Easy math puzzles

Quote: gordonm888

without much thinking:

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136

Since the sum of two randomly selected numbers in a range would be a probability distribution centered on the highest number (the maximum) of the range. 22 + 114 = 136

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Quote: ThatDonGuy

Quote: gordonm888

without much thinking:

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136

Since the sum of two randomly selected numbers in a range would be a probability distribution centered on the highest number (the maximum) of the range. 22 + 114 = 136

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I had a similar thought...

...but my line of reasoning was, since there are 21 possible numbers below the range, there are probably 21 possible numbers above it, so the total = 114 + 21 = 135.

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Quote: ChesterDog

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I'll guess about 135 tanks. Also, serial numbers from about five tanks were taken.

If we have N sequential integers, 1 through N, and choose n of them at random, the minimum and maximum will have expected values.

If we set the formula for the expected value of the minimum to 22 and the formula for the expected value of the maximum to 114, we can solve for N and n.

Instead, I did a short simulation and found that n is about 5 and N is about 135.

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Quote: Jeopardy

This is the name of an animal common to Texas, which German immigrants gave the name panzerschwein.

Quote: Jeopardy

This is the name of an animal common to Texas, which German immigrants game the name panzerschwein.

Quote: Wizard

Here is a tangential question. Suppose the only bit of information that survived was the largest Uberpanzers serial number recorded was 114. What is the best estimate of the total number of Uberpanzers?

Quote: Jeopardy

This is the name of an animal common to Texas, which German immigrants gave the name panzerschwein.

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Quote: aceside

Quote: Wizard

Here is a tangential question. Suppose the only bit of information that survived was the largest Uberpanzers serial number recorded was 114. What is the best estimate of the total number of Uberpanzers?

Quote: Jeopardy

This is the name of an animal common to Texas, which German immigrants gave the name panzerschwein.

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If the serial numbers are in the range of [22, Max] and 114 is the largest sample among two random samples, the calculation for Max is

114=(2xMax+22)/3.

Solving this equation, we have
Max=160.

This result is based on the theory that the largest number is at 2/3 of the [22, Max] interval.
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Quote: Wizard

Here is a tangential question. Suppose the only bit of information that survived was the largest Uberpanzers serial number recorded was 114. What is the best estimate of the total number of Uberpanzers?

On another topic, there was a Final Jeopardy question a while back that came down to something like this:

Quote: Jeopardy

This is the name of an animal common to Texas, which German immigrants gave the name panzerschwein.

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Quote: unJon

Edit: oh, Wiz said we know 2 things. 114 and that lower numbers were found but we don’t know what they are.
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Quote: Wizard

...Based on Gordon's reply...What is the most likely total number of uber-tanks made?...

Quote: DogHand

Chess Answer

1. Nd2

which leaves Black three legal replies:

1. ... c4
2. Ne4#

1. ... e4
2. Nc4#

1. ... Kxd5
2. Qd7#

I'd try the other, but I still haven't had my morning coffee.

Dog Hand
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Quote: charliepatrick

card problem

Yes it's a variation of the mixing wine with water question. Since there are 52 cards in eash pile, then the total number of Reds in Pile 1 (R1) and the total number of Blacks in Pile 1 (B1) = (R1+B1)=52. Also the total number of Reds in both piles (R1+R2)=52. Hence R2=B1. Similarly R1=B2. This will always be the case so you don't need to turn any cards over.

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card joke

chess joke

Quote: aceside

Do you mean 6 squares and the one rectangular adding together coverage?
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Quote: Gialmere

Quote: aceside

Do you mean 6 squares and the one rectangular adding together coverage?
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Yes.
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Quote: ChesterDog

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I get two-thirds.

The answer is so simple that it's probably right.

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Quote: charliepatrick

yes I get the simple answer!

Ignoring the centre rectangle for the moment, the squares and the large hexagon have six way symmetry. Thus let's look at one of the sides and what lies within the equilateral triangle.

Half one of the squares and half of the other (M and N) lie within it, and as the angle of the square has been split, the angle is 45. Thus this leaves two traingles (15 90 75) O and P. At the inside of the squares there are two identical triangles (Q and R) and this leaves a small equilateral triangle.

All six of these equilateral triangles form a small hexagon, which can be seen in the parts A thru F. The rectangle is CDEF, and as all triangles shown have the same area this is 2/3rds of the smaller hexagon.

Now look at MN (which can form a square) and adding the four triangles, which forms a larger square. What we need to find out is the ratio of the square (MNOPQR) to square MN. Thus we need to add Sin 15 and Cos 15, as this will give the length of the larger square.

Sin 15 = (SQRT(3)-1)/(2*SQRT(2)) and Cos 15 = (SQRT(3)+1)/(2*SQRT(2)); so Sin 15 + Cos 15 = SQRT(3)/SQRT(2). So the large square is 3/2 of the small square, or that can be rewritten as the small square is 2/3 of the large one.

The large square has an area equal corresponding to MNOPQR (six of these make up the large hexagon minus the inner hexagon). The small square corresponds to the area of one "square" (six of these make up all the squares on the original puzzle).

Therefore the six squares are 2/3 of the inner hexagonal part, and the rectangle is 2/3 of the small hexagon.

Hence the six squares and the rectangle are 2/3 of the area of the large hexagon.

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What's a porn star's favorite shape?

A rectangle.

Quote: aceside

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My guess is 7, 18 because they are very related to American history

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Quote: Gialmere

What number pair should be in the second position?
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Quote: Gialmere

The above is called a calculation maze. The object is to begin in the upper left red space with the start value shown and end at the lower right yellow space with the specified target value. Your value changes according to the calculation on the space you enter.

For this example maze your starting value is 2 and the target value is 6. If you went down two spaces then right two spaces your running calculation value would be: start=2, 1, 2, 6, 7. Oops. That's 1 too many.

You must follow the lines between spaces, traveling either up/down or left/right. This means you may move backwards to the space you just came from, however, you may never move diagonally. You may enter and re-enter a space as many times as you wish.


If you re-enter the start space your value will immediately revert to the starting value, essentially making you start over. (You're trying to get through the maze in as few moves as possible.) If you enter the end space with the wrong total, treat it as any other space by adjusting your value and continue on.

Example Maze Solve

Start Value = 2
Target Value = 6



start=2, 1, 3, 5, 4, 5, 6=end

--------------

Puzzle #1

Starting Value = 0
Target Value = 21

Puzzle #2

This maze has the added restriction that you must leave a space along a different path from the one you used to enter it.

Start Value = 0
Target Value = 10

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Quote: Wizard

You have two candles.

The red candle is 5" tall and takes 6 hours to burn out.
The blue candle is 3" tall and takes 10 hours to burn out.

You light them both at the same time.

In how many hours will they be the same height?


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Quote: Wizard

You have two candles.

The red candle is 5" tall and takes 6 hours to burn out.
The blue candle is 3" tall and takes 10 hours to burn out.

You light them both at the same time.

In how many hours will they be the same height?


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Quote:

This one should be easy.

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We set time=x, that is, in x hours the two candles will be the same height. We have this equation

5-5x/6=3-3x/10.

Solving this, we get x=3.75 hours.

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Quote: aceside

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On second thought, I found this puzzle does not have a solution. However, if we change the bottom right corner number to +0, then here is my path to solve this puzzle:

=0, down+1, right x4, right x2, down -2, down x2, left -2, left +2, down +0, right +2, up -2, up +2, right -2, right +2, up -2, right -2, down x2, down +1, down +0 = 21

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Quote: charliepatrick

puzzle #1 5x5 maze

Assuming you can go backwards and forwards (otherwise it's impossible). There is a wall of multipliers in the SE corner (x3 x4 x2). You have to go through one (or more) of these. If you come through the x4 or x2 then you'll be even, hence you can only come in via x3. From there you can add multiples of 4 by going to and fro the 3 and 1. To get to 21 you need to find a multiple of 3 that can be turned into 21 by adding a number of 4s. This number is 9. Hence you need to enter the "x3" with 3.
Now use similar logic since the starting position has a similar wall of multipliers (x3 x4 x2) and all the numbers between the two walls are even. Hence you have to come through x3.
Thus the route starts 0 1 x3 and ends x3 3 1 3 1 3 1, so you now need to enter the Southern x3 with 3. Thus you need the same number of +2s as -2s.
One notices that going the direct route (0 1 x3 2 -2 2 -2 2 -2 2) results in 5, so one needs to use the circle of -2s to get a lower number. Since that loop subtracts 4, you now need to add 2. So you also go via (2 0).
Hence the route is

A5 +0 0
B5 +1 1
C5 x3 3
D5 +2 5
E5 -2 -3
E4 -2 -1
D4 -2 -3
D3 +2 -1
C3 -2 -3
B3 +2 -1
A2 +2 1
A1 +0 1
B1 +2 3
C1 x3 9
D1 +3 12
E1 +1 13
D1 +3 16
E1 +1 17
D1 +3 20
E1 +1 21

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Quote: Gialmere

You're hot on the trail with your logic but seem to have missteps....

Quote: MichaelBluejay

"A town has only two colours of car: 85% are blue and 15% are green. A person witnesses a hit-and-run and says they saw a green car. If witnesses identify the colour of cars correctly 80% of the time, what are the chances the car is actually green?
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Quote: MichaelBluejay

"A town has only two colours of car: 85% are blue and 15% are green. A person witnesses a hit-and-run and says they saw a green car. If witnesses identify the colour of cars correctly 80% of the time, what are the chances the car is actually green?

(From this Twitter post.)

Not the solution

あなたのお母そんはそんなに太っているから、自分の郵便番号があります。

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