jul 10, 2011 1:32 pm

24 * 4 * 4 * 1 = 384?

StrangerCoug · Post Post #178 (isolation #35) » Mon Jul 11, 2011 11:36 am

I'm having trouble understanding your question, PersonalIdiot04...

StrangerCoug · Post Post #193 (isolation #36) » Tue Jul 12, 2011 11:41 am

I say 5 sqrt(2) actually.

StrangerCoug · Post Post #201 (isolation #37) » Tue Jul 12, 2011 12:57 pm

Anyway, you're still up.

StrangerCoug · Post Post #233 (isolation #38) » Wed Jul 13, 2011 11:38 am

Going in the first post as a clarification: Do not post a puzzle if it is not your turn unless the person who won says it is OK. (I'll allow exceptions for if it takes an unreasonably long time to put up a puzzle, but "an unreasonably long time" has yet to be defined.) I know that, in theory, one person could be the puzzle maker forever, but in practice, somebody somewhere knows or can figure it out.

StrangerCoug · Post Post #236 (isolation #39) » Wed Jul 13, 2011 11:45 am

OK, then:

If you were to add up all the numbers of each row of Pascal's Triangle, what pattern would result?

StrangerCoug · Post Post #238 (isolation #40) » Wed Jul 13, 2011 12:58 pm

yabbaguy wrote:A 2^n series. The nth row's sum is 2^n, that is, where the first row is designated 0.

That is correct!

StrangerCoug · Post Post #253 (isolation #41) » Thu Jul 14, 2011 10:12 am

They're the same height.

StrangerCoug · Post Post #303 (isolation #42) » Sun Jul 17, 2011 10:07 am

Zero?

StrangerCoug · Post Post #389 (isolation #43) » Tue Jul 19, 2011 11:16 am

The stressed syllables are all a possible pronunciation of the letter A?

StrangerCoug · Post Post #391 (isolation #44) » Tue Jul 19, 2011 11:19 am

Heh. I thought "abnegate" wasn't a word, but Firefox's spell check isn't flagging it.

StrangerCoug · Post Post #393 (isolation #45) » Tue Jul 19, 2011 11:30 am

If you're declared correct, then you can pass it to me.

StrangerCoug · Post Post #396 (isolation #46) » Tue Jul 19, 2011 1:09 pm

Arrange these 10 cards:

...into the empty spaces of this grid so that no two cards in the same row, column, or corner-to-corner diagonal are the same suit or rank:

StrangerCoug · Post Post #397 (isolation #47) » Tue Jul 19, 2011 1:14 pm

I just realized that the two unused suits can be swapped, so I'm going to add something to make sure there's only one solution: The bottom left corner has a club.

StrangerCoug · Post Post #400 (isolation #48) » Tue Jul 19, 2011 1:53 pm

That's right!

StrangerCoug · Post Post #409 (isolation #49) » Wed Jul 20, 2011 12:17 pm

Do not post a puzzle if it is not your turn unless the person who won says it is OK. (I'll allow exceptions for if it takes an unreasonably long time to put up a puzzle, but "an unreasonably long time" has yet to be defined.)

StrangerCoug · Post Post #412 (isolation #50) » Wed Jul 20, 2011 12:20 pm

DonJosh gave you the all-clear, so post what you have.

StrangerCoug · Post Post #514 (isolation #51) » Sat Jul 23, 2011 3:24 pm

JDGA wrote:How does jester's solution not work? :/

yabbaguy wrote:In the prison is a switch room, which contains two light switches labeled A and B, each of which can be in either the 'On' or the 'Off' position. I am not telling you their present positions.

I think we've surpassed three already, but I don't know how well that's being enforced.

StrangerCoug · Post Post #517 (isolation #52) » Sat Jul 23, 2011 3:33 pm

If it's about [O]PTIMIZE, I thought a reply wasn't necessary, but as I said, I

REALLY

don't want to overcommit. If you don't have someone more qualified, I can take it.

StrangerCoug · Post Post #553 (isolation #53) » Tue Jul 26, 2011 11:49 am

But I had one ready

StrangerCoug · Post Post #598 (isolation #54) » Wed Jul 27, 2011 2:27 pm

Nine, I think.

StrangerCoug · Post Post #601 (isolation #55) » Wed Jul 27, 2011 2:32 pm

That overruns the three-guess limit.

StrangerCoug · Post Post #603 (isolation #56) » Wed Jul 27, 2011 2:39 pm

Go ahead and give it.

StrangerCoug · Post Post #606 (isolation #57) » Wed Jul 27, 2011 2:46 pm

One off. Damn it.

Put up another, then.

StrangerCoug · Post Post #609 (isolation #58) » Wed Jul 27, 2011 2:51 pm

Clarification questions:

1.) Are the hats placed on their heads before or after they are lined up?
2.) Are the hat colors determined independently?

StrangerCoug · Post Post #641 (isolation #59) » Thu Jul 28, 2011 11:31 am

You're only allowed one shot, Max.

I think I have the way to compute this down pat; I just need the odds of each win.

StrangerCoug · Post Post #644 (isolation #60) » Thu Jul 28, 2011 11:37 am

I know.

StrangerCoug · Post Post #646 (isolation #61) » Thu Jul 28, 2011 11:57 am

The reason I disallow multiple guesses is to allow other people a chance at it.

StrangerCoug · Post Post #649 (isolation #62) » Thu Jul 28, 2011 12:04 pm

Max, up to you.

StrangerCoug · Post Post #651 (isolation #63) » Thu Jul 28, 2011 12:06 pm

I've got one.

StrangerCoug · Post Post #652 (isolation #64) » Thu Jul 28, 2011 12:17 pm

Your best friend is tied up and gagged in a bank vault and you must rescue him. The vault is kept shut by an electronic lock and you must enter an eight-digit combination to open it. Any eight-digit string from 00000000 to 99999999 is valid.

You do not know the vault's combination and the only way to break into the vault is to take advantage of a security flaw in the lock—the error beep indicating an incorrect combination is a noticeably different pitch depending on whether the combination is too high or too low. You do not know which pitch means you should be going which direction, but you do know that if you enter the correct combination, the vault opens automatically on the eighth keypress.

It takes you two seconds to enter an eight-digit combination and one second for the error beep to play, during which time the keypad will not accept any presses. You must open the vault within one minute and thirty seconds or your friend will suffocate.

How do you guarantee your friend's survival?

StrangerCoug · Post Post #654 (isolation #65) » Thu Jul 28, 2011 12:23 pm

No. Remember, you do not know which beep indicates an error in which direction.

StrangerCoug · Post Post #656 (isolation #66) » Thu Jul 28, 2011 12:25 pm

Getting warmer...

StrangerCoug · Post Post #658 (isolation #67) » Thu Jul 28, 2011 12:30 pm

You almost got it; you just need one minor tweak. (The tweak's not what I thought it was, but your answer did make me realize something.)

StrangerCoug · Post Post #660 (isolation #68) » Thu Jul 28, 2011 12:36 pm

If you hear the same pitch with 00000000 and 50000000, you should not be going down.

StrangerCoug · Post Post #663 (isolation #69) » Thu Jul 28, 2011 12:43 pm

That's better. Your way only requires 29 tries (and thus 1:26) at most to get your friend out. My way required a maximum of 30 (thus 1:29), but I realized that you don't need to try both 00000000 and 99999999. Just get 99999999 for the "go lower" pitch, and if you hear something different during the binary search process, it'll be the "go higher" pitch.

Your turn.

StrangerCoug · Post Post #666 (isolation #70) » Thu Jul 28, 2011 2:20 pm

I needed more specific information, though.

StrangerCoug · Post Post #667 (isolation #71) » Thu Jul 28, 2011 2:50 pm

Basically, any of the three will work. Here's how our methods work (the combinations are all the same but have been generated by random.org)

Spoiler: My way

Spoiler: Empking's way

Spoiler: Bowser's way

StrangerCoug · Post Post #703 (isolation #72) » Fri Jul 29, 2011 11:25 am

I wasn't hoping for riddles, but this is what we've got...

A vowel? xD

StrangerCoug · Post Post #730 (isolation #73) » Fri Jul 29, 2011 3:32 pm

105m?

OK, where the hell did I get that?

StrangerCoug · Post Post #778 (isolation #74) » Tue Aug 09, 2011 10:29 am

Give me time to think of one.

StrangerCoug · Post Post #779 (isolation #75) » Tue Aug 09, 2011 11:28 am

On January 1, 2011, six people—Alice, Bob, Charlie, David, Emily, and Felicia—decided to try making money by trading in the foreign exchange market. On January 1, 2011, each of them converted 100 American dollars (referred to by the code USD in forex trading) to different currencies. Exchange rates at the time were:

Currency	From USD	To USD
EUR	0.746798	1.33905
GBP	0.640553	1.56115
CAD	0.997699	1.00231
AUD	0.977326	1.02320
JPY	81.15	0.012322

Alice invested all 100 USD of her money in euros (EUR on the table), while Bob invested all his in pounds sterling (GBP). Charlie split his evenly between those two currencies, while David split his evenly across all five currencies on the table. Emily converted 25 USD to euros, 25 USD to pounds, and 50 USD to yen (JPY), and Felicia split her money up 40 USD to Canadian dollars (CAD), 40 USD to Australian dollars (AUD), and 20 USD to yen. When computing how much everybody got in foreign currency, round to the nearest 0.01 EUR, 0.01 GBP, 0.01 CAD, 0.05 AUD, and 1 JPY.

By July 1, 2011, the rates had become the following:

Currency	From USD	To USD
EUR	0.689289	1.45077
GBP	0.622330	1.60686
CAD	0.960788	1.04081
AUD	0.929930	1.07535
JPY	80.82	0.012374

Assuming they made no other trades, who had made the most profit at this point?

StrangerCoug · Post Post #781 (isolation #76) » Tue Aug 09, 2011 1:56 pm

That is correct

StrangerCoug · Post Post #957 (isolation #77) » Mon Aug 15, 2011 10:39 am

7-7-7

3-4-5

My bad for not paying attention—that's a yes.
3-5-4
4-3-5
4-5-3
5-3-4
5-4-3

StrangerCoug · Post Post #958 (isolation #78) » Mon Aug 15, 2011 10:44 am

If yes to all of mine, if the first number is A, the second number is B, and the third number is C, A + B > C?

StrangerCoug · Post Post #997 (isolation #79) » Wed Aug 17, 2011 1:21 pm

Give me a moment...

StrangerCoug · Post Post #998 (isolation #80) » Wed Aug 17, 2011 1:25 pm

White to move and get mated in two. Yes, from the starting position.

StrangerCoug · Post Post #1000 (isolation #81) » Wed Aug 17, 2011 1:54 pm

Yes!

StrangerCoug · Wed Aug 17, 2011 11:55 pm

The French tarot deck is a superset of the regular 52-card deck. To each suit is added a knight. In addition, the deck contains a fool and 21 trump cards. Calculate the odds of drawing five cards—one trump, one spade, one heart, one diamond, one club—without replacement.

StrangerCoug · Thu Aug 18, 2011 10:37 am

Accepting Lowell's answer. 19,208 chances in 502,645, to be exact.

StrangerCoug · Sat Aug 27, 2011 10:32 am

This doesn't appear to be a semaphore problem, but rather a pattern problem. Unless I messed up, I can't make anything out of NAAAMOONMANAONAONOOM.

StrangerCoug · Post Post #1094 (isolation #85) » Fri Sep 02, 2011 4:36 pm

So -. .- .- .- -- --- --- -. -- .- -. .- --- -. .- --- -. --- --- -- is supposed to mean something, I guess...

6:35?

StrangerCoug · Post Post #1099 (isolation #86) » Sat Sep 03, 2011 2:33 pm

So my guess is that one of this...

-.
.-
.-
.-
--
---
---
-.
--
.-
-.
.-
---
-.
.-
---
-.
---
---
--

...or this, perhaps with a slight variant of deleting the colon, plays some sort of significance:

..... ---... ...-- .....
-.... ---... ...-- .....
-.... ---... ...-- .....
-.... ---... ...-- .....
--... ---... .---- .....
----. ---... ..... -----
----. ---... ..... -----
..... ---... ...-- .....
--... ---... .---- .....
-.... ---... ...-- .....
..... ---... ...-- .....
-.... ---... ...-- .....
----. ---... ..... -----
..... ---... ...-- .....
-.... ---... ...-- .....
----. ---... ..... -----
..... ---... ...-- .....
----. ---... ..... -----
----. ---... ..... -----
--... ---... .---- .....

It may also be the case that one time represents a dot, one time represents a dash, one time represents a letter space, and one time represents a word space, but then why the vertical arrangement?

StrangerCoug · Post Post #1103 (isolation #87) » Sun Sep 04, 2011 9:33 am

-.
.-
.-
.-
--
..
..
-.
--
.-
-.
.-
..
-.
.-
..
-.
..
..
--

I still suspect a pattern problem, and we know that the next one is not .- or --, so it's -. (5:35) or .. (9:50). The thing is, .. looks wrong to me when stuck at the bottom.

5:35?

StrangerCoug · Post Post #1127 (isolation #88) » Fri Oct 28, 2011 4:36 pm

I was thinking maybe where there are turns...

StrangerCoug · Post Post #1141 (isolation #89) » Tue Nov 15, 2011 1:18 pm

We've established that much ><

StrangerCoug · Post Post #1143 (isolation #90) » Wed Nov 16, 2011 1:06 am

Tell me something I don't know already. I'm clueless.

StrangerCoug · Post Post #1145 (isolation #91) » Sat Nov 19, 2011 4:58 am

*shoots himself in the head*

StrangerCoug · Post Post #1148 (isolation #92) » Sat Nov 19, 2011 6:12 am

Now I'm thinking a Vigenère cipher...

StrangerCoug · Post Post #1149 (isolation #93) » Sat Nov 19, 2011 6:18 am

Well, poop. None of FYYHKDPTMHXXTMJQZWAEBTNUAG, FGSTCBFXWHJBZARSVGSUVRJIGI, or VCCTQXLHOTDDHORKBEAWZHNGAU spell anything.

StrangerCoug · Post Post #1153 (isolation #94) » Fri Nov 25, 2011 5:47 am

If that is right, then

PLEASE

put up something that doesn't require us to be Ken Jennings to even have the slightest clue...

StrangerCoug · Post Post #1155 (isolation #95) » Fri Nov 25, 2011 5:50 am

I interpreted "coded" in the hint as "encoded", not "decoded".

StrangerCoug · Post Post #1160 (isolation #96) » Fri Nov 25, 2011 2:01 pm

In post 1153, StrangerCoug wrote:
PLEASE
put up something that doesn't require us to be Ken Jennings to even have the slightest clue...

StrangerCoug · Post Post #1163 (isolation #97) » Fri Nov 25, 2011 2:43 pm

That's basically an elaborate way of saying "dumb it down a little". I want an easier one.

StrangerCoug · Post Post #1204 (isolation #98) » Sat Apr 28, 2012 5:39 pm

Give me until tomorrow morning and I'll do one.

StrangerCoug · Post Post #1284 (isolation #99) » Sun Jun 03, 2012 3:47 am

In post 1282, JDGA wrote:I stopped paying attention when I saw colour written without a u.

Oh, quit being so nationalistic. It's not like I have a clue what I'm supposed to be doing either.

StrangerCoug · Sun Jun 24, 2012 6:17 am

1.) 50%
2.) 50%
3.) C

StrangerCoug · Sun Jun 24, 2012 6:33 am

Okay, let's try this:

A.) 100%
B.) 50%
C.) They're both equally likely.

StrangerCoug · Sun Jun 24, 2012 6:33 am

You know what I meant.

StrangerCoug · Sun Jun 24, 2012 6:42 am

Can't be 2/3 for #1. Perhaps 25%?

StrangerCoug · Fri Jul 13, 2012 5:19 am

Let me think of one...

StrangerCoug · Fri Jul 13, 2012 5:41 am

Decipher this code to get a famous quotation:

XKVEQ UQZUB IZBFV XKVMD AAUBB WKWVL SIWGO VHTNG SXZRT KMRFZ GWSFH OJDMW WOWXG JMKLM BETGK RVEUX

StrangerCoug · Fri Jul 13, 2012 5:45 am

Who said it was a straight substitution?

StrangerCoug · Fri Jul 13, 2012 6:01 am

An example of what I mean by a "straight substitution" is that all instances of A decode to Q, all instances of B decode to F, etc. Read "straight" as "simple" if you don't get it. It may or may not be a Vigenère cipher.

StrangerCoug · Fri Jul 13, 2012 4:44 pm

HINT:

I picked a very weak key.

StrangerCoug · Sat Jul 14, 2012 2:19 pm

HINT #2:

Even if everybody were to delete their signatures, the key would still be somewhere on this page.

StrangerCoug · Sun Jul 15, 2012 4:51 am

HINT #3:

It is an autokey cipher.

StrangerCoug · Sun Jul 15, 2012 5:25 am

No, but that's the site I used to encode the quotation.

StrangerCoug · Sun Jul 15, 2012 7:08 am

That is correct

StrangerCoug · Sun Jul 15, 2012 7:35 am

You didn't.

StrangerCoug · Wed Aug 01, 2012 2:02 pm

I'm figuring the 1, 19, 186 part is some sort of pattern, but I don't know where that takes me.

StrangerCoug · Fri Aug 03, 2012 12:57 pm

The numbers in numerical order are 1, 6, 11, 19, 84, 114, 120, 170, 186, 202, 209, 237, 268, 279, 299, 350, 356, 433, 461, 552, 748, and 914, but I have the sneaking suspicion that I'm thinking about

the rugs clue

too hard and it's the

ORIGINAL

order that matters.

I've also played with the idea of it being a fractal code, but five of the numbers being shorter than three digits is throwing me off.

StrangerCoug · Sun Aug 05, 2012 4:30 am

DNI AQ IQGU NKCRV NFTO RQ PO VQ AFEC NOJXJOVV, MALF? >=(

StrangerCoug · Thu Dec 06, 2012 5:51 pm

Did the train run over the woman?

StrangerCoug · Fri Dec 07, 2012 7:52 am

There goes the theory I had that the woman that screamed was a murder witness...

StrangerCoug · Fri Dec 07, 2012 9:15 am

Did anyone else cause her death?

StrangerCoug · Mon Jul 22, 2013 8:56 am

I'm going to revive with my own puzzle:

x > 0, y > 0, and z > 0.
x + y + z = 17.
z – x = z^(1/y).
yz / x = 3.

Solve for x, y, and z.

StrangerCoug · Mon Jul 22, 2013 9:14 am

StrangerCoug · Mon Jul 22, 2013 9:36 am

In post 2094, Who wrote:
x=6, y=2, z=9

A step-by-step guide:
Step 1: Assume they're all integers.
Step 2: Find the equation which would be the most restriction by said assumption (Eq2)
Step 3: Find the variable which matters the most to said assumption in said equation (y)
Step 4: Make a guess for said variable, start with the lowest possible (Cannot be 1 since they are all greater than 0 and this would force x to be 0, thus start with 2)
Step 5: Find the lowest thing z could be while remaining an integer (Cannot be 1 since that would force x to be 0, thus 4)
Step 6: Find x
Step 7: Check 2,2,4 with EQ1
Step 8: Raise z to the next lowest thing and use it as a new guess (9)
Step 9: Find x
Step 10: Check 6,2,9 with EQ1
Step 11: Check 6,2,9 with EQ3
Step 12: Shoot yourself for doing math this way.

Yeah. I think making everything positive made things much easier.

StrangerCoug · Mon Jul 22, 2013 10:09 am

In post 2099, serrapaladin wrote:That equation can't be algebraically solved except in terms of omega functions, so I doubt the set can be solved.

WolframAlpha confirms the y=2 solution, but also has a solution at about y=1.321, which leads to something like x=4.79 and z=10.88

No idea how to find that answer by hand.

...That seems to work, too. I should mention that I'm looking for integer solutions next time.

StrangerCoug · Sat Sep 14, 2013 4:57 pm

Here's a fresh one:

What are the next three numbers of the sequence 1, 2, 6, 12, 60, 60, 420...?

StrangerCoug · Sun Sep 15, 2013 6:43 am

In post 2133, Scigatt wrote:...840, 2520, 2520,...,lcm(1,...,n),...

Yes.

StrangerCoug · Thu Oct 03, 2013 1:39 pm

80113871

StrangerCoug · Fri Oct 04, 2013 12:55 pm

In post 2139, xtopherusD wrote:"We only hire Software Engineers who can use Google."

Oddly enough, I'm going into software engineering... and looked up a base converter on Google.

New one later.

StrangerCoug · Fri Oct 04, 2013 4:33 pm

Decipher this code to get a famous

QUOTATION:

0103401624 0100100102 0700903331 4100102908 4300604003 0104900618 6719200102 6705192424 1101104207 4401100413 1302500636 2305201418 6602201518 2602402721 1800100118 6721140915 6714242424 0100901715 2403102342 6709141921 6718052424 6704011305 6719200903 6720180114 6717210912 6709202524 0103003033 4000200212 4101501809 4401001519 6704050605 6714190524 2301900613 0402401111 6719080524 1302703414 1301801013 6701140424 0701801010 1911910005 0104902602 2001000606 4700301717 6720152424 4700300514 4200103801 5700100113 1910901303 6704152424 1300902233 2800300605 0200600405 6100300101 6703151419 6720092021 6720091514 6500101114 5900100105 0104900618 6719200102 6705192424 5200401510 6701130518 6709030124

StrangerCoug · Sat Oct 05, 2013 1:14 pm

Actually, you don't need to know very much math at all.

StrangerCoug · Sat Oct 05, 2013 2:23 pm

Well, it's wrong xD

HINT:

The weakest points of the code are the numbers starting with 67. Try those first if you don't know where to start.

StrangerCoug · Mon Oct 21, 2013 12:44 pm

I think everyone gives up at this point anyway.

It was a book cipher with the King James Bible as the corpus. You had to divide up most of them into chunks of two, three, three, and two numbers. 0103401624 told you to look at the first book (01xxxxxxxx) of the bible, which is Genesis, then go to the 34th chapter (xx034xxxxx), then the 16th verse in that chapter (xxxxx016xx), then the 24th (xxxxxxxx24) word in the verse, which was "we". That would get you as far as WE THE PEOPLE OF THE UNITED.

There are only 66 books of the Bible, so for the ones starting with 67, which encoded words not in the Bible, you were supposed to throw out the initial 67 and break the rest of it into four blocks of two numbers, which was a simple substitution cipher. That's why I told you to try those first. 6719200120 (and not 6719200102 as I posted, and I made this typo both times—my bad) becomes 67-19-20-01-20, then 19-20-01-20. STAT is a word, but not the one we're looking for; you need the next block after that to get STAT ESXX. You were supposed to delete the filler X's and the space to get STATES.

With the fixes, you should get

WE THE PEOPLE OF THE UNITED STATES, IN ORDER TO FORM A MORE PERFECT UNION, ESTABLISH JUSTICE, INSURE DOMESTIC TRANQUILITY, PROVIDE FOR THE COMMON DEFENSE, PROMOTE THE GENERAL WELFARE, AND SECURE THE BLESSINGS OF LIBERTY TO OURSELVES AND OUR POSTERITY, DO ORDAIN AND ESTABLISH THIS CONSTITUTION FOR THE UNITED STATES OF AMERICA.

New question incoming.

StrangerCoug · Mon Oct 21, 2013 1:13 pm

Give the reduced row-echelon form of the following augmented matrix:

Code: Select all

┌             ╷     ┐
│  4 -2  1  3 ╎  45 │
│             ╎     │
│  3 -2 -1 -5 ╎  12 │
│             ╎     │
│  1  0  7 -2 ╎  13 │
│             ╎     │
│ -6  2  8  4 ╎ -16 │
└             ╵     ┘

Spoiler: Equivalent problem for people who are unfamiliar with matrices

StrangerCoug · Mon Oct 21, 2013 1:39 pm

Use whatever it takes.

StrangerCoug · Mon Oct 21, 2013 5:39 pm

In post 2152, Who wrote:

Code: Select all

┌         ╷   ┐
│ 1 0 0 0 ╎ 5 │
│         ╎   │
│ 0 1 0 0 ╎-7 │
│         ╎   │
│ 0 0 1 0 ╎ 2 │
│         ╎   │
│ 0 0 0 1 ╎ 3 │
└         ╵   ┘

That is correct

StrangerCoug · Tue Nov 05, 2013 12:50 pm

The thread has been dead for awhile, so I'll breathe some more life into it with this one:

Either find a rational number of radians

θ

such that at least two of sin

θ

, cos

θ

, tan

θ

, cot

θ

, sec

θ

, and csc

θ

are also rational or prove that none exists.

StrangerCoug · Tue Nov 05, 2013 1:29 pm

And that's the only solution as per the Lindemann-Weierstrass theorem. Sin 0 = 0, cos 0 = 1, tan 0 = 0, cot 0 is undefined, sec 0 = 1, and csc 0 is undefined.

StrangerCoug · Wed Nov 06, 2013 1:10 pm

In post 2170, Who wrote:Last I checked sin0=0, cos0=1, tan0=0, cot0=undef, sec0=1, csc0=undef.

Fixed.

StrangerCoug · Sun Feb 09, 2014 10:45 am

It's been a long time. Do a new one.

StrangerCoug · Sun Feb 09, 2014 11:09 am

I had the first part, but had my head in the gutter for the second and had no clue on the third.

StrangerCoug · Sun Feb 09, 2014 3:06 pm

In post 2184, Axxle wrote:
In post 2181, serrapaladin wrote:I realise it's axxle's turn, but the problems reminded me too much of high school maths competitions not to take a stab at.
I didn't realize we take turns. Doesn't matter to me, go ahead.

Enforcement has been on the lax side lately.

StrangerCoug · Mon Feb 10, 2014 10:00 am

I'm pretty sure that 1 is, but I can't think of how to do a general proof formally.

StrangerCoug · Wed Feb 12, 2014 5:44 am

Let's throw in a new one:

Prove that, for any real constant c, the derivative of ce^x is itself.

StrangerCoug · Wed Feb 12, 2014 8:27 am

I wasn't looking for the "by definition" answer for ce^x. I wanted work shown.

Edit:

To clarify, the definition of e^x is not taught in precalculus that way; an approximate value of 2.818 for e is taught before derivatives are taught. I was looking for an answer based on applying the rules of derivatives without that assumption.

StrangerCoug · Wed Feb 12, 2014 11:19 am

In post 2215, serrapaladin wrote:The only definition of e that makes sense from a precalc point of view is the one resulting from compound interest, for which the limit of (1+1/n)^n is the relevant definition.

This definition of e is fine.

StrangerCoug · Wed Feb 12, 2014 11:28 am

You know what, we'll give it to Who unless he can be proven incorrect.

StrangerCoug · Wed Feb 19, 2014 5:35 am

Let's try another derivative question:

A student was asked to compute the second derivative of sec(x) and got sec³(x)tan(x). He's wrong. What mistake did he make and what should the second derivative of sec(x) be?

StrangerCoug · Wed Feb 19, 2014 5:48 am

The mistake as regards the product rule is correct, but the second derivative of sec(x) is not sec(x)tan(x)+sec^3(x), either.

StrangerCoug · Wed Feb 19, 2014 5:56 am

That's better.

StrangerCoug · Sat Mar 08, 2014 9:07 am

In post 2232, Dessew wrote:I thought you were supposed to do exponentation before substraction.

You are. The order is parentheses, then exponents, then multiplication and division, then addition and subtraction.

StrangerCoug · Mon Mar 24, 2014 11:20 am

Here's something else that should be pretty easy:

Either find a nonzero integer x such that the decimal representation of 1/x neither repeats nor terminates or prove that no such integer exists.

StrangerCoug · Mon Mar 24, 2014 1:16 pm

That's a good starting point, and you're on the right track in terms of what I want. Can you be a bit more vigorous?

StrangerCoug · Mon Mar 24, 2014 2:23 pm

Very good

StrangerCoug · Tue Apr 01, 2014 10:43 am

OK, guys, let's have a little fun...

Determine whether 0.999... is equal to 1 and prove that your determination is correct.

StrangerCoug · Tue Apr 01, 2014 1:05 pm

Those are all good.

StrangerCoug · Wed Apr 02, 2014 2:03 am

Spoiler: Inelegant, but...

StrangerCoug · Wed Apr 02, 2014 12:19 pm

In post 2260, Scigatt wrote:
In post 2259, StrangerCoug wrote:
Spoiler: Inelegant, but...
First, draw your curve. That's pretty obvious.

Next, randomly select a point within the rectangle bounded by (0, 0), (a, 0), (0, b), and (a, b), allowing any point on those four corners or the lines connecting them to be valid. We'll put a green dot on the point if it is under the curve and a red dot on it if it is over the curve. Repeat a large number of times—we'll say 1,000,000.

If the area of the rectangle from which we are allowed to choose a point is 1, then the ratio of the number of green dots to the number of randomly chosen points is the approximate area under the curve.
There's a name for that.

I looked at the hint and went from there

New problem incoming.

StrangerCoug · Wed Apr 02, 2014 2:03 pm

I have a daughter named Michelle, who is in high school. This year, she is taking six classes: algebra, biology, English, French, history, and physical education in some order. Her teachers for them are Mrs. Burton, Mrs. Harris, Mr. Lambert, Mrs. Reeves, Mrs. Smith, and Mrs. White, who are in rooms 100, 105, 127, 212, 216, and 233 in some order. Michelle has no free periods.

Here are some facts about Michelle's schedule:

Mrs. Reeves teaches physical education, which is after biology, and has Michelle after Mrs. Harris does.
Mrs. Smith, who does not teach English, is in a room starting with a 2 and has Michelle for an even-numbered period, while the algebra teacher, who is also in a room starting with a 2, has Michelle for an odd-numbered period.
History, which is neither 5th period nor taught by Mr. Lambert, is in room 100.
Mrs. Harris is in room 127, while 4th period is in room 216. 6th period is in an odd-numbered room.
Michelle has Mr. Lambert in the first half of the day and English in the second half of the day.
Michelle has Mrs. Burton, who does not teach biology, 1st period and French, which is in a room starting with a 1, 2nd period. 1st period is in a higher-numbered room than 2nd period is.

Using this information, can you reconstruct my daughter's schedule?

StrangerCoug · Wed Apr 02, 2014 2:11 pm

Simply "after". Not necessarily the next one after.

StrangerCoug · Wed Apr 02, 2014 4:22 pm

That is correct

StrangerCoug · Wed Apr 09, 2014 4:33 am

Here's a fresh one.

You and your classmates are at a quiz bowl competition and asked a math question. You are not allowed any tools other than paper and pencil for such questions. You are asked to give the sum of all the numbers between 1 and 1,000,000 inclusive. As soon as the question is finished, your teammate buzzes in with the correct answer despite not knowing it in advance. What was the correct answer and how was she able to produce it so quickly?

StrangerCoug · Wed Apr 09, 2014 6:46 am

In post 2277, Scigatt wrote:(1,000,000*1,000,001)/2, bc. the sum is the 1,000,000th triangular number.

You have the method. Now compute it.

StrangerCoug · Wed Apr 09, 2014 7:23 am

Without using operator symbols, please.

StrangerCoug · Wed Apr 09, 2014 7:25 am

StrangerCoug · Mon Apr 21, 2014 12:42 pm

Reminder to myself to put up a decent problem when I have time.

StrangerCoug · Mon Apr 21, 2014 1:06 pm

f(x)	f'(x)	int f(x)
sin(x)	cos(x)	?
cos(x)	-sin(x)	?
tan(x)	?	?
cot(x)	?	?
sec(x)	?	?
csc(x)	?	?

Fill in the remainder of the table and show your work.

Spoiler: Hint for the int f(x) column

StrangerCoug · Mon Apr 21, 2014 1:22 pm

That's all I care about, yes.

StrangerCoug · Mon Apr 21, 2014 2:31 pm

int tan(x) is slightly off.

StrangerCoug · Mon Apr 21, 2014 3:09 pm

int cot(x) is fixed, now that you bring it up,

but int tan(x) is still wrong.

Actually, ln|sec(x)|+C is an acceptable alternative to -ln|cos(x)|+C. Very good.

StrangerCoug · Sat May 10, 2014 3:40 am

I think it's

true

. As there are infinitely many triangular integers and infinitely many square integers, there's got to be plenty that coincide, since their definitions are not mutually exclusive.

StrangerCoug · Fri Jul 04, 2014 6:06 am

OK, here's another one:

There are 52! possible ways to order a deck of cards and nCk(52,13) possible Bridge hands you could be dealt. Thus, since the order of cards in a player's hand doesn't matter, how many different ways could you deal a hand of Bridge if rotations are:
a.) Counted?
b.) Ignored?

StrangerCoug · Fri Jul 04, 2014 7:54 am

Very good, Dessew

StrangerCoug · Fri Jul 04, 2014 8:47 am

It's not real rigorous, but here goes my attempt:

For two points on the same radius, it's trivial—follow that radius.

If the two points are on the same circle, follow the shorter arc length between them.

Give me some more time to think about how to word the rest of the solution.

StrangerCoug · Fri Jul 04, 2014 9:22 am

For two points not on the same radius or circle, I believe the general case works something like this:

Find the intersection of the circle with the point closer to P (we'll call that A) and the radius of the point further from P (we'll call that B) and call the intersection C. The shorter arc length between A and C is arc AC; the line between C and B is line CB.

Define points D and E as follows:

If arc AC is shorter than line CB, then D is on the same radius as A, and line AD is defined as having the same length as arc AC. Draw an arc from D until line CB and call the intersection E.
If line CB is shorter than arc AC, then D is on the same circle as B, and arc BD is defined as having the same length as line CB. Draw a line from B until arc AC and call the intersection E.

Drive as follows:

	Point D is connected to your starting point	Point D is not connected to your starting point
You are starting on A	Drive one block on arc AC, then turn as appropriate at every intersection until you get to E, then drive the length of line BE.	Drive the length of arc AE plus one block, then turn as appropriate at every intersection until you get to B.
You are starting on B	Drive one block on line CB, then turn as appropriate at every intersection until you get to E, then drive the length or arc AE.	Drive the length of line BE, then turn as appropriate at every intersection until you get to A.

StrangerCoug · Fri Jul 04, 2014 9:28 am

Actually, I think the definitions of D and E are a bit messed up, but I'll wait for judgment.

StrangerCoug · Fri Jul 04, 2014 10:03 am

By "turning appropriately" I mean in the direction that would get you closer to the destination, and the "one block" is because you disallow straight lines unless you're traveling on the radius, and circles of shorter radii have shorter circumferences.

If the roads are infinitely dense, then I don't get how driving as straight a line as the roads let you is suboptimal.

StrangerCoug · Fri Jul 04, 2014 10:13 am

I'm pretty far off the mark anyway, so forget it.

StrangerCoug · Sat Jul 05, 2014 6:10 am

Eight?

StrangerCoug · Sat Jul 05, 2014 8:20 am

...I

HAD

it, but then I reread the question and my method is 72 moves, not 64.

StrangerCoug · Fri Jul 11, 2014 8:19 am

A fresh one, and one that's relatively simple: What shortcut can you use to expand (x+y)ⁿ?

StrangerCoug · Fri Jul 11, 2014 8:26 am

Very good

StrangerCoug · Wed Jul 16, 2014 6:10 am

A number is divisible by 3 if the sum of its digits is divisible by 3, and a number is divisible by 9 if the sum of its digits is divisible by 9. However, it does

NOT

follow that a number is divisible by 27 if the sum of its digits is necessarily divisible by 27.

Give any number the sum of whose digits is divisible by 27, but is not itself divisible by 27.
Give any number the sum of whose digits is
NOT
divisible by 27, but
IS
itself divisible by 27.
Give any correct method of determining divisibility by 27 and prove that it works.

StrangerCoug · Wed Jul 16, 2014 6:14 am

Extra credit: Prove or disprove (as appropriate) that there is

ANY

integer n > 2 such that, if the sum of the digits of an integer m is 3ⁿ, then m is also divisible by 3ⁿ.

StrangerCoug · Wed Jul 16, 2014 6:34 am

Looks good to me for the main problem. That leaves the extra credit.

StrangerCoug · Wed Jul 16, 2014 6:49 am

Checks out

StrangerCoug · Wed Jul 16, 2014 6:58 am

There are multiple ways of doing the EC; I came up with two myself on my own.

StrangerCoug · Wed Jul 16, 2014 7:13 am

Spoiler: My own ways, if you're curious

Edit:

Left out an important bit.

StrangerCoug · Wed Jul 23, 2014 10:50 am

Fill out the table. The trivial cases have been done for you.

In terms of:	sin(x)	cos(x)	tan(x)	cot(x)	sec(x)	csc(x)
sin(x)=	sin(x)					1/csc(x)
cos(x)=		cos(x)			1/sec(x)
tan(x)=			tan(x)	1/cot(x)
cot(x)=			1/tan(x)	cot(x)
sec(x)=		1/cos(x)			sec(x)
csc(x)=	1/sin(x)					csc(x)

StrangerCoug · Wed Jul 23, 2014 11:30 am

They check out OK on Wolfram Alpha for 0 < x < π/2 to me. Could you make the functions piecewise-defined that need to be?

ETA:

I've only checked the first three rows.

StrangerCoug · Wed Jul 23, 2014 11:39 am

I think we're good.

StrangerCoug · Sun Aug 10, 2014 5:06 pm

Let's do a fresh one.

You have a standard 52-card deck. Obviously, if you never put back cards you draw from it, then the odds that you will never draw the same card twice in 52 draws is 100%. What do those odds become if you shuffle each card drawn back in the deck before drawing again?

StrangerCoug · Sun Aug 10, 2014 5:16 pm

That is correct

StrangerCoug · Tue Aug 12, 2014 7:36 am

Since I like puzzles related to cards...

Take 50 cards from a standard 52-card deck and group them to form one each of the ten different five-card hands in poker from high card to royal flush. (Note the phrasing: you are not allowed to use a card in more than one hand.)

StrangerCoug · Tue Aug 12, 2014 8:19 am

inspiratieloos' solution has been validated. As there are multiple possible answers, I will check Mitillos' as well.

StrangerCoug · Tue Aug 12, 2014 8:25 am

And Mitillos has a valid solution as well.

StrangerCoug · Tue Aug 12, 2014 8:41 am

I'm in the mood for more solutions, actually, so let's add a constraint to the puzzle. For these four, do the same thing as #2457, but make the straight flush:

queen-high.
jack-high.
ten-high.
nine-high or lower
AND
of a different suit than the royal flush (since a nine-high straight flush of the same suit as the royal has been given already).

StrangerCoug · Tue Aug 12, 2014 8:53 am

Hint: It's possible to do each of the four with only minor changes to each solution.

StrangerCoug · Tue Aug 12, 2014 11:34 am

An actual deck of cards was how I was verifying the problem

Parts A and B I feel safe ruling correct off the top of my head. For parts C and D, I'm having trouble following—are the steps supposed to be followed in order? If so, from the source image or from the previous part?

StrangerCoug · Wed Aug 13, 2014 6:32 am

Original C is valid, but as you said, not original D (as you have two flushes and no straight flush, in violation of the original constraints). Revision C and D are both valid.

StrangerCoug · Sat Aug 16, 2014 8:37 am

An easy calculus problem:

You type in a positive number in a calculator and hit the square root key over and over. Each time you press the square root key, the answer produces becomes closer to what number? (That is, for any positive x, what is the limit of x^(1/(2^n)) as n approaches infinity?)