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#1 Exercises » A Medley of Melodramatic Problems » 2017-03-19 14:39:18

Mathegocart
Replies: 1

1.If gif.latex?abcd%3D25%21, then what is the minimum of a+b+c+d?
2.   What is the minimum of gif.latex?%28x+1%29%5E2+%28x-1%29%5E2+%28x+2%29%5E2+%28x-2%29%5E2+%28x+4%29%5E2+%28x+5%29%5E2
3. Find the least positive integer n such that n and n+1 have prime factorizations of exactly 5(not necessarily distinct) prime factors.
4.Ted flips five fair coins. The probability of Ted getting more heads than tails is m/n where m and n are relatively prime. Find m+n.
5. What is gif.latex?2%5E%7B2245%7D+3%5E%7B2%5E%7B15%7D%7D+5%5E%7B14%7D mod 14?
6.gif.latex?%5Cfrac%7B1%7D%7B2%5E2%7D+%5Cfrac%7B2%7D%7B3%5E2%7D+%5Cfrac%7B3%7D%7B4%5E2%7D+%5Cfrac%7B4%7D%7B5%5E2%7D+%5Cfrac%7B5%7D%7B6%5E2%7D+%5Cfrac%7B6%7D%7B7%5E2%7D
7. Alice chooses 1 positive integer from the set [1,1000]. She chooses another number from that set. What is the probability that the Harmonic Mean + the Arithmetic Mean of these 2 numbers is greater than 510?

#2 Re: Help Me ! » Geometry Angle Problem » 2017-03-15 06:57:35

bobbym wrote:

Hi;

Get better fast and when you grow up move to a warmer climate like Florida.

bit of trig and geometric innovation

Unnecessary! You can solve for the angle at A easily. Lots of times when those types pose problems the only way they can make it difficult is by providing misleading diagrams. You can defeat that by drawing a good diagram. Either you can call forth the powers of Giotto within or you can use the Gebra.

My headache is alleviating a little bit, atan(10/4).

#3 Help Me ! » Geometry Angle Problem » 2017-03-15 00:41:09

Mathegocart
Replies: 4

KlDhezQ.png..
I am supposed to find angle A with a bit of trig and geometric innovation.. but as my flu has only abated now, I have only made these observations..

#4 Re: Help Me ! » Harry Potter and the Age of Man » 2017-02-26 10:25:43

Humans are within 67% in sentence 14,896.

#5 Re: Help Me ! » Harry Potter and the Age of Man » 2017-02-26 09:50:59

I would, I've seen the mistake committed in here.

#7 Re: Help Me ! » Harry Potter and the Age of Man » 2017-02-26 09:32:08

Humans appear on year 4,539,900,000, so (4,539,900,000)/(304,780) = 14895.662445    i.e, the 14,895th sentence

#8 Help Me ! » Harry Potter and the Age of Man » 2017-02-26 01:07:14

Mathegocart
Replies: 9

Hello my fellow beneficiaries... I'd like to confirm my solution to this problem
The best estimate for the age of the Earth put it at 4.54 billion years. Modern humans have been here for 100,000 years. Harry Potter and the Deathly Hallows has 784 pages, and each page has 19 sentences. If the Earth's existence, all 4.54 billion years of it, were on a 784 page book with 19 sentences per page, on what sentence and page would humans appear on?
Through a proportion:


we obtain that humans appeared on the 14,895th sentence and the 783rd page.
Correct?

#9 Re: Help Me ! » Toothpick problem: » 2017-02-16 06:11:59

thickhead wrote:

Perhaps I meant- why doesn't my strategy work?

#10 Re: Help Me ! » Toothpick problem: » 2017-02-15 22:50:18

bobbym wrote:

Hi;

I do not understand the question. How do you know how many rows there are? Or how many columns?

#11 Help Me ! » Toothpick problem: » 2017-02-15 14:52:01

Mathegocart
Replies: 8

Anyhow I solved this problem:
We connect dots with toothpicks in a grid as shown in Figure 1.14. If there are 10 horizontal toothpicks in each row and 20 vertical ones in each column, how many total toothpicks are there?
ee8309c2c96c143eddcf6169511ac63ad9979235.png
The ans says 430 toothpicks, but I my method does not concur, for some peculiar reason.
Here's how I did it: The first row of squares contains 31 toothpicks, additional ones add 21 toothpicks. Since we add 9 more, 21*9=189, so 189+31=220 toothpicks. I feel like I'm missing something crucial, but I can't figure it out currently.

#14 Re: Help Me ! » Beach volleyball - how many combinations of 5 pairs can I make from 10 » 2017-02-08 14:17:14

bobbym wrote:

Hi;

I coach beach volleyball. We field 5 pairs of athletes. How many possible combinations of 5 pairs I can make from 10 athletes?

Another GF?

#15 Re: Help Me ! » Intriguing Combinatorics problem » 2017-02-05 00:49:18

My code was:

                which generated these quintuplets..

#16 Re: Help Me ! » Intriguing Combinatorics problem » 2017-02-04 15:33:44

THat should be enthralling, let me code it tomorrow. ahhahah an A8-6410....

#17 Re: Help Me ! » Intriguing Combinatorics problem » 2017-02-04 15:08:10

What are the specifications of your desktop? I am still currently thinking of the solution(late here), did you use generating functions?

#18 Re: Help Me ! » Intriguing Combinatorics problem » 2017-02-04 12:17:57

nice? quite inelegant but gets the job done.

#19 Help Me ! » Intriguing Combinatorics problem » 2017-02-04 11:23:54

Mathegocart
Replies: 10

1. How many quintuplets of numbers (a,b,c,d,e) where a,b,c,d and e are either -1,0, or 1 such that a + b^2 + c^3 + d^4 + e^5 = 1? Any quick way to do that? had ~5 mins to do this problem and I don't think I've got all these quintuplets.

#20 Re: Help Me ! » Combinatorics problem » 2017-02-04 11:14:54

bobbym wrote:

Hi;

This will be a long reply and I still will only scratch the surface.

Depends on how you define simple. The reason for the generating function is because it eliminates the need to reason about the problem. Also, it can do much more difficult problems than this one without increasing the complexity of the solution. We turn a combinatorics problem into a computational one. Computers do not reason, they can not assist us when we reason as you did. But they can compute, which is one reason to favor the gf approach. Your computer can help you!

As for why I use them on even smaller ones is because of the teakettle principle. Why is it considered simpler to know dozens of tricks for each type of problem when the gf will do them all?

And I don't see how come it's mandatory, unless I'm missing something...

I stress it first because the gf approach is a general method for solving counting problems. The same method does them all, difficult or simple, it does not matter.

Because they can be done with a computer the answers are error free and quick. They fit nicely into a framework we call Experimental Math. Not only do we get the answer, we have proof and we know we did not make a mistake. Of course, I can use both methods and so can lots of other people, but my weapon of choice is the gf.

When I lived in Vegas I was a professional player for 20 years. I found that it was necessary on a daily basis to solve tough combinatorics and probability problems. Problems that even experienced mathematicians would sometimes get wrong. I found that a programmer who could hardly code beyond a beginners level could often solve such problems easily and quickly when math reasoning failed. It was the beginning of my education into EM.

So to sum it up, when you have to solve a quadratic which is better? Completing the square, factoring or the formula. Well, if you have been following along the answer is simple - the quadratic formula!

I have a preference for algebra instead of ugly combinatorics, gfs sound interesting to learn as a general approach.

#21 Re: Help Me ! » Combinatorics problem(board) » 2017-02-04 11:09:54

Thought process: we choose 2 blocks from 49, so 49 choose 2. Counting vertical and horizontal squares, we find 42+42=84. so 84/(49,2)

#22 Re: Help Me ! » Combinatorics problem(board) » 2017-02-04 11:04:51

bobbym wrote:

Hi;

haidi wrote:

First off, try not to think about formulas, use just plain logic. I assume bobbym is asking you to do the same.

Well not exactly, he has the right answer. It all depends on how he interprets the word adjacent... I was just trying to get him to explain a bit. Or maybe even show a bit of EM...

Adjacent isdefined as two squares bordering a side.

#23 Help Me ! » Combinatorics problem(board) » 2017-02-02 22:55:10

Mathegocart
Replies: 11

There is a 7x7 board. I choose 2 points randomly. What is the probability that     
these 2 points are adjacent?
  There are 49 choose 2 ways to choose the 2 points, and there are 42+42 possibilities for adjacent sides- so the probability is 84/(49,2)?

#25 Re: Help Me ! » Combinatorics problem » 2017-02-02 09:27:46

Yes, I had to insert another dongle... frustrating..
So I presume GFs are used for a myriad of things, including combinatorics and discrete maths in general?

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