Math and Logic Puzzles: Redux

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Harambey180
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Post Post #25  (ISO)  » Tue May 01, 2018 10:29 am

Third one is February 12, 2020. It is the 10,176th plate.
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Post Post #26  (ISO)  » Tue May 01, 2018 10:31 am

In 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.

Since I am pretty close, I'm going to guess at my error and go with

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.
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Post Post #27  (ISO)  » Tue May 01, 2018 10:32 am

Last one is March 25, 2022. And it is a Friday.
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Post Post #28  (ISO)  » Tue May 01, 2018 1:16 pm

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  (ISO)  » Tue May 01, 2018 3:23 pm

The error in the last date was mine, so Harambey180 has corrected all his mistakes.
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Post Post #30  (ISO)  » Wed May 02, 2018 7:01 am

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 #31  (ISO)  » Wed May 02, 2018 8:03 am

Monday -- Ring -- from Vivienne
Tuesday -- Phone -- from Eric
Wednesday -- Watch -- from Richard
Thursday -- TV -- from Bradley
Friday -- Car -- from Jessica
Saturday -- Wallet -- from Michelle

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Post Post #32  (ISO)  » Wed May 02, 2018 9:48 am

That is correct :D
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Post Post #33  (ISO)  » Wed May 02, 2018 10:07 am

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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Post Post #34  (ISO)  » Wed May 02, 2018 12:11 pm

I mean,

0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
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Post Post #35  (ISO)  » Wed May 02, 2018 12:22 pm

1 3 2 4
2 4 1 3
3 1 4 2
4 2 3 1

All horizontal lines add up to 10.
All vertical lines add up to 10.
Both diagonals add up to 10.
All 2x2 squares add up to 10.

Beat this.
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Post Post #36  (ISO)  » Wed May 02, 2018 12:26 pm

1 4 3 2 5
2 5 3 1 4
3 3 3 3 3
4 1 3 5 2
5 2 3 4 1

I'm just too good at these magic squares...
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Post Post #37  (ISO)  » Wed May 02, 2018 12:37 pm

4,047-2,9532,047
-0,9531,0473,047
0,0475,047-1,953


I couldn't help doing this (with a generator)
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Post Post #38  (ISO)  » Wed May 02, 2018 12:54 pm

I can't tell the tone in your posts. But just in case you're not joking... we're looking for consecutive positive integers only from 1 -16 for 4x4, 1-25 for 5x5, and 1-36 for 6x6.

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Post Post #39  (ISO)  » Wed May 02, 2018 5:18 pm

Prove that the derivative of an odd function is an even function and that the derivative of an even function is an odd function.
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Post Post #40  (ISO)  » Thu May 03, 2018 2:52 am

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?

32
30
24

63
54
51

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?

6115
5?4
2163


6318
4?9
2126


These should be easier than previous puzzles I posted here.
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Post Post #41  (ISO)  » Sun May 06, 2018 7:05 pm

Puzzle A: (top left + top right) * (bottom left + bottom right); (3+2)*(2+4)=30; (6+3)*(5+1)=54; (3+1)*(8+2)=40.

Puzzle B; first square you multiply both numbers minus the green number: 6*3-3=15*2-15=16*1-1=5*4-5=15. Second square it's just a multiplication, result is 36.

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Post Post #42  (ISO)  » Mon May 21, 2018 10:16 am

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 #43  (ISO)  » Mon May 21, 2018 11:59 am

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.
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Post Post #44  (ISO)  » Mon May 21, 2018 12:32 pm

Looks good :)
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Post Post #45  (ISO)  » Thu May 24, 2018 10:06 am

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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Post Post #46  (ISO)  » Thu May 24, 2018 1:31 pm

Spoiler: Attempt
Let a_k be the number of patterns for k spikes, and b_k be the number of patterns for k spikes so that the last marble is red. Then we have: a_k = 2a_{k - 1} + b_{k - 1}, where b_j = a_j - a_{j - 1}.

Explanation for a_k: For every pattern for k - 1 spikes, we can colour the top of the kth spike green, and the last marble either green or red. If we colour the top of the kth spike red, the penultimate bottom marble and the last one both have to be red.

Explanation for b_j: The number of patterns which end in red is equal to the total number of patterns, minus the ones which end in green, i.e. the ones in which the last top and bottom marbles are green. The colours of all the previous marbles in such a pattern are therefore not restricted by the jth top marble, hence there are a_{j - 1} of them.

Simplifying the expression we get a_k = 3 * a_{k - 1} - a_{k - 2}, with a_0 = 1 and a_1 = 2. Recall that the Fibonacci sequence goes f_j = f_{j - 1} + f_{j - 2}, and usually starts with f_0 = 1, f_1 = 1, and therefore has f_2 = 2. Then f_j = 2f_{j - 2} + f_{j - 3} = 2f_{j - 2} + f_{j - 2} - f_{j - 4} = 3f_{j - 2} - f_{j - 4}. In other words, a_k matches exactly f_{2k}, so the number of patterns for n spikes is the (2n)th Fibonacci number.
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Post Post #47  (ISO)  » Fri May 25, 2018 12:55 am

Yes! You're a bit off with the indices, since a_1=5=f_4, a_2=13=f_6, so in general a_i=f_{2i+2}, but aside from that your solution is correct. :)

(In the hypothetical k=0 case you'd still have one marble, which you can either colour or not, so a_0=2.)

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Post Post #48  (ISO)  » Fri May 25, 2018 2:28 am

Oh, yeah, I was counting bottom marbles, not top marbles. Mea culpa.
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Post Post #49  (ISO)  » Sat May 26, 2018 3:03 pm

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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