Easy math puzzles

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August 10th, 2020 at 11:54:53 AM permalink

Here's an easy problem with what may be a trick answer:

In the late 1970s, "biorhythms" were all the rage.
Each person has three "cycles": a 23-day Physical cycle, a 28-day Emotional cycle, and a 33-day Intellectual cycle.
Each cycle is a sine wave with positive and negative values, and a "critical point" at the zero line (so, for example, the Physical cycle goes through a critical point every 11 1/2 days).
All three cycles begin at the critical point, headed in the positive direction.
When all three cycles converge at a critical point, this is Not A Good Thing. How often does this happen?

Ace2

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August 10th, 2020 at 11:58:42 AM permalink

Quote: ThatDonGuy
My simulations show that Player A wins about 60% of the time, which is a 20% advantage.

I am working under the following assumption:
(a) Player A always has a Small bet and a Tall bet in play, except when either someone rolls a 7 on a comeout (in which case, both bets lose, and neither can be placed again until the shooter misses a point) or one of the two bets is won (in which case that bet cannot be placed again until the shooter misses a point);
(b) Player B always has the All bet in play (again, unless someone rolls a 7 on a comeout, in which case that bet is lost and the player cannot place it again until the shooter misses a point).

You may be assuming an independence that is not there. For example, if a Small bet is won, then the current All bet must have 2, 3, 4, 5, and 6 made, plus whatever numbers in the current Tall bet have been made.

Bets are made before every new shooter�s first roll. Always then and only then

It�s all about making that GTA

rsactuary

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August 10th, 2020 at 12:17:20 PM permalink

Quote: ThatDonGuy
Here's an easy problem with what may be a trick answer:

In the late 1970s, "biorhythms" were all the rage.
Each person has three "cycles": a 23-day Physical cycle, a 28-day Emotional cycle, and a 33-day Intellectual cycle.
Each cycle is a sine wave with positive and negative values, and a "critical point" at the zero line (so, for example, the Physical cycle goes through a critical point every 11 1/2 days).
All three cycles begin at the critical point, headed in the positive direction.
When all three cycles converge at a critical point, this is Not A Good Thing. How often does this happen?

Since none of the three cycles have a common divisor, it should just be the product of the three, then divided by 2 because the sine waves cross the critical point twice in a cycle? 23x29x33/2 = 10626 days or 29.09 years.

ETA: Looked at Wiki after I posted this and it says I'm wrong, but I'm not convinced.

ThatDonGuy

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August 10th, 2020 at 1:52:39 PM permalink

Quote: rsactuary
Since none of the three cycles have a common divisor, it should just be the product of the three, then divided by 2 because the sine waves cross the critical point twice in a cycle? 23x29x33/2 = 10626 days or 29.09 years.

ETA: Looked at Wiki after I posted this and it says I'm wrong, but I'm not convinced.

No, that is correct, but next time, put answers in a spoiler box.

The trick is that it's not 23/2 x 29/2 x 33/2; you have to think of these in integral numbers of "half-days."

Wikipedia is using the coincidence of the "full cycles," but that doesn't take into account the possibility that the Physical and Intellectual cycles can be ending a full cycle at the same time that an Emotional cycle can be halfway through one.

rsactuary

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August 10th, 2020 at 5:29:30 PM permalink

sorry, I get so excited that I can actually answer a question, and then I forget to spoiler it.

I found an online calculator and proved that the lines do indeed all intersect at 29.09 years and at 58.18 years.

Gialmere

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August 11th, 2020 at 8:06:23 AM permalink

On a standard pair of six-sided dice there is one way of obtaining a 2, two ways of obtaining a 3, and so on, up to one way of obtaining a 12. Find another pair of six-sided dice such that:

a) The set of dots on each die is not the standard {1,2,3,4,5,6}.

b) Each face has at least one dot.

c) The number of ways of obtaining each sum is the same as for the standard dice.

Have you tried 22 tonight? I said 22.

rsactuary

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August 11th, 2020 at 8:58:54 AM permalink

deleted.. I really didn't read this well.

Last edited by: rsactuary on Aug 11, 2020

Gialmere

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August 11th, 2020 at 9:02:25 AM permalink

Quote: rsactuary
one die is the standard six faced die with numbers 1 to 6. The other die is 1,1,1,6,6,6

Sorry, no.

Note criteria a.

Have you tried 22 tonight? I said 22.

rsactuary

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August 11th, 2020 at 9:14:24 AM permalink

oooooh

charliepatrick

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August 11th, 2020 at 12:44:05 PM permalink

One of the dice has an 8

I've seen this before but worked it out remembering the top were 4 and 8. 1 3 4 5 6 8 ; 1 2 2 3 3 4.

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August 11th, 2020 at 2:16:07 PM permalink

My answer agrees with Charlie's, but I know he gets the credit for being first.

My solution was basically trial and error. It took about 20 minutes to stumble on the answer.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

Gialmere

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August 11th, 2020 at 5:06:50 PM permalink

Quote: charliepatrick
I've seen this before but worked it out remembering the top were 4 and 8. 1 3 4 5 6 8 ; 1 2 2 3 3 4.

Quote: Wizard
My answer agrees with Charlie's, but I know he gets the credit for being first.

My solution was basically trial and error. It took about 20 minutes to stumble on the answer.

Correct!

Some versions of this puzzle will send you on a snipe hunt by asking you to find all pairs of dice that qualify. In fact, the only alternative dice that meet the conditions are {1,2,2,3,3,4} and {1,3,4,5,6,8}.

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

Dice. My favorite Dice rhyme: "Jack 'n Jill went up the hill, both wit a buck 'n a quarter..."

... Jill came down wit two fifty

Have you tried 22 tonight? I said 22.

rsactuary

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August 11th, 2020 at 6:59:08 PM permalink

silly me tried to do it without a number going above 6.... that probably should have been stated in the question....

ThatDonGuy

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August 12th, 2020 at 6:57:39 AM permalink

Quote: Gialmere
Some versions of this puzzle will send you on a snipe hunt by asking you to find all pairs of dice that qualify.

In order to have exactly one way to roll a 2, each die needs exactly one 1.
Because of this, in order to have exactly two 3s, either both dice have one 2, or one has two 2s and the other has none.

At the other end, in order to have exactly one 12, the highest numbers on the two dice have to be (6,6), (5,7), or (4,8), and they have to be unique on their respective dice, as otherwise there would be multiple ways to roll 12.
(3,9) is not possible as, on the die where 3 is the highest, there can only be one 3 and one 1, so the die would be {1,2,2,2,2,3}, but that would have four ways to roll a 3.
This leaves eight possibilities (well, nine, but {1 x x x x 6} {1 2 2 x x 6} is the same as {1 2 2 x x 6} {1 x x x x 6}):
{1 2 2 x x 6} {1 x x x x 6}
{1 2 x x x 6} {1 2 x x x 6}
{1 2 2 x x 5} {1 x x x x 7}
{1 2 x x x 5} {1 2 x x x 7}
{1 x x x x 5} {1 2 2 x x 7}
{1 2 2 x x 4} {1 x x x x 8}
{1 2 x x x 4} {1 2 x x x 8}
{1 x x x x 4} {1 2 2 x x 8}
In each case, x is between 3 and (the highest number on that die - 1)
This means the last three are:
{1 2 2 3 3 4} {1 x x x x 8}
{1 2 3 3 3 4} {1 2 x x x 8}
{1 3 3 3 3 4} {1 2 2 x x 8}
However, the last two are impossible as each has (at least) four ways to roll a 4, and (at least) three ways to roll an 11.

With {1 2 2 3 3 4} {1 x x x x 8}, there are already 2 ways to roll 10 (28 28), so there can be only one more; this has to be (46)
{1 2 2 3 3 4} {1 x x x 6 8} - there are 3 ways to roll 9 (18 36 36), so there can be only one more; this has to be (45)
{1 2 2 3 3 4} {1 x x 5 6 8} - there are 4 ways to roll 8 (26 26 35 35), so there can be only one more; this has to be (44)
{1 2 2 3 3 4} {1 x 4 5 6 8} - there are 5 ways to roll 7 (16 25 25 34 34), so there can be only one more; this has to be (43)
{1 2 2 3 3 4} {1 3 4 5 6 8} is a solution
Only six other combinations to go...

Also note that, in order for the probabilities to match, the sum of the numbers on the two dice must be 42.

Gialmere

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August 13th, 2020 at 7:33:31 AM permalink

Twenty-seven identical cubes are painted white. They are assembled into a larger 3x3x3 cube, the outside of which is then painted black. Next, the cube is disassembled and the smaller cubes thoroughly shuffled in a bag. A blindfolded man then reassembles the pieces back into a 3x3x3 cube.

What is the probability that the outside of this cube is completely black?

Have you tried 22 tonight? I said 22.

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August 13th, 2020 at 8:29:22 AM permalink

Quote: Gialmere
Twenty-seven identical cubes are painted white. They are assembled into a larger 3x3x3 cube, the outside of which is then painted black. Next, the cube is disassembled and the smaller cubes thoroughly shuffled in a bag. A blindfolded man then reassembles the pieces back into a 3x3x3 cube.

What is the probability that the outside of this cube is completely black?

The 27 cubes consist of one "center" cube with no black sides, 6 "face center" cubes with one black side, 12 "edge center" cubes with two black sides, and 8 "corner" cubes with three black sides.
The center cube has to be in the center as otherwise it will expose at least one white face.
The corner cubes have to be in the corners as otherwise a cube in the corner will expose at least one white face.
The edge center cubes have to be in the edge centers as, since the corner cubes have to be in the corners, any other cube in an edge center will expose at least one white face.
This leaves the face center cubes for the face center positions.

Of the 27! possible ways to arrange the cubes, there are 12! 8! 6! ways to arrange them so that each cube is in a correct space.
Each corner cube has 8 corners, one of which has to be in the corner of the large cube in order for all 3 black faces to be exposed. Note that the orientation of the cube is irrelevant.
Each edge center cube has 12 edges, one of which has to be the edge of the two exposed faces; again, orientation is irrelevant.
Each face center cube has 6 faces, one of which has to be the exposed face.
The probability is (12! 8! 6!) / (27! 12^12 8^8 6^6) = (12! 8! 6!) / (27! 2^54 3^18) = (12! 8! 6!) / (27! 24^18).
According to the JB calculator, this is 1 / 5,465,062,811,999,459,151,238,583,897,240,371,200

Gialmere

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August 13th, 2020 at 10:57:02 AM permalink

Quote: ThatDonGuy

The 27 cubes consist of one "center" cube with no black sides, 6 "face center" cubes with one black side, 12 "edge center" cubes with two black sides, and 8 "corner" cubes with three black sides.
The center cube has to be in the center as otherwise it will expose at least one white face.
The corner cubes have to be in the corners as otherwise a cube in the corner will expose at least one white face.
The edge center cubes have to be in the edge centers as, since the corner cubes have to be in the corners, any other cube in an edge center will expose at least one white face.
This leaves the face center cubes for the face center positions.

Of the 27! possible ways to arrange the cubes, there are 12! 8! 6! ways to arrange them so that each cube is in a correct space.
Each corner cube has 8 corners, one of which has to be in the corner of the large cube in order for all 3 black faces to be exposed. Note that the orientation of the cube is irrelevant.
Each edge center cube has 12 edges, one of which has to be the edge of the two exposed faces; again, orientation is irrelevant.
Each face center cube has 6 faces, one of which has to be the exposed face.
The probability is (12! 8! 6!) / (27! 12^12 8^8 6^6) = (12! 8! 6!) / (27! 2^54 3^18) = (12! 8! 6!) / (27! 24^18).
According to the JB calculator, this is 1 / 5,465,062,811,999,459,151,238,583,897,240,371,200

Correct! Down to the last digit.

I wondered if the JB calculator might get used for this one.

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

I placed a sugar cube on the kitchen floor for an ant.
When he left to get the rest of the ants, I quickly removed the sugar cube.

Now everyone thinks he lied.

Have you tried 22 tonight? I said 22.

Gialmere

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August 13th, 2020 at 1:09:06 PM permalink

An older brother was sent to find his younger brother who had gone bicycle riding at the park. When the older brother entered the park, he spotted his younger brother's tire treads on a dirt trail crossing his path. (See above image.)

Was the younger brother traveling left to right, or right to left?

How do you know?

Have you tried 22 tonight? I said 22.

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August 13th, 2020 at 1:45:31 PM permalink

Here's a puzzle that is probably too easy, even for this thread. But I was playing with numbers and stumbled on this.

Decompose the cube of an integer, n, into a sum of two squares - one of which must be n².

Ex: 17*17*17 = 4913. And 4913 = 52² + 47² but also = 68² + 17²

a) Give a few examples of other integers (other than the trivial example of "1") that meet this criteria
b) Give a generic mathematical formula for identifying such integers.

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

ThatDonGuy

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August 13th, 2020 at 2:15:24 PM permalink

Quote: gordonm888
Decompose the cube of an integer, n, into a sum of two squares - one of which must be n².

Ex: 17*17*17 = 4913. And 4913 = 52² + 47² but also = 68² + 17²

a) Give a few examples of other integers (other than the trivial example of "1") that meet this criteria
b) Give a generic mathematical formula for identifying such integers.

8 = 2³ = 2² + 2²
125 = 5³ = 5² + 10²
1000 = 10³ = 10² + 30²

The general solution is, the cubes of numbers that are 1 more than squares (e.g. (1² + 1)³ = 8, (2² + 1)³ = 125, (3² + 1)³ = 1000).
Proof:
n³ = n² + k², where k is a positive integer
n³ - n² = k²
n² (n - 1) = k²
By the Fundamental Theorem of Arithmetic, since every prime factor in k² appears an even number of times, and every prime factor in n² appears an even number of times, then every prime factor in (n - 1) appears an even number of times, so n - 1 is a perfect square.

("Fundamental Theorem of Arithmetic"?
Every integer > 1 is either prime or has a unique set of (not necessarily distinct) prime factors whose product is that integer.
Corollary: every perfect square > 1 has all of its prime factors appear an even number of times (since if prime factor p appears k times in n, then it must appear 2k times in n²).)

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August 13th, 2020 at 3:51:18 PM permalink

Quote: ThatDonGuy
Quote: gordonm888
Decompose the cube of an integer, n, into a sum of two squares - one of which must be n².

Ex: 17*17*17 = 4913. And 4913 = 52² + 47² but also = 68² + 17²

a) Give a few examples of other integers (other than the trivial example of "1") that meet this criteria
b) Give a generic mathematical formula for identifying such integers.

8 = 2³ = 2² + 2²
125 = 5³ = 5² + 10²
1000 = 10³ = 10² + 30²

The general solution is, the cubes of numbers that are 1 more than squares (e.g. (1² + 1)³ = 8, (2² + 1)³ = 125, (3² + 1)³ = 1000).
Proof:
n³ = n² + k², where k is a positive integer
n³ - n² = k²
n² (n - 1) = k²
By the Fundamental Theorem of Arithmetic, since every prime factor in k² appears an even number of times, and every prime factor in n² appears an even number of times, then every prime factor in (n - 1) appears an even number of times, so n - 1 is a perfect square.

("Fundamental Theorem of Arithmetic"?
Every integer > 1 is either prime or has a unique set of (not necessarily distinct) prime factors whose product is that integer.
Corollary: every perfect square > 1 has all of its prime factors appear an even number of times (since if prime factor p appears k times in n, then it must appear 2k times in n²).)

CORRECT!

I did say this was probably too easy, even for this thread. ThatDonGuy solved this in what must be near-record time.

I characterized this as finding the zeroes for n³ - n² - m² = 0 where m is a real positive integer.

If you recognize that if n = k² + 1, then for any k there is a solution where m = k *(k² + 1).

So, for k=4, n=17 and m= 4*17 =68.

Its handy to know that for 50653, which is 37³, that k=6, and therefore 50663 decomposes as 37² + 222². And we know this because 6*37 = 222!

I'm easily amused.

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

charliepatrick

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August 13th, 2020 at 5:44:46 PM permalink

Quote: gordonm888
Here's a puzzle ...Decompose the cube of an integer, n, into a sum of two squares - one of which must be n².

I immediately saw the following
17*17*17 = (17*17)*(16+1) = 17^2*16 + 17^2 = (17*4)^2 + 17^2.

So clearly this works because 17 is one more than a square.

So (N²+1)³ will create (N²+1)² x (1 + N²) = (N²+1)² + (N²+1)² x N² = (N²+1)² + ( (N²+1) x N ) ².

ThatDonGuy

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August 13th, 2020 at 6:06:43 PM permalink

Quote: Gialmere

An older brother was sent to find his younger brother who had gone bicycle riding at the park. When the older brother entered the park, he spotted his younger brother's tire treads on a dirt trail crossing his path. (See above image.)

Was the younger brother traveling left to right, or right to left?

How do you know?

Which line designates what? For that matter, where did the older brother enter (left side, or right - or does that matter)?

I am assuming that one line is the older brother's path, and the other is the dirt trail.

charliepatrick

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August 13th, 2020 at 6:13:55 PM permalink

Quote: ThatDonGuy
I am assuming that one line is the older brother's path, and the other is the dirt trail.

I had assumed one was the front wheel and the other back.

I'm guessing you have to look at which line could be the back wheel given the cyclist pedals and can create force to propel the cycle via the back wheel. My feeling, guess here, is this means it's the blue line is. That would marry up with coming in from the left, pointing the front wheel to the right, dragging the back wheel outwards, before turning the front wheel into the curve left. But this is an educated guess at this stage.

Gialmere

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August 13th, 2020 at 6:49:19 PM permalink

To clarify:

Yes, the lines represent the front and rear tire tracks of the bicycle.

The scale of the picture is roughly 10 feet from side to side.

The older brother approaches the tracks from the bottom of the picture on a path heading to the top of the picture.

The younger brother's meandering tire tracks are on a (roughly) left/right trail that crosses the path.

Have you tried 22 tonight? I said 22.

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August 13th, 2020 at 7:05:08 PM permalink

Quote: charliepatrick
Quote: ThatDonGuy
I am assuming that one line is the older brother's path, and the other is the dirt trail.
I had assumed one was the front wheel and the other back.
I'm guessing you have to look at which line could be the back wheel given the cyclist pedals and can create force to propel the cycle via the back wheel. My feeling, guess here, is this means it's the blue line is. That would marry up with coming in from the left, pointing the front wheel to the right, dragging the back wheel outwards, before turning the front wheel into the curve left. But this is an educated guess at this stage.

Based on these being front and back tire pathways with readable tread marks:

the scale is difficult to judge, but if these are the back and front tires then there is a a considerable "lag" between them. I would guess that the red is the front tire because it executes the sharpest turn, but that the entry was from the right side of the chart.

Basically, if the motion originates from the left of the chart and the red wheel is leading, then the blue lines should not be crossing the red line at the far right and then continuing to diverge.

I think I have a 25% chance of being right.

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

ssho88

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August 13th, 2020 at 10:04:00 PM permalink

Quote: Gialmere
To clarify:

Yes, the lines represent the front and rear tire tracks of the bicycle.

The scale of the picture is roughly 10 feet from side to side.

The older brother approaches the tracks from the bottom of the picture on a path heading to the top of the picture.

The younger brother's meandering tire tracks are on a (roughly) left/right trail that crosses the path.

My guess, younger brother is travelling to the LEFT.

Reasons :-

1) Red line is for the front tire track because front tire can make sharper curve track .
2) Blue line curve peak should be "behind" red line curve peak.

Just my two cents. LOL

Gialmere

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August 14th, 2020 at 3:59:23 PM permalink

Quote: gordonm888
the scale is difficult to judge, but if these are the back and front tires then there is a a considerable "lag" between them. I would guess that the red is the front tire because it executes the sharpest turn, but that the entry was from the right side of the chart.

Basically, if the motion originates from the left of the chart and the red wheel is leading, then the blue lines should not be crossing the red line at the far right and then continuing to diverge.

I think I have a 25% chance of being right.

Quote: ssho88

My guess, younger brother is travelling to the LEFT.

Reasons :-

1) Red line is for the front tire track because front tire can make sharper curve track .
2) Blue line curve peak should be "behind" red line curve peak.

Just my two cents. LOL

Correct!

Obviously, in the real world, you can quickly differentiate between the front and rear tires because the rear track will be superimposed over the front one. Even without this, however, if you've ridden a bike you can probably guess which track is the front tire. As gordonm888 and ssho88 point out, the front tire will make more pronounced turns while the rear tire track is a calmer, straighter version of the front track.

To prove this mathematically, observe that regardless of how erratic the rider is turning the handlebars, the rear tire will always be pointed at the front tire. So a tangent line drawn on the curve of the rear track will always intersect the track of the front tire. In the upper left picture, the tangent drawn on the red line will never intersect the blue line. Therefore the red line MUST be the front tire track.

For direction, consider that the front and rear tires are always the exact same distance apart (again regardless of how wobbly the rider steers). In the upper right picture, two tangents are drawn along the blue (rear) line. Note that CD and FG, which intersect the red line to the right, are different lengths. BC and EF, however, which intersect the red line to the left, are both exactly the same length.

Therefore, the older brother, when he encounters the bicycle tracks, should turn left and follow them down the trail to find his little brother and hopefully get him home before dinner gets cold.

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

What�s the difference between a well dressed man on a bike and a poorly dressed man on a unicycle?

Attire!

Last edited by: Gialmere on Aug 14, 2020

Have you tried 22 tonight? I said 22.

Gialmere

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August 14th, 2020 at 3:59:23 PM permalink

Double posted.

Last edited by: Gialmere on Aug 14, 2020

Have you tried 22 tonight? I said 22.

Gialmere

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August 15th, 2020 at 1:23:46 PM permalink

In how many ways, counting (dead heat) ties, can eight horses cross the finishing line?

(For example, two horses, A and B, can finish in three ways: A wins, B wins, A and B tie.)

Have you tried 22 tonight? I said 22.

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August 15th, 2020 at 2:09:29 PM permalink

8! + 6!28 + 5!56 + 4!70 + 3!56 + 2!28 + 1 = 69,273

8! ways no ties + 6! no ties * c(8,2) ways to choose 2 tying horses + 5! no ties * c(8,3) ties etc

Ties only counted for first place

It�s all about making that GTA

Gialmere

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August 15th, 2020 at 2:40:01 PM permalink

Quote: Ace2
8! + 6!28 + 5!56 + 4!70 + 3!56 + 2!28 + 1 = 69,273

8! ways no ties + 6! no ties * c(8,2) ways to choose 2 tying horses + 5! no ties * c(8,3) ties etc

Ties only counted for first place

Well done Ace, but the puzzle asks for every possible combination.

Have you tried 22 tonight? I said 22.

charliepatrick

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August 15th, 2020 at 3:28:25 PM permalink

Interesting puzzle. I was brought up in Lewes (it's slightly inland from Brighton near the South Coast and lies on the Greenwich Meridian) and the racetrack had a triple dead heat for first and deat heat for fourth. ( http://www.greyhoundderby.com/Lewes%201962.htm ). It was also where I had my first winning bet!

I couldn't see a quick way other than just going through all the ways a horse race of eight runners could end in various dead heats and then perm the horses within each group.

Update: The table below is wrong as I forgot to multiply in Perms*Perms etc. e.g. 3-3-2 should be 3 * 56 * 10 = 1680 not 560.

Ways of the dead heats	Perms	Factors	Perms
(8)	1	( 1 )	1
(71)	2	( 7 1 )	7
(62)	2	( 28 1 )	28
(611)	3	( 28 2 1 )	56
(53)	2	( 56 1 )	56
(521)	6	( 56 3 1 )	168
(5111)	4	( 56 3 2 1 )	336
(44)	1	( 70 1 )	70
(431)	6	( 70 4 1 )	280
(422)	3	( 70 6 1 )	420
(4211)	12	( 70 6 2 1 )	840
(41111)	5	( 70 4 3 2 1 )	1 680
(332)	3	( 56 10 1 )	560
(3311)	6	( 56 10 1 )	560
(3221)	12	( 56 10 3 1 )	1 680
(32111)	20	( 56 10 )	560
(311111)	6	( 56 5 4 3 2 1 )	6 720
(2222)	1	( 28 15 6 1 )	2 520
(22211)	10	( 28 15 6 2 1 )	5 040
(221111)	15	( 28 15 4 3 2 1 )	10 080
(2111111)	7	( 28 6 5 4 3 2 1 )	20 160
(11111111)	1	( 8 7 6 5 4 3 2 1 )	40 320
(Total)		( )	92 142

Last edited by: charliepatrick on Aug 15, 2020

ChesterDog

ChesterDog

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August 15th, 2020 at 4:10:31 PM permalink

Quote: Gialmere
...In how many ways, counting (dead heat) ties, can eight horses cross the finishing line?...

I get 545,835.

There are 22 ways to arrange 8 horses into various groupings: 8, 71, 62, 611, 53, 521, 5111, 44, 431, 422, 4211, 41111, 332, 3311, 3221, 32111, 311111, 2222, 22211, 221111, 2111111, and 11111111.

Example calculation with the grouping 332: ways = combin(8,2)*combin(6,3)/fact(2) = 280. Then there are fact(3) ways for these three groupings to finish, for 280*6 = 1680 ways.

charliepatrick

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August 15th, 2020 at 4:54:20 PM permalink

I had forgotten to multiple the Perms*Perms (you can see the error in the previous post) so agree with your 332 figure. Correcting this and a few other things I agree.

Ways of the dead heats	Perms	Factors	Perms
(8)	1	( 1 )	1
(71)	2	( 8 1 )	16
(62)	2	( 28 1 )	56
(611)	3	( 28 2 1 )	168
(53)	2	( 56 1 )	112
(521)	6	( 56 3 1 )	1 008
(5111)	4	( 56 3 2 1 )	1 344
(44)	1	( 70 1 )	70
(431)	6	( 70 4 1 )	1 680
(422)	3	( 70 6 1 )	1 260
(4211)	12	( 70 6 2 1 )	10 080
(41111)	5	( 70 4 3 2 1 )	8 400
(332)	3	( 56 10 1 )	1 680
(3311)	6	( 56 10 2 1 )	6 720
(3221)	12	( 56 10 3 1 )	20 160
(32111)	20	( 56 10 3 2 1 )	67 200
(311111)	6	( 56 5 4 3 2 1 )	40 320
(2222)	1	( 28 15 6 1 )	2 520
(22211)	10	( 28 15 6 2 1 )	50 400
(221111)	15	( 28 15 4 3 2 1 )	151 200
(2111111)	7	( 28 6 5 4 3 2 1 )	141 120
(11111111)	1	( 8 7 6 5 4 3 2 1 )	40 320
(Total)		)	545 835

Taking 3-3-2 as an example, put the horses in strict 1st thru 8th (8! ways = 40320). Consider 3a-3b-2, then use similar logic for 3a-2-3b, then 2-3a-3b.

For 3-3-2 It would be the same result with 12345678 13245678 etc. So the 40320 permutations include, six times, where the same horses are in the first group of three. Similarly the second group of three have been counted six times for each group; and finally the third group of two, two times. So for each unique result it has been counted 6*6*2 times.

Similarly for 3-2-3 and 2-3-3.

So the number of combinations is 8! / (3!*3!*2!) * 3 = 40320 / 72 * 3 = 1680.

This seems a nicer way to work each out.

Last edited by: charliepatrick on Aug 15, 2020

Gialmere

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August 15th, 2020 at 6:46:07 PM permalink

Quote: ChesterDog
I get 545,835.

There are 22 ways to arrange 8 horses into various groupings: 8, 71, 62, 611, 53, 521, 5111, 44, 431, 422, 4211, 41111, 332, 3311, 3221, 32111, 311111, 2222, 22211, 221111, 2111111, and 11111111.

Example calculation with the grouping 332: ways = combin(8,2)*combin(6,3)/fact(2) = 280. Then there are fact(3) ways for these three groupings to finish, for 280*6 = 1680 ways.

Quote: charliepatrick
I had forgotten to multiple the Perms*Perms (you can see the error in the previous post) so agree with your 332 figure. Correcting this and a few other things I agree.
Ways of the dead heats Perms Factors Perms

(8)
1
( 1 )
1

(71)
2
( 8 1 )
16

(62)
2
( 28 1 )
56

(611)
3
( 28 2 1 )
168

(53)
2
( 56 1 )
112

(521)
6
( 56 3 1 )
1 008

(5111)
4
( 56 3 2 1 )
1 344

(44)
1
( 70 1 )
70

(431)
6
( 70 4 1 )
1 680

(422)
3
( 70 6 1 )
1 260

(4211)
12
( 70 6 2 1 )
10 080

(41111)
5
( 70 4 3 2 1 )
8 400

(332)
3
( 56 10 1 )
1 680

(3311)
6
( 56 10 2 1 )
6 720

(3221)
12
( 56 10 3 1 )
20 160

(32111)
20
( 56 10 3 2 1 )
67 200

(311111)
6
( 56 5 4 3 2 1 )
40 320

(2222)
1
( 28 15 6 1 )
2 520

(22211)
10
( 28 15 6 2 1 )
50 400

(221111)
15
( 28 15 4 3 2 1 )
151 200

(2111111)
7
( 28 6 5 4 3 2 1 )
141 120

(11111111)
1
( 8 7 6 5 4 3 2 1 )
40 320

(Total)
)
545 835

Taking 3-3-2 as an example, put the horses in strict 1st thru 8th (8! ways = 40320). Consider 3a-3b-2, then use similar logic for 3a-2-3b, then 2-3a-3b.

For 3-3-2 It would be the same result with 12345678 13245678 etc. So the 40320 permutations include, six times, where the same horses are in the first group of three. Similarly the second group of three have been counted six times for each group; and finally the third group of two, two times. So for each unique result it has been counted 6*6*2 times.

Similarly for 3-2-3 and 2-3-3.

So the number of combinations is 8! / (3!*3!*2!) * 3 = 40320 / 72 * 3 = 1680.

This seems a nicer way to work each out.

Correct!
----------------------------

1 1 was a racehorse. 2 2 was 1 2.
When 1 1 1 1 race, 2 2 1 1 2.

Have you tried 22 tonight? I said 22.

Gialmere

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August 18th, 2020 at 10:11:59 AM permalink

There is a 50m long soldier�s unit advancing forward.

The last soldier in the line needs to give a message to the officer leading the unit at the front end of the column. So, while the unit is marching, he runs to the front, reaches the officer and hands over the message to him. Then, without stopping, he runs and comes back to his original position at the rear of the column. In the mean time the whole unit has moved ahead by 50m.

How much distance did the soldier cover in that time?

Assume that he ran the whole distance with uniform speed.

Have you tried 22 tonight? I said 22.

ThatDonGuy

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August 18th, 2020 at 1:38:36 PM permalink

Quote: Gialmere

There is a 50m long soldier�s unit advancing forward.

The last soldier in the line needs to give a message to the officer leading the unit at the front end of the column. So, while the unit is marching, he runs to the front, reaches the officer and hands over the message to him. Then, without stopping, he runs and comes back to his original position at the rear of the column. In the mean time the whole unit has moved ahead by 50m.

How much distance did the soldier cover in that time?

Assume that he ran the whole distance with uniform speed.

Let d be the distance the line moved before the soldier got to the front.
In the time it took the line to move d, the soldier moved (d + 50), so the soldier/unit rate ratio = (d + 50) / d.
After turning around, the soldier moved d (i.e. the (d + 50) originally moved, minus the 50 that the line moved by the time the soldier got back to the end of the line) in the time it took the line to move (50 - d), so the soldier/unit rate ratio = d / (50 - d).
The ratios are the same, so (d + 50) / d = d / (50 - d)
(d + 50)(50 - d) = d² = 2500 - d²
d = 50 sqrt(2)
OOPS it can't be more than 50, as the unit would have moved more than 50m by the time he returned to the back
That should be:
d = 25 sqrt(2)
The total distance = (50 + d) + d = 50 + 50 sqrt(2)

Last edited by: ThatDonGuy on Aug 18, 2020

Ace2

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August 18th, 2020 at 2:03:07 PM permalink

Assume group speed of 50 meters per minute. Runner speed s in meters per minute, time t in minutes, s-50 is runner speed relative to group going forward, s+50 going back.

(s-50)t = 50
(s+50)(1-t) = 50

s = (1/.5^.5 +1)50 = 120.7 meters/min for 1 minute is 120.7 meters

Last edited by: Ace2 on Aug 18, 2020

It�s all about making that GTA

Gialmere

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August 18th, 2020 at 6:14:50 PM permalink

Quote: Ace2
Assume group speed of 50 meters per minute. Runner speed s in meters per minute, time t in minutes, s-50 is runner speed relative to group going forward, s+50 going back.

(s-50)t = 50
(s+50)(1-t) = 50

s = (1/.5^.5 +1)50 = 120.7 meters/min for 1 minute is 120.7 meters

Correct!

120.71 for those wanting to be precise.

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

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�Five beers please.�

Have you tried 22 tonight? I said 22.

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August 18th, 2020 at 7:12:36 PM permalink

Quote: Gialmere
120.71 for those wanting to be precise.

I think the only way to express the answer �precisely� is (1/.5^.5 +1)50. Since it contains .5^.5 (an irrational number) any decimal number truncates infinite digits from the exact answer

Last edited by: Ace2 on Aug 18, 2020

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Gialmere

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August 18th, 2020 at 7:19:27 PM permalink

[shrug] Here's the official solve from the puzzle site.

It is given that the platoon and the last person moved with uniform speed. Also, they both moved for the identical amount of time. Hence, the ratio of the distance they covered � while person moving forward and backward - are equal.

Let's assume that when the last person reached the first person, the platoon moved X meters forward. Thus, while moving forward the last person moved (50+X) meters whereas the platoon moved X meters. Similarly, while moving back the last person moved [50-(50-X)] X meters whereas the platoon moved (50-X) meters.

Now, as the ratios are equal,
(50+X)/X = X/(50-X)
(50+X)*(50-X) = X*X
Solving, X=35.355 meters

Thus, total distance covered by the last person
= (50+X) + X
= 2*X + 50
= 2*(35.355) + 50
= 120.71 meters

Note that at first glance, one might think that the total distance covered by the last person is 100 meters, as he ran the total length of the platoon (50 meters) twice. TRUE, but that's the relative distance covered by the last person i.e. assuming that the platoon is stationary.

Have you tried 22 tonight? I said 22.

Ace2

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August 18th, 2020 at 7:28:32 PM permalink

50 * .5^.5 = 35.3553390593274..........

It does not equal exactly 35.355

It�s all about making that GTA

Gialmere

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August 18th, 2020 at 7:35:21 PM permalink

Quote: Ace2
50 * .5^.5 = 35.3553390593274..........

It does not equal exactly 35.355

Correct Again!

Have you tried 22 tonight? I said 22.

Gialmere

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August 19th, 2020 at 8:35:56 AM permalink

What is the fewest number of knights you must place on a chess board so that every square is either occupied by, or under attack by, at least one knight?

Have you tried 22 tonight? I said 22.

ThatDonGuy

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August 19th, 2020 at 11:50:39 AM permalink

Quote: Gialmere
What is the fewest number of knights you must place on a chess board so that every square is either occupied by, or under attack by, at least one knight?

Doing a modified Monte Carlo (where the first four knights each have to be attacking a different corner), after 700 million placements, I have a 13-knight solution.

I have a feeling the answer is 12. I know it has to be at least 8, as, since a knight can cover 9 spaces, 7 knights cover only 63 of the 64.

charliepatrick

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August 19th, 2020 at 12:23:38 PM permalink

If you just consider the four squares nearest the corner, then it must take three different Knights to cover them. One will get the corner square (A), one can get the two on the edge (B), and the third (C) will get the one that's one row and column from the corner. Also these Knights will have to be in, for example, the SW quarter.

b . C a
c A B .
b c . a
a b a b

By symmetry there needs to be three other Knights in each of the other quarters. Therefore the minimum is 12 knights.

I think B has to be where it is, but there's flexibility whether A can be West or South of B, but by symmetry they're the same; so assume West.

Now C has to cover the fourth one, so could be North of B or West of B. Putting it North and filling in the SE corner using similar rotated logic leads to...

b . C a f . e d
c A B f E D . f
b c e a b F e d
a b a b d e d f

Note how E and F cover two squares in the SW quadrant and B covers one square in the SE quadrant.
You can see that the other quadrants will subsequently fill in the spaces not covered in the SW and SE areas.

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August 19th, 2020 at 2:12:57 PM permalink

I found one with 12, after only 4,697,727,345 attempts:

 b c b a e d e a
 c f d b c A d e
 f B C a D E g a
 c h F b a k a e
 b c b c e G e j
 l f H I l J K g
 h i L j g i g j
 l h i h i k j k

The red letters are the knights; the black letters indicate that an empty space is covered by that knight.

charliepatrick