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# New Maths Problems

Bikerman
OK time for a few more.
Here's the first

A creeper plant is climbing up and around a cylindrical tree trunk in a helical manner. The tree trunk has a height of 450 inches and a circumference of 40 inches.

If the creeper covers a vertical distance of 75 inches in one complete twist around the tree trunk, what is the total length of the creeper?
kelseymh
 Bikerman wrote: A creeper plant is climbing up and around a cylindrical tree trunk in a helical manner. The tree trunk has a height of 450 inches and a circumference of 40 inches. If the creeper covers a vertical distance of 75 inches in one complete twist around the tree trunk, what is the total length of the creeper?

This is a great problem! Someone with "a little knowledge" could probably solve it with calculus, setting up a function for the position along the creeper and integrating from 0 to 2pi, but there's a wonderfully elegant solution that requires no analysis at all, just "plug and play."

I'll send you a note with my answer, rather than spoiling the thread.
Peterssidan
I was going to post the answer but ...
 Quote: I'll send you a note with my answer, rather than spoiling the thread.

... I guess I should do the same.
Bikerman
I can confirm that two people - kelseymh and Peterssidan - have the correct solution.
Well done to them - it doesn't require any fancy maths but it is a nice little puzzler...
Bikerman
Another problem.
A group of 4 card-game players are always arguing about who should go first. One of their number is a mathematician and he arrives at a solution by designing four 12 sided dice. The dice will give a definitive order, no ties, and do it fairly.*
The first dice has the following numbers: 1, 8, 11, 14, 19, 22, 27, 30, 35, 38, 41, 48
The second dice has the following numbers: 2, 7, 10, 15, 18, 23, 26, 31, 34, 39, 42, 47

What numbers are on the faces of the other two dice?

*regardless of what subset of dice are rolled against one another, each player will always have an equal chance of rolling the high number (and thus being the "starting player"). Put another way, two, three, or four players may each take one of the dice, roll it, and each will always have a 1/2, 1/3, or 1/4 chance, respectively, of having the highest result.
dharmin
This a question from the Common admission test (CAT) exam for entrance to IIMs...

I cant figure out how to solve this, or these type of questions..
if you know how to solve, please show me the methord plz!

What is the remainder when 1! + 2 × 2! + 3 × 3! + 4 × 4! + … + 12 × 12! is divided by 13?

A. 1
B. 12
C. 11
D. 0
likeabreeze
 dharmin wrote: This a question from the Common admission test (CAT) exam for entrance to IIMs... I cant figure out how to solve this, or these type of questions.. if you know how to solve, please show me the methord plz! What is the remainder when 1! + 2 × 2! + 3 × 3! + 4 × 4! + … + 12 × 12! is divided by 13? A. 1 B. 12 C. 11 D. 0

when you find out that N*N! = (N+1-1)*N! = (N+1)! - N!, everything is clear!
----------------------
1 * 1! = 2! - 1!
2 * 2! = 3! - 2!
3 * 3! = 4! - 3!
4 * 4! = 5! - 4!
.....
12 * 12! = 13! - 12!
----------------------
Add them together, we can get 1! + 2 × 2! + 3 × 3! + 4 × 4! + … + 12 × 12! = 13! - 1
so, (1! + 2 × 2! + 3 × 3! + 4 × 4! + … + 12 × 12!) MOD 13 = (13! -1) MOD 13 = 12
The answer is B.
likeabreeze
 Bikerman wrote: I can confirm that two people - kelseymh and Peterssidan - have the correct solution. Well done to them - it doesn't require any fancy maths but it is a nice little puzzler...

a nice little puzzler?
but I have to use the Pythagorean theorem.
Bikerman
Yes, that seems to be the best (and only) solution...
Afaceinthematrix
 kelseymh wrote: This is a great problem! Someone with "a little knowledge" could probably solve it with calculus, setting up a function for the position along the creeper and integrating from 0 to 2pi, but there's a wonderfully elegant solution that requires no analysis at all, just "plug and play." I'll send you a note with my answer, rather than spoiling the thread.

I solved this problem using a little bit of calculus. I parametrized a function in R3 for a helix with those specifications but you have to integrate from 0 to 12pi (since you have to go around the tree 6 times to make it to the top) or multiply your answer by 6. I also solved it using a little basic geometry that you'd learn as a child.

I'd be interested in seeing your "plug and play" solution. How do you "plug and play" with something like this? I sent Bikerman my solutions; hopefully they were correct.

I have a problem, now. I am posting it because I saw it on a website; it didn't have a solution, and I seriously cannot figure it out.

 Quote: In the middle of the ocean there is a small island where everybody is part of this strange religion. One of the rules the people with this religion have to follow is the following : if you ever find out that you have blue eyes, you have to kill yourself that day at midnight. Because of this rule, people never talk about other people eyes and there are no mirrors in the island. So they can all live happily without knowing what the color of their own eyes is. They do see other people eyes, moreover, they all meet every day at noon where they have the chance to look at each other. One day, a foreigner comes to the island and at the noon meeting he tells everybody: "at least one of you has blue eyes". The questions is what happens in the island thereafter. Assume, for example, that in the island there are 100 people with blue eyes and 100 people with brown eyes. Another important assumptions is that everybody in the island is a perfect logician.

This doesn't make sense. If one person has blue eyes then that person can see that no one else does and then kill himself or herself. If two or more people have blue eyes then everyone is aware of someone else with blue eyes anyways and so that foreigner is saying absolutely nothing that everyone doesn't already know. I seriously do not get this. It's bugging me.
kelseymh
Afaceinthematrix wrote:
 kelseymh wrote: This is a great problem! Someone with "a little knowledge" could probably solve it with calculus, setting up a function for the position along the creeper and integrating from 0 to 2pi, but there's a wonderfully elegant solution that requires no analysis at all, just "plug and play." I'll send you a note with my answer, rather than spoiling the thread.

I solved this problem using a little bit of calculus. I parametrized a function in R3 for a helix with those specifications but you have to integrate from 0 to 12pi (since you have to go around the tree 6 times to make it to the top) or multiply your answer by 6. I also solved it using a little basic geometry that you'd learn as a child.

I'd be interested in seeing your "plug and play" solution. How do you "plug and play" with something like this? I sent Bikerman my solutions; hopefully they were correct.

You don't need calculus! Bikerman gives you the circumference, so you can unwind the problem to be just a simple right triangle: Short side is 40 in (circumference), long side is 75 in (vertical height). The length of the creeper over one winding is sqrt(40^2+75^2) = 85 in. The whole trunk is 450/75 = 6 windings, so the total length of the creeper is 85*6 = 510 in.
Afaceinthematrix
kelseymh wrote:
Afaceinthematrix wrote:
 kelseymh wrote: This is a great problem! Someone with "a little knowledge" could probably solve it with calculus, setting up a function for the position along the creeper and integrating from 0 to 2pi, but there's a wonderfully elegant solution that requires no analysis at all, just "plug and play." I'll send you a note with my answer, rather than spoiling the thread.

I solved this problem using a little bit of calculus. I parametrized a function in R3 for a helix with those specifications but you have to integrate from 0 to 12pi (since you have to go around the tree 6 times to make it to the top) or multiply your answer by 6. I also solved it using a little basic geometry that you'd learn as a child.

I'd be interested in seeing your "plug and play" solution. How do you "plug and play" with something like this? I sent Bikerman my solutions; hopefully they were correct.

You don't need calculus! Bikerman gives you the circumference, so you can unwind the problem to be just a simple right triangle: Short side is 40 in (circumference), long side is 75 in (vertical height). The length of the creeper over one winding is sqrt(40^2+75^2) = 85 in. The whole trunk is 450/75 = 6 windings, so the total length of the creeper is 85*6 = 510 in.

That's exactly what I did! I said:

I solved this problem using a little bit of calculus. I parametrized a function in R3 for a helix with those specifications but you have to integrate from 0 to 12pi (since you have to go around the tree 6 times to make it to the top) or multiply your answer by 6. I also solved it using a little basic geometry that you'd learn as a child.

I just thought that you'd have a different cool solution because I don't really consider this to be "plug and play." Oh well, disappointing :( I love seeing multiple solutions to problems and solutions that I didn't see. I did exactly what you did and got the same answer. I may have visualized it differently than you, though. I imagined the tree being unrolled - like you'd unroll and rolled up carpet. That will create a right angle between the ground the a fixed line segment. The helix connects two points, one at the top and one at the bottom, and is essentially stretched out - connecting the two ends of my lines and creating a right triangle. Then I used the Pythagorean Theorem and multiplied by 6... So essentially what you did... I also did:

To parametrize a helix we have:

x(t) = (20/pi)cos(t)
y(t) = (20/pi)sin(t)
z(t) = (75/2pi)t

My reasoning is that the radius is 20/pi and that the pitch is 75 and z(t) = bt where 2pi*b is the pitch.

Now, we need to find an arc length from 0 to 2pi and then multiply that by 6. (We can also find arc length from 0 to 12pi but that's a little extra work). To find the arc length, we need to integrate the following from 0 to 2pi.

|X'(t)| where || denotes norm and X is the vector containing (x(t), y(t), z(t)).

Well, that is simple. We do:
sqrt(400/(pi*pi) + 5625/(4*pi*pi))*2pi (the last 2pi is from integrating the norm - which is just a number - from 0 to 2*pi.

We can turn that last 2pi into sqrt(4*pi*pi) in order to bring it into the big square root function.

So now we have sqrt(1600 + 5625) which is 85. Multiply that by 6 and we get the same answer - 510. Q.E.D.
shashwatblack
first we'll need to calculate the total number of turns the creeper makes,
this is easily calculated by dividing the total height of trunk by the vertical distance the creeper covers in one turn,
that is, number of turns (n) = 450/75 = 6

now we find the length of creeper in one turn,
this as it turns out is calculated by, length² = circumference² + height²

imagine straightening out the curve of the trunk to get a 2d figure,
you get a right angled triangle with the creeper as the hypotenuse, the circumference of cross section of trunk as base and the height covered in one turn as perpendicular.
as Pythagoras says h² = p² + b²
so we get length = 85 inches

so now total length of creeper = 85 * 6 = 510 inches..
mazito
i use the same aproach that shashwatblack, and same result, i like this one for my high school students as a gift problem in one exam or for a challenge in class or homework, here in Mexico the math level is so poor, i recibe students that hardly read properly, so if they (i think 45 % of each class) cant read how i expect that they learn calculus.

the educational system in Mexico is so hard, the teachers cant give reprovatory notes in junior high or before, then when the students face the high school they came with a vicious way of see the school, then try to convince they is hard and in the school that i teach is just two years with a decent math program, so i have a + 75 % of students that cant aprove, but when i find some students that actually want to learn to have a chance to go to the college is great, and then when i meet after they go to college is a good thing that they make it.

well i went for other side, sorry for the off topic

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