Easy math puzzles

Quote: djtehch34t

Quote: ThatDonGuy

Here's a new problem - bad news: it's a "proof" problem.

Interplanetary Express serves 65 different locations throughout the solar system. Every pair of locations is served by a route that is one of four colors - red, blue, green, or yellow. It turns out that, given any three locations, the three routes that connect them to each other are not all the same color.

Somebody wants to add a 66th location, with routes to each of the other 65, and maintaining the "no single-color triangles" policy.
Prove that this is impossible.

(Hint: the fact that you can with 65 is irrelevant; in fact, I am not entirely sure that such a mapping exists.)
link to original post

Show Spoiler

Assume we have a graph that satisfies the condition (call this 4-colorable). Take the 66th vertex and look at its edges. The color C1 that has the maximum number of edges must have at least ceiling(65/4) = 17 edges. Look at the subgraph with the vertices we get from those edges. This must not have any edges with C1 (aka be 3-colorable), or else we get a cycle with color C1.

Pick an arbitrary vertex in the subgraph of size 17 and repeat the argument. Color C2 must have ceiling(16/3) = 6 edges. So, this creates a 2-colorable subgraph of size 6. Repeat the argument once more to get a 1-colorable subgraph of size ceiling(5/2) = 3. But, this is a contraction, as that's the same as a cycle of the same color.

link to original post

Quote: Wizard

To reply to all posts since my puzzle on the 13 cards, I can't give anyone credit for a proper answer yet.

To clarify, yes, both players may look at their cards.
link to original post

Quote: Mental

Either you simply don't like the form of my answer or there is something missing from your description.
link to original post

Quote: ThatDonGuy

Answer

Note that the lines touching x = 5 and x = 7 are parallel, and the lines touching x = 12 and x = 17 are parallel, so the four triangles that are visible are all similar.

Let x be the length of the side of the square, a the distance from the square to (5,0), and b the distance from the square to (17,0)

(12 - 5) / a = (17 - 5) / (a + x) => 7 a + 7 x = 12 a => a = 7/5 x
(17 - 7) / b = (17 - 5) / (b + x) => 10 b + 10 x = 12 b => b = 5 x
Pythagorean Theorem: (a + x)^2 + (b + x)^2 = (17 - 5)^2
(12/5 x)^2 + (6 x)^2 = 144
1044/25 x^2 = 144
x^2 = 144 * 25 / 1044 = 100 / 29
Since x is a side of the square, the area is x^2, or 100 / 29

link to original post

Quote: Wizard

There are ten lillipad (is that the right spelling?) in a row. A frog is sitting on the first one from one side. He can hop to the next pad or jump over it to the next one. How many ways can he get to the last lillipad? Note, he must land on the last one, no jumping over it.
link to original post

Quote: charliepatrick

lillipad

55.

I think you can prove it's a fibonacci series...
Let's assume the starting position is called #9, and the home called #0.
Firstly consider the number of ways of getting from #1 to #0, we'll call this f(1) - obviously there is only one way. So f(1)=1.
Next consider getting from #2 to #0. You can either go #2>#1, then f(1), or #2>#0 = 1. So f(2)=f(1)+1 = 2.
Next consider getting from #3 to #0. You can either go from #3 to #2, then f(2); or #3 to #1, then f(1). So f(3)=f(2)+f(1)

f(1)	1
f(2)	2
f(3)	3
f(4)	5
f(5)	8
f(6)	13
f(7)	21
f(8)	34
f(9)	55

link to original post

Quote: charliepatrick

Show Spoiler

Let's call the camels A B C and D and, to help describe things, assume there are "mile markers" at 1-day, 2-days, 3-days, 4 days from the oasis.

(i) Camel A goes to MM4, delivers the message and returns to MM3 (at the end of day 5).
(ii) After two days have passed, Camel B goes to MM3 (to reach it at the end of of day 5). Since there are now 2 units of water left, this enables A and B to return (from MM3) to MM2 (at the end of day 6).
(iii) Similarly Camel C goes to MM2 to meet A and B; this uses up 2 units, so they have 3 units to allow A, B and C to travel to MM1 (at the end of day 7).
(iv) Similarly Camel D goes to MM1 to meet A, B and C; this uses up 1 unit, so they have 4 units to allow all four to travel back to the oasis.

link to original post

Quote: charliepatrick

rubber band approach

On the first day the ant walks 1 centimeter (of the 1m long band), and before the second day the band is enlarged to 2m. So on the second day the ant walks 1 centimeter (of the 2m long band). This is equivilant to walking 1/2 centimeter on a 1m long band. Similarly on other days the ant can be considered as walking 1/N centimeters on a 1m long band.

Thus the question changes to how long does the series 1 + 1/2 + 1/3 ... need to get to 100cm.

Well this is a very large number!! (e.g. End after 10000000000 numbers PrevTot = 23.603066594897502 ThisTot = 23.6030665949975 , so a simple program isn't going to find the answer)

Your hint suggests googling the Euler constant and I saw ( https://en.wikipedia.org/wiki/Harmonic_series_(mathematics) ) a formula.
H(n) = ln (n) + Euler's constant + 1/2n - e(n) (where e(n)<1/8n²).

Using excel and assuming e(n)=the value, this gives 15 092 688 622 100 000 000 000 000 000 000 000 000 000 000 (or about 1.5E43)

link to original post

Quote: avianrandy

Also don't forget alextrebek stamp being issued July 22
link to original post

Quote: Wizard

I was hoping for more traction on the previous puzzle, but maybe integral calculus isn't everybody's cup of tea.
link to original post

Quote: Wizard

You have a flashlight that requires two batteries. You also have eight batteries, of which four are good and four are bad. As usual, you can't tell the good from the bad ones by appearance. The flashlight requires two good batteries to work.

What is the least number of tests required to ensure you have two good batteries in the flashlight?


link to original post

Quote: PowerSupplyGuy

- There's a room with 3 lights inside. Outside the room there are 3 switches. One switch for each light. If you can only walk in to room once, how do know which switch is for which light?

- There's a boat in a lake with a bowling ball on it. The ball rolls off the boat into the lake. What happens to the level of the lake? Does it go up,down or stay the same?

-psg
link to original post

Quote: Joeman

Pearls

Take two pearls weigh them against any other two pearls. If the scale does not balance, the fake is among the group of pearls currently on the scale. Otherwise, the fake is in the untested group of 4. Let's call the pearls in "fake" group A, B, C, & D.

Take A and weigh it against B. Then, take A and weigh it against C. If both of these trials are unbalanced, A is the fake. If the first trial is unbalanced and the second balances, B is the fake. If the first trial is balanced and the second is unbalanced, C is the fake. If both trials are balanced, D is the fake.

link to original post

Quote: Wizard

You have a balance scale and 8 pearls. One of the pearls is fake and weighs differently, either more or less, than the seven identically-weighted real pearls. You may use the balance scale three times to isolate the fake pearl. How do you do it?

P.S. All due apologies if this has been asked before. I think it's a classic.
link to original post

Quote: Wizard

Here is another classic that may have been asked before.

Alex, Bob and Cindy are in a truel. The probability of each making a deliberate shot are:

A: 30%
B: 50%
C: 100%

They must shoot in alphabetic order. All surivors will keep taking turns in that order until only one person remains.

What's is optimal strategy for Alex initially?

Assuming all follow optimal strategy, what is each players probability of winning?
link to original post