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**Mathegocart**- 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?

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.

Bob

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..

Correct? Incorrect?

**Mathegocart**- Replies: 33

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.

Explanation:

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

thus...

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.

bobbym wrote:

Hi;

I am getting something around 46..

**Mathegocart**- Replies: 5

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.

thickhead wrote:

Verified, correct.

bobbym wrote:

Hi;

Correct, verified.

bobbym wrote:

8) You mean 2 of the 5.

Indeed.

**Mathegocart**- 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:

(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.

What is a TH player?

I will be developing a new thread.

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

bobbym wrote:

Hmmm, I am 96 years old.

I doubt the veracity of this statement..

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

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?

(5)

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.)

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!

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.

thickhead wrote:

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

bobbym wrote:

Hi;

Hello, I vastly disagree with you on 1, I could not get a negative solution.

Well done, and a is indeed an integer.

**Mathegocart**- Replies: 36

1. The sum of the first 2 terms of an geometric series is 7. The sum of the first 6, however, is 91. What is the sum of the first 4 terms?

2. Find the sum of .(Please put your solution in fractional forme.)

3. By adding the same constant A to 20, 50 and 100, a geometric progression results. Find the common ratio and the constant A.

4. The first three numbers of a set A form a arithmetic progression. The last three form a geometric progression. The sum of the first and last elements of A is 16. The sum of the middle two numbers is 12.Find these 4 elements, but be warned that there is more than one solution. BONUS to anyone who finds both of them.

All elements of A are integers.

5. The sum of the first 2011 terms of a GEOMETRIC PROGRESSION is 200. The sum of the first 4022, however, is 380, Find the sum of the first 6033 terms.

6. Find the coefficient of the term that contains x^4 in the expansion of

7. Find the least positive integer whose digits add to a multiple of 27 yet the number itself is not a multiple

of 27. For example, 87999921 is one such number.

8.How many non-congruent isosceles triangles (including equilateral triangles) have positive integer side

lengths and perimeter less than 20?

9. The Bobbym Lizard Garden Store sells grass seed in ten-pound bags and fifteen-pound bags. Yesterday half

of the grass seed they had was in ten-pound bags. This morning the store received a shipment of 27 more

ten-pound bags, and now they have twice as many ten-pound bags as fifteen-pound bags. Find the total

weight in pounds of grass seed the store now has.

10. The zeros of the function f(x) = x^2-ax+3a are integers. What is the sum of the possible values of a. If there are NONE, respond with NONE.

thickhead wrote:

(2)

No.of 0's in unit place=9999 div 10=999

No. of 0's in tens place=10*(9999 div 100)=990

No. of 0s in hundred place=100*(9999 div 1000)=900 Total=2889

No of any other digit=( 9999 div 10+1)+10*(9999 div 100+1)+100*(9999 div 1000+1)+1000*(9999 div 10000+1)=4000

Strange, because with casework, I have gotten the same solution as bobbym has. I will check if I made a naive mistake, but..

thickhead wrote:

Wait, what? One of his acquaintances is named Liz?

bobbym wrote:

Hi Mathegocart;

The battle is, indeed, described in this manner.