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curlious
09-02-2004, 03:49 PM
How many digits are in the number (8^15)*(5^37) no calculator.

The radius of the earth is 3960 miles. What length of string would you need to wrap it around the earth at the equator?

Imagine you hang the string on a pole 6 feet high. How much more string would you need to encircle the earth at the equator?

Suppose you have a square with its perimeter equal in length to the circumfrence of the earth. If you draw a new larger square, whose sides are 6 feet away from the inner square. Would the perimeter of the new square be greater than, less than or equal to the total amount of string need to encircle the earth 6 feet above the equator as in the previous question?

Suppose you had nine eggs with one containing a jewel in the center and hence heavier. Given two balance scales of which you can make one measurement on each, how would you determine wich egg contained the jewel.


A square cake measures 15 inches by 15 inches by 3 inches tall and is frosted on the sides and on the top. How could you cut the cake into 5 pieces so that each piece has exactly the same amount of cake and frosting.?

A damsal in distress is stuck in a tower in a square castle surrounded by a square moat. The moat is 20ft across. The night has two planks 19ft long by 8 inches wide but no nails, screws, saws or superglue. Is it possible to rescue the damsal in distress.

Two rooms are connected by a hallway that has a bend in it so that it is impossible to see one room while standing in the other. One room has three light switches. Exactly one of the light switches turns on the light in the other room and the remaing two switches are not connected to any lights. What is the fewest number of times you would walk into the other room to figure out which switch turns on the light and how? (no guessing)

This was the beginning of a 1 cedit class that will be focusing on calc practice. It is supposed to be a fun way to hone your calc skills. None of these brain teasers require calc but if people are interested and we continue to do interesting problems like this in the future I will post each week.

PJYelton
09-02-2004, 06:06 PM
Suppose you had nine eggs with one containing a jewel in the center and hence heavier. Given two balance scales of which you can make one measurement on each, how would you determine wich egg contained the jewel.

Measure three eggs vs three eggs on one scale to narrow it down to only three (the three that tip the scale or the three left out if balanced). Then measure one vs one and the answer is the one that tips the scale or the one not weighed if they balance.


Two rooms are connected by a hallway that has a bend in it so that it is impossible to see one room while standing in the other. One room has three light switches. Exactly one of the light switches turns on the light in the other room and the remaing two switches are not connected to any lights. What is the fewest number of times you would walk into the other room to figure out which switch turns on the light and how?
Feel like I'm missing something here, couldn't you just flip one switch, walk in, walk back, switch another and walk in? If lit either time you know, and if still not lit its the one you never flipped.

XSquared
09-02-2004, 06:10 PM
Two rooms are connected by a hallway that has a bend in it so that it is impossible to see one room while standing in the other. One room has three light switches. Exactly one of the light switches turns on the light in the other room and the remaing two switches are not connected to any lights. What is the fewest number of times you would walk into the other room to figure out which switch turns on the light and how?
Turn switch #1 on for 5 minutes, then turn it off, and turn on #2. Walk into the other room. If the light is on, it's #2. If none of the lights are on, feel the bulb. If it's on, then switch #1 is the one. Otherwise, it's #3.

JaWiB
09-02-2004, 07:05 PM
A damsal in distress is stuck in a tower in a square castle surrounded by a square moat. The moat is 20ft across. The night has two planks 19ft long by 8 inches wide but no nails, screws, saws or superglue. Is it possible to rescue the damsal in distress.


Put one across the corner and another on that:
______
| /
|/\
|_ \|
|__|


(Not to scale :p )

Edit: Hmm my picture wont show up right...
Edit2: Ok I think that's close enough

PJYelton
09-02-2004, 07:28 PM
Suppose you have a square with its perimeter equal in length to the circumfrence of the earth. If you draw a new larger square, whose sides are 6 feet away from the inner square. Would the perimeter of the new square be greater than, less than or equal to the total amount of string need to encircle the earth 6 feet above the equator as in the previous question?

Without doing the math, its greater. The perimeter of a square grows by a factor of 8 relative to the "radius" while the circumference of a circle only grows by a factor of 2pi.

Magos
09-03-2004, 05:01 AM
3960 miles
6 feet high

Omg, convert to metric system world man :)

Sang-drax
09-03-2004, 05:37 AM
The radius of the earth is 3960 miles. What length of string would you need to wrap it around the earth at the equator?Don't know about miles, but the length of a string around the earth from the north pole to the equator would be 10,000.000 km give or take a micrometer.


Imagine you hang the string on a pole 6 feet high. How much more string would you need to encircle the earth at the equator?
12pi feet.
The square would require 6*8 feet.



(8^15)*(5^37)
8^15 * 5^37 = 2^45 * 5^37 = 10^37 * 2^8 = 10^37 * 256 = 2.56 * 10^39
So, 40 digits!

PJYelton
09-03-2004, 01:50 PM
The string around the earth and the square problem is very similar to a brain teaser I heard a lot during the first day of class. The question went something like two farmers have a wooden fence around their circular properties, one has a very small ranch the size of an acre, and the other has an enormous ranch the area the size of California. Now both want to push out their fences by 10 meters, which farmer needs more fencing to accomplish this task?

Most students would obviously think the bigger ranch would need more fencing when it turns out they both need exactly the same amount. The original circumference doesn't matter.

Another brain teaser a lot of people get wrong: One family has two children, the youngest one being a boy. A second family also has two children, at least one of which is a boy. Now which family is more likely to have a daughter?

curlious
09-03-2004, 02:38 PM
Wow good job people we solved the problems in groups of three and it took about an hour for all of them. Many where stumped by the cake problem. Yes its funny my calc teacher is a native german but he accomodates us and uses the english measurement system. I prefer metric.

XSquared
09-03-2004, 03:13 PM
A square cake measures 15 inches by 15 inches by 3 inches tall and is frosted on the sides and on the top. How could you cut the cake into 5 pieces so that each piece has exactly the same amount of cake and frosting.?

I think this works.

curlious
09-03-2004, 04:36 PM
Yes that is the answer we came up with but proving the iceing on top between all the regions is the same would require more effort.

Sang-drax
09-04-2004, 05:22 AM
Starting with the slices with a side of 12:
A1 = 12 * 7.5 / 2 = 45
Total area = 15*15 = 225 = 5 * 45

If the slice with sides 6 and 6 also has an area of 45, all slices does.
We separate the slice into two regions, one of which has an area of 6*6/2 = 18
The other slice has the area (using vectors and determinants):


[ 1.5 7.5 ]
A = det [ -7.5 -1.5 ] / 2 = 27

The total area of that slice is thus 45, and the area of evey slice must be 45, because of the symmetry.

XSquared
09-04-2004, 08:28 AM
And if you don't know what the hell determinants are (like me), then you can just connect the diagonals of the square, and then you will get 8 different triangles, all with height 7.5. Then, once you calculate the areas of the individual triangles, you find that the areas are all 45.

Sang-drax
09-04-2004, 03:04 PM
And if you don't know what the hell determinants are (like me), then you can just connect the diagonals of the square, and then you will get 8 different triangles, all with height 7.5. Then, once you calculate the areas of the individual triangles, you find that the areas are all 45.Yes, that'd be better :D