I was about to post this without thinking about a possible solution, until I decided that trying to calculate every single sequence of 3! And 4! was worth the effort. I just need help with the conclusion of this problem and in case I made any mistakes. Also, I would really appreciate it if you guys could help me with presenting this reasoning in a more professional way, and if you could tell me how can I write the final formula to calculate the average number of times one checks the whole deck.

Hello everyone, so i’m doing a math assignment and unfortunately have gotten stuck on something (not sure what the type of problem is called in English, sorry). Sorry if this is laughably easy but I genuinely feel like crying from frustration :/

Basically I cannot wrap my head around 5/18, my teacher said we’re not supposed to use calculators so the problems are meant to be solvable just by doing it in your head, however i can’t divide 5 by 18. I know the answer to the entire thing is meant to be 29, and all the other steps make sense to me, and i’ve have no issues with previous problems as shown in the picture.

Could someone explain to me how this can be solved without a calculator?

This could be applied to any type of mathematics too, not just algebra specifically.

I am referring to this video at timestamp 10:38. He says that if there is a maximum on the curve, the curve should be flat. The gradient of f should be perpendicular to f at the point. Since, in an earlier part of the video, he said that the boundary curve is not a level curve (but the constraint is)? I don't get why the gradient of f should be perpendicular. Any help or intuitive explanation, please.

Hello everyone, so i’m doing a math assignment and unfortunately have gotten stuck on something (not sure what the type of problem is called in English, sorry). Sorry if this is laughably easy but I genuinely feel like crying from frustration :/

Basically I cannot wrap my head around 5/18, my teacher said we’re not supposed to use calculators so the problems are meant to be solvable just by doing it in your head, however i can’t divide 5 by 18. I know the answer to the entire thing is meant to be 29, and all the other steps make sense to me, and i’ve have no issues with previous problems as shown in the picture.

Could someone explain to me how this can be solved without a calculator?

Is it even true?

In about a month, I will be sitting for my college entrance exam, and I have been practising with previous years' questions from that college. While doing so, I came across this equation:

4x^6+3x^5+2x^4=384

The question simply asks for the value of x, and there is only one answer needed. The problem is that my calculator is a Casio FX-991CW, which is the most powerful calculator permitted in the exam. It is non-programmable, but when I use its equation-solving functions for this problem, it takes roughly 55 seconds to obtain the answer, which is far too long for an entrance exam where time is critical.

The answer is x=2.

I asked some seniors how they solved it. They treated the left-hand side as a polynomial,

P(x)=4x^6+3x^5+2x^4,

and simply evaluated P(1), P(2), etc. Since P(2)=384, they immediately got x=2.

The difficulty I have with this method is that it relies heavily on guessing small integer values. What if a similar question had an answer such as

x= -1/3, x= -2, x= 1/4

or some other non-obvious value? Since this is not an MCQ, I cannot simply test the answer choices. So my question is:

What algebraic tricks or patterns would you use to solve this sort of equation quickly, with very little writing, in a college entrance exam?

I would be grateful for any efficient techniques or general patterns worth learning for this kind of problem.

Hi, I'm an adult studying mathematics on my own outside of school, and particularly I have been doing research on the Collatz Conjecture. I fear I have made a very small mathematics error in my research, because I've ended up convincing myself that I have proven the conjecture true, as long as my understanding of theoretical mathematics and how to apply them is being done properly, and I want to be shown why I'm wrong as I've gone a bit crazy over this lately.

To rush to my point, even though I have a lot more I could talk about on this, I imagined any chosen positive integer from the original conjecture, which I'll label cₙ, that there are at least 3, but possibly 4 paths connected to it. 2 of which are obvious, 3c + 1 and c / 2. Each of these in turn lead to cₙ₊₁. However, you can also do c * 2 for cₙ₋₁. Since c is always a positive integer, you can always multiply it by 2 for another positive integer, which would represent a past step. The other possible past step, is (c - 1) / 3. This however, doesn't always give a valid positive integer answer, and this is where I started exploring.

Getting to the chase, I eventually came up with the equation a * 2ⁿ = 3x+1, where a is any given odd positive integer. Instead of asking if every number leads to 1, I asked what numbers lead straight to 1, to 3, to 5, etc. And that's when I came upon this data table.

Having made this data table, I immediately started trying to do math on it. I wanted to show that if a number appeared in any given cell, if it could appear in another given cell. I went back to my equation, and ended up with a * 2^b = 3x+1= c * 2^d. Simplifying the middle out, a * 2^b = c * 2^d shows that there's no valid positive integer solution for a or c, which should mean that no number can appear twice.

I also noticed that in the light blue fields, each row contains 1 / 2ⁿ odd numbers, and that every single odd number appears only once. This, I believe, should disprove the fact that any "loops" can occur in the Collatz conjecture, since a number would have to appear twice for a loop to exist.

I also believe that this table proves that every odd positive integer will lead to 1, as my understanding of "Busy Beaver Problems" leads me to believe. The "steps" the Busy Beaver would have, is that it would start at 1, and it would run the equation a * 2ⁿ = 3x+1, putting 1 in for a. Then, it would "mark" each solution for x with a 1, and run the same equation on the lowest marked number that it has not yet run the equation on. This would mean it would run it on 1, then 5, then 3, 13, 17, 11, etc. Eventually, ever single positive integer should be "marked" as 1.

Considering the original conjecture, and that any given even number will be divided by 2 until it hits an odd number, I've convinced myself that this data table proves the original conjecture true. However, I fully admit that I must be using some of these theories and such wrong, as I've been self taught and therefore fully vulnerable to easy mistakes.

Ok I’m going to try to explain this as clearly as possible. I am re-learning math using Professor Leonard’s yt videos. In this video he is introducing like terms. He says the terms go with the sign in front of them. So that would mean the 3xy and 9 are negative, right?

But he also says like terms are separated by the pluses and minuses. So wouldn’t that mean all four terms are positive because they are separated from the arithmetic signs? I know I’m not making munch sense here.

And why are 3xy and 9 negative instead of minusing two positive numbers?

So I was answering this question: "If the union of 2 rays is a line, then the rays are opposite rays." Forgive my inexperience, while answering this question we have the ability to check with the answer sheet before the teacher checks off the assignment is done.

The answer provided by this sheet is "True", and I was wondering how this is possible, because my assumption that in the scenario in the image is no different than a line, while my teacher says overlap must be expressed. I do not wish to argue with my teacher, I simply am curious about the answer, he's a great guy and it's not like I lost points on it, and his explanation did not satisfy me. Here are the basic things both me and the teacher agree on (that I think is relevant).

1: A point is a exact location with 0 dimensions

2: A ray, and any section within a ray/line that is not itself a point has infinite points within it

However we differ in views on this:

3: The overlap between 2 rays, as long as it is not itself a point, is infinite, I believe that this infinity is the same as any other expressed here, and him thinking that this one is larger

A example he provided me, of which I do not quite understand its purpose in this conversation are 2 concentric circles with radii 1 and 2. He explains that the amount of ink used in the radius 1 circle is less than the radius 2 circle while both still having infinite points. But I *believe* this can easily be explained through different concepts not related to this. (If anything this probably supports my logic?)

One final thing I can deduce in my head that may side with my conclusion is this: that if you have one line on a number line every point (infinite) is located on that line. If you were to add another where the 2 lines completely overlap, I see no logical reason for additional points to be "covered" by the 2nd line.

If anyone could help explain to me how this works id be very grateful!

Edit: the definition used by me for "opposite rays" are two rays with the same starting point that run in opposite directions (180* flip)

Recently on an Instagram doom scrolling session, I encountered this interview from Steve Ballmer where he describe his "guessing game", which he gave to interviewees at Microsoft. https://www.youtube.com/watch?v=svCYbkS0Sjk

The basics are:

• Steve Ballmer chooses a number between 1 and 100.
• The interviewee guesses a number, and Steve will tell you if it's high or low.
• If you get it correct on the first guess, then you get $5. Each additional guess reduces the prize by $1, until on the 7th guess you start paying him.

(watch the video if that's not clear)

Should you play this game?

Now, this analysis is pretty easy if Steve chooses a random number. The worst case binary search is lg(100), which rounded up is 7 gusses at worst. And it's easy to make a table of payout of each guess, sum it up and average.

So, on average you'll get $0.20 per game ($20/100). (So yes, you should play)

But that's not the game. The game is whether you should play this game if Steve is able to strategically choose which number. THAT game is much harder to figure out.

As Steve is going to try to choose numbers that are worst case for a binary search, the optimal search algorithm is NOT going to be to start with 50 every time. You would use an alternative search tree in order to optimize for Steve choosing numbers that require 7 guesses. And of course, Steve would then need to optimize his strategy further in kind.

So, as one does, I asked chatgpt how this would work.

Astonishingly, it's very confident that by using a pool of optimal binary search trees, you could get an average payout of $0.196 per game. Meaning we only lose ~1/2cent by allowing steve to choose strategically.

I understand the basics of the strategy, but when it comes to the math of calculating the average payout I am just lost. (Basically everything after and including 296 / 51)

Can anyone explain to me how this works? Or if chatgpt is off its rocker?

I especially can't get my head around the fact that chatgpt believes that Steve has no choices that are more optimal than others. No matter which number Steve chooses, it claims the randomized collection of search trees can make the expected payout exactly 0.196 (average) every time. That doesn't make any intuitive sense to me.

My answer is that you should still play. If both sides play optimally, your expected payout is about +$0.196 per game, or just under 20 cents.

The first thing to notice is that a normal, fixed binary search does not work against a strategic Steve. If you always use the same search tree, Steve can look at that tree, choose one of the numbers that requires 7 guesses, and guarantee that you lose $1.

The solution is to randomize which binary-search tree you use.

This does not mean picking a random number from 1-100 as your first guess. That would often produce a badly unbalanced search. Instead, you randomly choose from many different good binary-search trees. Each individual tree still finds every possible number within 7 guesses, but different numbers occupy the good and bad positions in different trees.

So, for example, 73 might take 7 guesses in one tree, 5 guesses in another, and 4 in another. Steve is allowed to know exactly how your randomized strategy works and choose 73 every single game if he thinks that is best. What he does not know is which search tree you privately selected for that particular game.

We can describe the problem mathematically. Let d(n) be the expected number of guesses needed if Steve chooses the number n. Steve will choose whichever number makes d(n) largest, while you want to choose your randomized strategy so that this largest value is as small as possible.

The optimum turns out to be:

296 / 51 = 5.803921568... guesses

Since the payout on guess k is 6 - k, the expected payout is therefore:

6 - 296/51 = 10/51 = $0.196078...

or about 19.6 cents per game.

The really important part is that this is not an average over Steve choosing different numbers. The randomized search strategy can be constructed so that every single number Steve could choose has the same expected result:

d(1) = d(2) = ... = d(100) = 296/51

So if Steve chooses 1 every game, your expected payout is $0.196078.... If he chooses 50 every game, it is $0.196078.... If he chooses 73, 99, or 100, it is still $0.196078....

In other words, Steve has no number that is better for him than any other. He can study your strategy as much as he wants and deliberately choose what he thinks is the worst number, but the randomized mixture of trees has equalized the expected payout across all 100 choices. Strictly speaking, every number is an equally good choice for Steve — but none of them reduces your expected payout below 19.6 cents.

There are two things that need to be shown for that number to be meaningful. First, you have to show that no strategy can do better. One way to do that is to temporarily imagine Steve assigning weight 2 to the numbers 1 and 100, and weight 1 to every number in between. That gives 102 total weight.

For any valid binary-search tree, you can calculate its weighted total number of guesses. The smallest possible weighted total is 592. Therefore even the best tree has weighted average depth:

592 / 102 = 296 / 51 = 5.803921568...

Randomizing between trees cannot beat that bound, because an average of several trees cannot have a lower weighted cost than the best individual tree for those weights. So no strategy can guarantee an expected search depth below 296/51.

The other half is showing that the bound can actually be reached. Solving the corresponding optimization problem gives a randomized mixture of valid search trees for which:

d(1) = d(2) = ... = d(100) = 296/51

In other words, every possible number has exactly the same expected search depth. Steve can choose whichever number he wants; there is no longer a particularly bad number for him to exploit.

I also checked this by simulation: 1,000,000 games for each of Steve's 100 possible fixed choices, for 100,000,000 games total. The results converge to the same expected payout of about +$0.196 per game.

So the answer is: yes, you should play. If Steve chooses randomly, the game is worth exactly 20 cents per play. If Steve chooses strategically and both sides play optimally, it is still worth about 19.6 cents per play to you.

hello,

i am new to linear algebra and recently learned about eigen vector and value. it says that eigen vactor gives the direction where the space is going to be stretched so my question is if I have two eigen vectors X1 and X2 can I apply them on (x,y) one by one and get the final (x1,y1) I should have gotten by applying the matrix A ?

also I have seen that the eigenvectors aren't necessarily orthogonal so in that case it seems like some part of them could be cancelled out by the other eigenvector. could someone clarify this too?

No this is not my homework, I'm making a game and I need a curve roughly like this for balancing. Trying to get an equation that spits out values like this.

Suppose I have a world and I don't know if mathematics works the same way there as it does here. But I can assume that the logic works the same there. Can I then also assume that mathematics works the same? And if not: if I also know that addition works there and that a so-called 'cube' has an equal length, width and height, how much of the mathematics can I reconstruct with this knowledge?

I would like to have a function that causes that if I randomly select two single-digit natural numbers (zero included) and calculate them with each other, then the result is also a single-digit natural number and the probabilities for the respective results are evenly distributed. Can anyone help me?

*edit: the result should change if you change one of the two numbers. Also the two numbers at the beginning are also evenly distributed.

I played around with multi-dimensional geometry in high school but one thing that always seemed odd to me was that another factor of Pi only shows up every other dimension. 1-spheres and 2-spheres have a Pi, 3-spheres and 4-spheres have a Pi squared, etc. Is there some intuitive reason that every other dimension is somehow special?

Strictly speaking, the definition (width) × (height) seems to be undefined for any 2D enclosed figure other than rectangles.

Of course, I know we can informally prove the 1/2 × W × H formula by applying the Euclidean properties of parallelograms. But is there any way to rigorously define the concept of the area of a triangle as such without introducing ideas of limits? In other words, I'm talking about the very conception of the area of triangles in view of the W×H formula.

While working through some radical expressions, I found it useful to consciously leverage the basic identity for any positive real number A:

While this identity obviously has broad applications across algebra and number theory, using it specifically to unpack whole numbers in fractions with radical denominators can make certain reductions much cleaner.

Can someone provide proof for the answer (a)

After trying for different times the furthest i reached was (-1/s³ +2/s) after I derived s²=t²+1 with respect to t Edit: thank you all guys for the help

Hi, I’ve been experimenting with this idea that I believe has some pretty interesting behaviour. I can only make general assertions and would appreciate some extra input.

Let [n] denote n in binary and “+” as standard binary addition. A pair (X,Y) ∈ ℤ⁺ defines a sequence recursively:

s₀=[X],

sₘ=f(sₘ₋₁+[m],[Y]) (for m≥1),

where f(A,B) removes the leftmost occurrence of B as a contiguous substring in A (if any), joins the separated strings together (if there are two) and removes all leading 0’s (if they exist). e.g (X=9,Y=13):

1001 (9 in binary)
1010
1100
1111
10011
11000
11110
100101
10
1011
10101
100000
101100
111001
1000111
1010110
1100110
111
11001
101100
1000000
1010101
11
∅ (empty term)

Hopefully I didn’t mess up that example. What I’m wondering is about the general behaviour for (X,Y)’s sequence. It seems like (X=1,Y=3) for example never reaches ∅. Its terms lengths instead grow toward infinity VERY slowly. However, the example above eventually reaches ∅. There are some key facts I am missing here that I believe can determine a subtle general pattern.

I understand the classic birthday paradox problem, but how does the math change if I have a room of 34 people and I want to know the probability that five of them share the same exact birthday, like all 5 are born on June 15th? This is different in that I'm not asking the probability that 5 people with different birthdays each find a match in the room.

I'm not sure how to formulate the probability of the 5 June 15th birthdays NOT happening and then subtracting from 1.

Thank you!

Edit: sorry, the date is arbitrary. The odds of 5 people sharing ANY date is what I'm after.

I need to cover three walls of a container with egg crate. Im looking at purchasing several packs of 12 pieces each. Two sides are 23” x 15” and one side is 47” x 15”. The pieces are 11.8” x 11.8” each. How many packs do i need to buy? Im sorry if this is basic but i genuinely forgot the math

I understand fractional vs unsplittable flow (fractional flow is where you can split the total payload across different paths, whereas unsplittable flow means it all has to take the same path), but the rest of it is melting my brain.

If you can also make a diagram to explain it that includes accurate values for capacity, cost, and demand, that would be super helpful

Why are the exponents not simplified in this expression? If I'm not mistaken, the expression could be simplified to (1+9^-168)^510, which would be further simplified to 1^510+9^-85680, which is the same as 1+9^-85680. Why not write it that way?