You and a friend are discussing how you choose four-digit PINs. You establish that neither of you would ever use the digit 0.

“I like to choose four different random digits,” you say.

“I like to choose three different random digits,” they reply, “so one of the digits is used twice.”

Which strategy gives the largest pool of possible four-digit PINs?

P12573
Why are you starting with 8?

P12573
>still feels a little counter-intuitive?
ig it makes sense
the first strategy you are simply picking 4 different numbers
the second, even though you ar picking only 3, you are also scrambling their positions, which gives you more possibilities than just picking an extra number

bump

I would just use password generator like [bold: apg] which is customizable and can generate 4 digit string.
The PIN needs to be just secure enough not to be brute-forced in 10 or so attempts before login ability gets blocked.

P23327
>Password generators
When you forget the generated password, you are in trouble. Locksmith might be the only solution...

Here's a nice paper about the disease that has infected academic statistics.

P22453
Seethe cuelet.
Physicists are expected to apply physics. In fact, what vindicated the study of physics in the first place was constructing buildings that didn't collapse, i.e., structural engineering. This is a better demonstration than tennis, as everyone with a 3-digit IQ will understand.
The challenge for psychologists revolves around their claimed skill set. You can't use psychology to build a building and see if it stands, but they do claim to be able to read their patients and to reverse engineer their research subjects' actions back to unobservable mental phenomena, so poker is an excellent test of such acumen.
>poker players use psychology
Poker players use their experience first and foremost. They discard as many psychological suggestions as they adopt. And what always wins is playing the game with other good players, not reading psych papers. It's almost like there's wisdom to the old tradition of playing poker with your friends.

P22454
The problem is that P(A) (the probability that the hypothesis is true, prior to the experiment) is usually unknown. Moreover, the fact that this is a random variable in the first place is a philosophical question itself.
At best, it could be used to precise prior estimates.

P22501
Why not simply choose 1/n for each of n hypotheses? If that's found later to be not reasonable, it's easy to correct.

P22512
Let's consider a simple yes-no question. With your proposition, we would have P(A)=1/2 and P(not A)=1/2
Let's say that the result of your experiment happens 90% of the time if your hypothesis is true and 50% of the time if it isn't. That is, P(B|A) = 0.9, P(B|not A) = 0.5 and P(B)=0.7
Using Bayes' formula, we find [tex: P(A|B) = \frac{0.9 \times 0.5}{0.7} = 0.64]

I guess it works, and it allows subsequent experiments to precise the value.
If another lab were to do another experiment with P(C|A) = 0.8, P(C|not A) = 0.4, they can use the previous P(A|B) to estimate P(C) = 0.66 (assuming the two experiments to be independent).
This gives [tex: P(A|B,C) = \frac{0.8 \times 0.64}{0.66} = 0.78

Now, someone with enough standing should convince people in social science to use this

It would still be *****d, but Bayesian methods would be an improvement over the null ritual simply because it forces you to state your research hypothesis. An even better improvement would be if they started reporting something like confidence/credible intervals instead of just yes/no results.

Caitlin wants to draw n straight line segments, without lifting her pencil off the paper and without retracing her path, so that each segment crosses exactly one other segment (not counting intersections at vertices) and she ends up back where she started.

a. Show how she can do this for n = 6. (Draw a picture!)
b. For which n can it be done, and for which n is it impossible? Prove your answer.

The problem cannot be solved because Caitlin is female and therefore lacks the cognitive ability required to complete the task.

P18170

I'm working on the case of 2 intersections for each line. It's easy when n is odd, but when n is even it appears to require picking up the pencil once.

P18610 JEWWWWWSSSSSS

I have found a construction that works for all n=2k with k odd.
Step 1: draw a regular k-agon
Step 2: draw a second regular k-agon inside and in the same orientation as the first one
Step 3: link all vertices by alternating between the small and the big polygon, thus making two complete rotations around the center of the figure.
This makes a regular k-pointed star.

Let's assume the n vertices have coordinate [tex: A_i=(cos(2\pi×i/k), sin(2\pi×i/k))] and [tex: B_i(a cos(2\pi×i/k), a sin(2\pi×i/k))], without loss of generality (this is just a frame transformation).[tex: [A_{i}B_{i+1}]] only crosses[tex: [A_{i+1}B_{i}]].

The polygons don't really have to be regular, nor in the same orientation, but it makes the explanation simpler. What is important is that one polygon is comprised inside the other and that all the lines are on the outside of the small polygon.

I don't know if another construction could work for other values of n

Picrel is an example for n=6 (k=3)

P19219
Looks good so far.

What is the value of
[tex: \lfloor (\sqrt{1.000003} - 3\sqrt{1.000002} + 3\sqrt{1.000001} - 1)^{-1/3} \rfloor]?
(Your calculator will probably do the problem, but it may not get it right.)

[tex: \lfloor x \rfloor] means the floor function, the greatest integer not exceeding x.

P17417
>Is there an algorithm to construct a function such as f, whose Taylor series is zero until order n? For f, n would be 3.
The approach I used to construct the original function was to use a linear combination and solve for the coefficients that would make the first few terms cancel.

Oh, [tex: f_n(x) = \sum_{i=0}^n (-1)^i \binom{n}{i} \sqrt(1+i\times x) ] seems to work. How come?

P17426
It's equivalent to repeatedly taking discrete differences of the original function (in this case, the square root function).
[tex: f_1(x) = f_0(x + \Delta x) - f_0(x)]
[tex: f_2(x) = f_1(x + \Delta x) - f_1(x) = f_0(x + 2 \Delta x) - 2 f_0(x + \Delta x) + f_0(x)]
[tex: f_3(x) = f_2(x + \Delta x) - f_2(x) = f_0(x + 3 \Delta x) - 3 f_0(x + 2 \Delta x) + 3 f_0(x + \Delta x) - f_0(x)]
Since
[tex: f_1(x) \approx f_0'(x) \Delta x]
[tex: f_2(x) \approx f_0''(x) \Delta x^2]
[tex: f_3(x) \approx f_0'''(x) \Delta x^3]
the result should shrink unless the repeated derivatives blow up.

So you can think of it as numerical differentiation amplifying noise from rounding errors. (I wonder how better numerical differentiation algorithms work. Seems like an interesting problem to try to compute derivatives with as little amplification of rounding errors as possible. I'll have to read up on it.)

P17267

Does 3\sqrt denote 3 times the square root of, or cube root?

P17772
3 times the square root.

A farmer sold 108 pounds of produce that consisted of z pounds of zucchini and c pounds of cucumbers. The farmer sold the zucchini for $1.69 per pound and the cucumbers for $0.99 per pound and collected a total of $150.32. Which of the following systems of equations can be used to find the number of pounds of zucchini that were sold?

z + c = 108
1.69z + 0.99c = 150.32
SAT questions are boring.

That makes no sense. Why would they have weighed the zucchinis and cucumbers together?

What do you think about supertasks (tasks with an infinite number of steps, where we're interested in what happens after you finish the infinite steps)? Why do they so often lead to paradoxes? Are mathematicians right to avoid them? Does Zeno's paradox about Achilles and the tortoise mean we have to grapple with supertasks anyway?

P16737

Infinity is a Jewish concept.

you do realize that math is a made-up concept that humans somehow lost control of, right?
paradoxes and shit like that are just natural consequences of the retarded machinations of the human mind

P16794
While "humans are retarded" may be a correct explanation for why we have paradoxes, it's not a very helpful one if your goal is to actually understand things. Ideally we want to figure out what's wrong with our made-up concepts and how we can choose made-up concepts that better model the world.

P16808

It's a model or be modeled world out there.

God-man will do this supertask!

Two regular blinkers begin with synchronized blinks at time 0, and afterwards there is an [bold: average] of one blink per minute from the two blinkers together. However, they never blink simultaneously again (equivalently, the ratio of their frequencies is irrational).

Prove that after the first minute (from time 0:00 to time 0:01) there is [bold: exactly one blink] in every interval between time t minutes (t an integer) and time t + 1!

Three sisters combine their money to buy a mystery box full of an unknown amount of comic books. The box arrives at the sisters’ house one night, and they agree to store it in their *****er brother’s room and not open it until the next morning.

But that night, one sister can’t sleep, so she decides to sneak in and open the box. Looking inside, she finds that she can split the comic books into three equal shares, plus one extra comic. She sees her brother awake and spying, so she hands him the extra comic, takes her own share of one-third of the remaining comic books, and goes back to sleep.

An hour later, the second sister does the exact same thing. She opens the box and sees that the comics can be split into three equal shares, plus one extra comic. She bribes the spying little brother with a comic, takes one-third of the remaining books, and goes back to sleep.

Then, an hour after that, the third sister does the same thing yet again, bribing the spying little brother with a comic, taking one-third of the remaining books, and returning to bed.

In the morning, none of the sisters want to reveal that they’ve already taken any comics. So they each pretend to be surprised when they look in the box. They dump out the comics and realize that the number of books can easily be split into three equal shares along with one extra comic, which they gave to their happy little brother.

What’s the smallest possible number of comic books that could have been in that mystery box when it first arrived? And how many comics does each sibling have?

>What’s the smallest possible number of comic books that could have been in that mystery box when it first arrived?
79
>And how many comics does each sibling have?
when 1st sister raided the stash (79 left): S1 has 26, LB has 1
when 2nd sister raided the stash (52 left): S2 has 17, LB has 2
when 3rd sister raided the stash (34 left): S3 has 11, LB has 3
morning (22 left): S1 has 33, S2 has 24, S3 has 18, LB has 4

i have 2 brute force solutions:

---

when it is said that X can be divided in 3 equal parts with 1 leftover, it means X%3=1
so after one sister raids the stash, bribe the lil bro and takes 1 third, there is (X-1)*(2/3) comics left
this is repeated 3 times, and each time the leftover must an integer number which modulus 3 must be 1
this can easily be brute forced, the first integer to satisfy these criteria is 79

---

lets invert the previous function: if there are currently X comics leftover, before the previous sister raided the stash there were (3X+2)/2 comics
let F(x) = (3x+2)/2
we know that the final number of leftover comics before the morning was an integer which modulus 3 equals 1
the smallest natural number which mod3=1 is 1
we also know that 3 raids happened
so we can simply brute force all possible final values of leftovers, starting from 1, and increasing by 3, by recursiing F(X) 3 times and the first integer it yields is the number of stating comics
meaning we want to know the smallest value of x for which F(F(F(x))) is an integer
turns out the smallest x is 22 (number of leftover comics in the morning) and the result is 79 (initial number of comics)

P16576
That's correct!
https://www.popularmechanics.com/science/math/a41335543/mystery-box-riddle-solution/

A baker’s dozen (thir*****) bagels have the property that any twelve of them can be split into two piles of six each, which balance perfectly on the scale. Prove that all the bagels have the same weight.

Nobody asked.

seems pretty obvious to prove tbh
if there is 6 on each side of the scale plus one leftover
then for the property to hold true, the leftover must have the same weigh as every other bagel
if that wasnt the case, if you swapped any bagel with the leftover, the scale would tip, meaning the property is false
and since the property must be true for any 12 bagels on the scale, its the same as saying it must be true for any one bagel left over
therefore all of them must have the same weigh

P16097
Looks good to me.

The hour and minute hands of a clock are indistinguishable. How many moments are there in a day when it is not possible to tell from this clock what time it is?

At least 11 times a day when the two hands are at the same angle.
Can't really think of any other moments. Can't proof it tho :(

P15988
That would make it impossible to distinguish the hour and minute hand, but it wouldn't stop you from telling the time. (Maybe you got it backwards? We're looking for the moments when it's [bold: not] possible to tell the time.)

>How many moments are there in a day when it is not possible to tell from this clock what time it is?
According to an abstract clock model, infinite. Because the model must have high enough resolution to encode the arm positions of _any_ clock, which means it must have infinite, in order to account for any possible clock in the world. The hour hand does not snap to any certain position either, in most clocks, if that's what you believed when you made this thread.

P16024
It does not follow from the fact that the clock has infinitely many possible positions that there are infinitely many moments at which you can't tell the time.

time doesn't exit

P15929
Assuming it's a tetrahedron. All triangles are equilateral. Treat all edges as length 1 for now.
Brute force linear algebra to solve for edges going up:
<e1,[0 1 0]> = 1/2 (right face)
<e2, [0 1 0]> = 1/2 (hidden face)
<e1, e2> = 1/2 (left face)
e1 - e2 = [1 0 0] (top edge horizontal, and perpendicular to bottom edge)
|e1| = 1, |e2| = 1
e1 = [a b c], e2 = [x y z]
b = 1/2, y = 1/2, c = z
a - x = 1
a^2 + 1/4 + c^2 = 1
x^2 + 1/4 + c^2 = 1
a^2 = x^2
choose a>0, a = -x
a = 1/2
-1/4 + 1/4 + c^2 = 1/2
c = 1/√2
So the answer is:
(pi * 10cm) * (2.5cm + (5cm * 1/√2) + 2.5cm) - 4 * 4/3 * pi * (2.5cm)^3
= 6.3523 cm^3
Felt harder than it should be.

Oops, it's actually:
(pi * (5cm)^2) * (2.5cm + (5cm * 1/√2) + 2.5cm) - 4 * 4/3 * pi * (2.5cm)^3
= 408.58 cm^3

P15954
That agrees with what I got.

If only two immortal humans distributed randomly on the Earth's landmass survived a collapse of the human population, what would be the most likely time frame for them to find each other?

Let's assume all radio systems were destroyed or they just didn't know how to use them or weren't listening to a long range radio for messages.

How would you estimate or predict the range of time it would likely take for them to find each other?

And what is your prediction?

I guess it would be some sort of calculation of movement speeds and the chances of them being within range to hear or see each other or signs of each other's existence.

Probably they wouldn't move randomly, but they'd move around bodies of water. Eventually they might use a ship or just swim across the oceans to look there too.

It could be millennia, huh?

Could go down significantly if they're both Muslims and undertake the Hajj.

Some other good places+times to look for people:
https://en.wikipedia.org/wiki/List_of_largest_peaceful_gatherings

A colony of chameleons currently contains 20 red, 18 blue, and 16 green individuals. When two chameleons of different colors meet, each of them changes his or her color to the third color. Is it possible that, after a while, all the chameleons have the same color?

Once any two of the groups have the same number of individuals, they can all pair up to eliminate both groups. So, can you make two of the groups the same size? [spoiler: No, because every time two chameleons meet, the difference between each pair of group sizes either stays the same (the two colors that meet) or changes by 3 (either color that meets decreases by 1, while the result color increases by 2). And none of the group sizes differs from another by a multiple of 3.]

P15496
Yep, you got it.

Red, Green, Blue are the only colors that exist.
[bold: The anwser is: OpenSUSE]

You want to connect the points A, B, C and D to each other by drawing the connection with a marker. But you want to use the minimum amount of ink of the marker.

What is the shape that you will draw?

P14037
Given the [tex: \geq 120^\circ] angles, I conjecture that for hexagons and above, going around the outside is optimal. I can say at least that it's a local minimum in the sense that it's not improved by changes like pic related (which was not true for the square).

That may be true, but your explanation isn't very convincing because there are topologies other than the one you used.
Picrel is [tex: 3\sqrt3 \approx 5.2>5] though

P14041
Yes, thus a conjecture. I haven't gotten anywhere on a general proof yet, but I think for the specific case of the hexagon, I think I can show the only cases we need to consider are these three. The first two options are P14041 and P14038, and the third is the second pic attached. Will post some more details later on how to calculate the points for each case and show that the points found are unique.

P14214
>second pic attached
ended up being the first pic

P14214
>Will post some more details later on how to calculate the points for each case and show that the points found are unique.

The solution has to be some sort of tree, and as P13897 pointed out, we can represent any tree as a tree in which each point except the terminal ones has three neighbors, with some points confounded. Every point except the terminal ones has to be at the Fermat point of its neighbors, which means that the lines to the three neighbors form 120° angles, except in the case where the three points form an angle greater than 120°, in which case the Fermat point is the vertex of that angle. This exceptional case can only occur when the vertex is one of the terminal points, since otherwise we obtain a smaller total length by changing the connections to the confounded points. For the regular hexagon, we don't have to worry about this case, since none of the points can be outside the convex hull of the points we're connecting, which limits any angle at one of the vertices of the hexagon to 120°.

The three possible ways of connecting six points by such a tree are shown in P14214. Let's consider the third case. We know point G which connects to A and B must lie on the circle centered at H, which by the inscribed angle theorem must be constructed to make m∠AHB = 360° - 2*m∠AGB = 120°. If we take the third line segment GJ and extend it in the opposite direction, it must intercept the circle at a point I located 120° away from A and B, again by the inscribed angle theorem. We can repeat this process with I and C to find a point L that must lie on the extension of JP. The same process for the lower half of the hexagon gives us another point R on the extension of JP. Drawing the line between them and finding their intersections with the circles gives us J and P. We can now connect J with I to find G, and P with O to find M.

F = (0, 0)
E = (1, 0)
A = (-1/2, √3/2)
B = (0, √3)
C = (1, √3)
D = (3/2, √3/2)
H = (-1/2, 5√3/6)
I = (-1, √3)
K = (0, 4√3/3)
L = (0, 2√3)
N = (3/2, √3/6)
O = (2, 0)
Q = (1, -√3/3)
R = (1, -√3)
J = (3/7, 5√3/7)
P = (4/7, 2√3/7)
G = (1/14, 11√3/14)
M = (13/14, 3√3/14)
AG = DM = CJ = FP = JP = 2√7/7
BG = EM = GJ = MP = √7/7
total length = 2√7 ≈ 5.3 > 5

So for the specific case of the regular hexagon, going around the outside is indeed the best.

Fif***** people are trapped aboard a ship that's going to sink in exactly 20 minutes. Their only chance for survival is the five-person life raft stowed on their vessel. To make matters worse, the waters around the ship are teeming with man-eating sharks, so swimming to safety is out of the question.

A round-trip to the nearest island and back to the ship takes nine minutes on the raft. How many people will live to see dry land?

There's time for 2 round trips before the boat sinks, which means 3 departures of the life raft. Each trip to the island carries 5 people, but someone has to bring the raft back each time, which brings the number of people saved to 13.

Was that the answer or is there some subtlety I missed?

P13680
P13809
Yes. It's correct.

P13809
Everyone survives, so long as the two people remaining on the boat are women
They're man-eating sharks, after all ;)

Is there no rope on this ship? If there is, it can be used to bring the raft back without anyone in it, therefore saving everyone.

Imagine a digital clock like the one shown below. How many times will the clock display three or more of the same number in a row over the course of one day?

In case you were wondering, the clock in this puzzle displays time on a 12-hour scale, not on military time.

34 total or am I missing something?

12:22
01:11
...
05:55
10:00
11:10
...
11:19

P13505
Yes.
P13578
Correct.

Don't these clocks display midnight as 00:00? That would make it 43

P12592
*42, I can't count

P13582
>the clock in this puzzle displays time on a 12-hour scale, not on military time
There is no 00:00 in 12-hour scale.

From which path should one learn?

What are you trying to learn about? Mesh networking?

Can you solve it?

P12019
>This one is famously impossible.
how? I remembered when I first got that toy (the wooden ones and way smaller than the one in the picture. So small that it fits in your waistcoat pockets), I would spend the whole day getting the numbers in order 1,2,3,4,5,6,7,8,9,10,11,12,13,14

and I did it! when I got those number blocks sorted, I would mess it up and start fresh.

It was fun as you have to move those number blocks until you get them right, you can't remove those blocks. I love how smooth it was when flicking those number blocks with your thumbs

P12020
And if some cruel prankster were to take your toy, disassemble it, swap exactly two number blocks as shown in the OP picture, and put everything back together, then it would be rendered an impossible puzzle.

P12030 and if that happened, I would cry and suck my thumb as usual.

P12019
This one made me realize I knew about even and odd permutations but didn't know how to prove elegantly that even permutations could only be done in an even number of swaps and odd permutations only in an odd number of swaps. Best I thought of myself was decomposing into cycles and doing a case analysis, which works but is not elegant.

Apparently the easy way of seeing it is by looking at whether the inversion number is even or odd, where the inversion number means how many pairs of elements in the list are in reverse order.

In this puzzle, if we consider the empty square to be a 16, we can count the number of pairs of squares which are in reverse order. In the OP picture there is exactly one such pair, the 15 and 14, so the inversion number is 1, an odd number. In the desired state, the inversion number is 0, an even number. Each move changes the inversion number from even to odd or vice versa, which is why the number of moves to swap only two blocks must be odd.

What's a fast way to look at a scrambled 15-puzzle and determine whether it is solvable or not?

How would you explain to an elementary school student why n + 0 = n?
How would you explain mathematical induction to a high school student?
And how can formal mathematical proofs be made more similar to the way a non-autistic person would explain things?

P11812
>How would you explain to an elementary school student why n + 0 = n?
i hold one pencil in my hand
i add another pencil
now i hold two
i add no more pencils
i still hold two
if you arent braindead you know the same is true for whatever amount of pencils im holding, they arent just gonna appear out of nowhere
the other question ill leave for the other anons

P11866
I would replace a pencil with a quill so instead of spending time thinking, I dip the quill into the inkwell over and over again.

P11866
Seems more like a scientific demonstration than a mathematical proof, though. It's based on empirical observations that pencils and other counted objects don't appear out of thin air. Of course that distinction may not be important for explaining it to an elementary schooler.

I guess the question is, what sort of mathematical postulates really deserve being treated as self-evident? Do any of them?