Math and Logic Puzzles: Redux
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Harambey180 He/himGoonHe/him
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brassherald he/himJack of All Tradeshe/him
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Since I am pretty close, I'm going to guess at my error and go withIn post 19, brassherald wrote:So, January 5, 2018 will be AFC-126, I believe.
December 31, 2018 will be BMD-532, I think.
If I'm right on those two, I'm pretty surprised, because I feel like I'm doing something wrong and will not do the other 2 questions for now.
AFC-125
BMD-531
Otherwise, Harambey is probably going to get it before me, but I was thinking I might have been 1 integer too high with my calculations.I've only made one good post, and don't you dare accuse me of doing it again.-
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Harambey180 He/himGoonHe/him
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StrangerCoug He/HimDoes not ComputeHe/Him
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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: Stranger Than You!
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StrangerCoug He/HimDoes not ComputeHe/Him
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The error in the last date was mine, so Harambey180 has corrected all his mistakes.STRANGERCOUG: Stranger Than You!
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StrangerCoug He/HimDoes not ComputeHe/Him
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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: Stranger Than You!
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BuJaber Mafia Scum
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StrangerCoug He/HimDoes not ComputeHe/Him
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That is correctSTRANGERCOUG: Stranger Than You!
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BuJaber Mafia Scum
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Okay a classic.
Construct a 4 x 4 Magic square
If you want a harder challenge try a 5 x 5 and/or a 6 x 6.
There are many solutions. Just try and have fun with it. Don't look up a solved one-
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D3f3nd3r he/himBest Social Gamehe/him
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Harambey180 He/himGoonHe/him
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Harambey180 He/himGoonHe/him
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Harambey180 He/himGoonHe/him
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BuJaber Mafia Scum
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Ircher He / Him / HisWhat A Grand IdeaHe / Him / His
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Harambey180 He/himGoonHe/him
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Alright, here are two new puzzles!
Puzzle A:
Use the four numbers in the four corners the same way each time to get to the number in the middle. What should that middle number be for the third?
323024
635451
31?82
Puzzle B:
This puzzle is the same as the puzzle in post 9. These are two individual puzzles that work the same but have no meaning for the other puzzle.
When you see the puzzle as having four lines, each of which are straight lines going through the middle, how do you use the numbers to get to the number in the middle and what should that number therefore be?
61155?42163
63184?92126
These should be easier than previous puzzles I posted here.hi-
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pinturicchio Mafia Scum
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StrangerCoug He/HimDoes not ComputeHe/Him
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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: Stranger Than You!
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Mitillos HeMafia ScumHe
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I think it can be done if we split the proof up into odd and even perfect numbers.
1) Let n be an odd perfect number with k divisors s_1 < s_2 < ... < s_k that is refactorable. Then n = s_1 + s_2 + ... + s_{k - 1} and n = ak for some integer a. Since n is odd, a and k have to be odd, and so do all s_i. But if k is odd, k - 1 is even. An even sum of odd integers is even, so we have that n is even, contradicting the hypothesis.
2) Let n be an even perfect number. Then n is of the form (2^k - 1)2^{k - 1} with 2^k - 1 prime, and n has 2k divisors. Observe that since 2^k - 1 is prime, so is k. Therefore, if 2k divides n = (2^k - 1)2^{k - 1}, then either k = 2, or k = 2^k - 1 (the only prime divisors of n). If k = 2, n = 6, which is not refactorable, and there is no prime k > 1 so that k = 2^k - 1.You don't have ambiguity; you haveoptions.-
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StrangerCoug He/HimDoes not ComputeHe/Him
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Looks goodSTRANGERCOUG: Stranger Than You!
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Gosrir Elmer Odels Goon
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Suppose I have a fence with n spikes. For example:
Code: Select all
/¤\ /¤\ / \ / \ / \ / \ ¤/ \¤/ \¤ | | | | | | | | |
The fence above has 2 spikes. The ¤'s represent marbles that are there for decoration, but currently they're all green! (as you can see) So I want to paint some of them red, to improve the looks. But there is a problem: something is wrong with the paint, so whenever I paint a marble on the top red, the paint runs down to the two marbles below, & colours those red too.
How many patterns are possible if my fence has n spikes?
(Example: for n=1 there are 5 patterns: all green; bottom left red; bottom right red; both at the bottom red; or all red.)
(Disclaimer: I intentionally tried to avoid very formal mathematical language while formulating this question.)-
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Mitillos HeMafia ScumHe
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Gosrir Elmer Odels Goon
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Mitillos HeMafia ScumHe
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Gosrir Elmer Odels Goon
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Here's a fun one, an EGMO question slightly modified (from last year maybe, but idfk.)
Let there be finitely many lines on the plane, no three of them concurrent. Turbo the Snail (the name is crucial) begins her journey at a point on exactly one of the lines. She moves along the line in one direction until she reaches an intersection. Then she starts moving along the other line, & so on in the same fashion. At the nth intersection she encounters she turns right if n is a prime, & left otherwise. Is there a segment that Turbo the Snail passes through in both directions?
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