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#1 Help Me ! » The Three Hat Logic Problem » 2016-10-08 00:59:46

Replies: 1

I have been bombarded with a particularly perplexing logic problem which states as follows.
Middle, Rear and Front are standing in a line. A eccentric millionaire challenges them to a game. Choosing arbitrarily from 3 white hats and 2 black hats, the millionaire puts a hat on each of the three men. The men cannot see their own hats. The millionaire then asks each of the three the color of their hats. Rear, who can see two other hats, declines to give an answer(he cannot determine his hat's colour.). Middle, who can see only Front's hat, also declines to give an answer(he cannot determine his hat's colour.). Finally, Front, who cannot see any hats, states the color of his own hat.
If somebody decides to link me to the MIF "Three Hats Puzzle", keep in mind we have THREE white hats and TWO black ones.
Now I disprove that scenarios with Front having a black hat could happen in these conditions(Either Middle or Rear would be able to clearly define the colour of their hat.), and thus Front has a white hat. Wikipedia concurs.
However, I have noticed that I end with a convoluted, case by case solution. Can anyone provide me with a simpler solution(not case by case..)?

#2 Re: Help Me ! » Ln e ????? » 2016-10-05 11:06:29

Zeeshan 01 wrote:

Ok but how this  If there is -x how we go to this step -xlne and second step is lne^-x       ln(c.e^-x)

That is a property of logarithms. 2log(10) = log(10^2).
Could you please clarify your last step? What is c.e^x????

#3 Re: Computer Math » Geogebra - The earth and the string » 2016-10-05 10:57:44

I am happy I could incite and spark debate about the enigma, the Earth and the String.

#4 Help Me ! » Confirmation on Transversals » 2016-10-05 10:54:23

Replies: 1

Hello fellow genial friends,
in the diagram 6lNgRB6.png, is < 1 = < 3 and <2 = <4?

#6 Help Me ! » Logical Bachelor Problem » 2016-09-24 04:37:24

Replies: 1

1. There are three people, A,B, and C. They are waiting in a barren place where a person asks which of the three has the treasure.
A: Not me
B: I have it
C: Not B.
Strangely, 2 of them always lie, and only 1 tells the fruitful truth.
It is obviously A through casework, but is there another way?

#7 Re: Help Me ! » The Earth and The String » 2016-09-18 22:32:39

bob bundy wrote:

hi Mathegocart and thickhead

I've searched their site but cannot find any dimensions for their trucks.  Strange!  Wouild you buy a vehicle without knowing how big it is? Pictures with a trucker in shot suggest they are only about 12 ft tall.  If a truck really was 17 ft tall it wouldn't fit under UK motorway bridges.


ps. LATER EDIT:  This is how to cut off the main route from the rest of England to the Dover continental ferry port and the Channel Tunnel:

What a calamity! They look as if to fit under most US bridges here..

#9 Help Me ! » The Earth and The String » 2016-09-18 10:55:56

Replies: 38

A string is wrapper around the Earth's equator(i.e, the circumference)  and the two ends of the string just touch. Now suppose that another string is tied to the original string so that it is 100 ft longer. If this new string is placed around the equator and pulled tight so it is suspended over the earth, how high will the string be above the ground?
a. An atom high
b. A bowling ball high
c. A Mack truck high
d. Exactly 100 ft.
I believe it is C.
Let CE = Earth's circumference and let CS = String's new circumference
Let RE= Radius of the Earth and let RS = Radius of the string
CE = 2πRE
and CR+100 = 2πRS.
We want to find the difference between the two radiuses(aka height.)
C/2π = RE
C+100/2π= RS
RS - RE = (C+100-C)/2π.
RS - RE = 100/2π = 50/π
50/pi ≈ 17.
Therefore Mack truck, aka C, is the solution.
I would LaTeX the aforementioned equations but I'm in a hurry.

#10 Re: Help Me ! » Ant and crumb problem » 2016-09-06 22:51:25

bobbym wrote:


I am getting something around 46..

#11 Help Me ! » Ant and crumb problem » 2016-09-06 12:40:25

Replies: 5

7JQCciT.jpg has been given to me as homework to be done. It asks for me to give a proper explanation why as well. Optimal path refers to the path that would minimize the length of the voracious ant's travel. I am thoroughly befuddled by the "it includes the floor and ceiling" hunt as the optimal path does not seem to go through as presupposed.

#12 Re: Exercises » Mathematics Challenges, #2 » 2016-09-03 12:06:59

thickhead wrote:

Verified, correct.

#13 Re: Exercises » Mathematics Challenges, #2 » 2016-09-03 12:05:47

bobbym wrote:


Correct, verified.

#14 Re: Exercises » Mathematics Challenges, #2 » 2016-09-03 02:21:17

bobbym wrote:

8) You mean 2 of the 5.


#15 Exercises » Mathematics Challenges, #2 » 2016-09-03 02:12:58

Replies: 8

These questions are a compilation of physics-related mathematics problems. Enjoy!

1. In bobbym's lavish home, he has a triangular grandstand shown by Diagram 1.0:
CCFXLMI.png(A and a are the same length.)
Find where the center of mass is with calculus.

2. Bobbym is riding a sled on a gargantuan hill, and his speed is given by 29 - t^2 + t. How far will the valiant bobbym travel from t(time) =0 to when he stops at v =0?

3. What is the least positive integer with the property that the product of its digits is 5! ?

4.Find the sum of all the integers N > 1 with the properties that the each prime
factor of N is either 2, 3, or 5, and N is not divisible by any perfect cube greater
than 1.

5.If you roll six fair dice, let p be the probability that exactly three different numbers
appear on the upper faces of the six dice. If p = m/n where m and n are

relatively prime positive integers, find m + n.
5a. Say p is the probability that exactly four different numbers appear on the upper faces of the dice. How much smaller or larger is this probability?

6.Find the sum of all the digits in the decimal representations of all the positive
integers less than 10000.
7 Bobbym has some square tiles. Some of the tiles have side length 5 cm while
the others have side length 3 cm. The total area that can be covered by the
tiles is exactly 20124 cm squared.
. Find the least number of tiles that Bobbym can have.
8.Bobbym needed to address a letter to 27432 Mathematica Road. He remembered the
digits of the address, but he forgot the correct order of the digits, so he wrote
them down in random order. The probability that Bobbym got exactly two of the
four digits in their correct positions is m/n
, where m and n are relatively prime
positive integers. Find m + n.
9.Find the number of subsets of {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 22} where the
elements in the subset add to 49.
10. There is a triangle where all sides of the triangle is 1, and Bobbym picks 3 points from the triangle's area. What is the probability that the area of the three points is greater than .4? All three points are distinct.

#18 Re: Exercises » Progression Into Mathematics, #1 » 2016-09-03 01:39:50

Bobbym, would you be fine with some adept calculus problems in my physics thread?

#19 Re: Help Me ! » Geometric Series » 2016-09-03 00:18:01

bobbym wrote:

Hmmm, I am 96 years old.

I doubt the veracity of this statement..

#20 Re: Exercises » Progression Into Mathematics, #1 » 2016-09-02 12:41:44

Note: I am planning to develop a formidable armada of physics questions.

#21 Re: Exercises » Progression Into Mathematics, #1 » 2016-09-02 12:35:27

thickhead wrote:
Mathegocart wrote:

Well done, thickhead, but I am getting a similar but different answer. Let me verify mine again..

Where do we differ?


#22 Re: Exercises » Progression Into Mathematics, #1 » 2016-09-02 04:15:49

bobbym wrote:

There are actually 6 values for a:

{-4, -4, 0, 12, 16, 16}

each one corresponding to a different x. I decided to use the double -4 and 16 in the sum. That is where I got 36. I could not deduce from the question whether doubles were to be considered or not because no where in the question does it say distinct a's. Now, looking at the question and having to guess what Mathegocart wants I would go with 24 too.

That was the intended solution(24.)

#23 Re: Exercises » Progression Into Mathematics, #1 » 2016-09-02 04:14:39

thickhead wrote:

I chose common ratio r=sqrt(3) since I thought it would suit the irrational personality of Mathegocart.
But bobbym thought otherwise and chose r=2*sqrt(-1) as he thought there is nothing real but only imaginary value present in Mathegocart.

Irrational? My profound eccentricity is only relative to yours!
I am not real nor imaginary, I am truly complex.
Also I see you have adapted my signature into a more adept one. Too bad there is no LaTeX implementation in the sigs, for it would make for quite a grandeur look!

#24 Re: Exercises » Progression Into Mathematics, #1 » 2016-09-01 06:30:44

bobbym wrote:

Hello, I vastly disagree with you on 1,

People who disagree with me are eaten by gators, which seems fair.

I am sticking with the answer for one. There is also another answer for one which I did not put down.

Bobbym, I know you fear your nefarious lizard neighbors so I will infest your house with lizards. tongue

#25 Re: Exercises » Progression Into Mathematics, #1 » 2016-09-01 06:28:15

thickhead wrote:

Well done, thickhead, but I am getting a similar but different answer. Let me verify mine again..

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