Recommended online casinos

Thread Rating:

Poll

21 votes (45.65%)
14 votes (30.43%)
6 votes (13.04%)
3 votes (6.52%)
12 votes (26.08%)
3 votes (6.52%)
6 votes (13.04%)
5 votes (10.86%)
12 votes (26.08%)
10 votes (21.73%)

46 members have voted

Gialmere
Gialmere
Gialmere
  • Threads: 45
  • Posts: 3047
Joined: Nov 26, 2018
November 27th, 2020 at 10:00:38 AM permalink
Today's puzzle is the simpler, 2d version of this week's toughie Tuesday question...



From the diagram above we can see that the number of tiles on the perimeter, 16, exceeds the number of tiles on the inside, 8.

How many rectangles exist for which the number of tiles on the perimeter are equal to the number of tiles on the inside?
Have you tried 22 tonight? I said 22.
Wizard
Administrator
Wizard
Wizard
  • Threads: 1520
  • Posts: 27118
Joined: Oct 14, 2009
Thanked by
Gialmere
November 27th, 2020 at 11:13:54 AM permalink
Quote: Gialmere

How many rectangles exist for which the number of tiles on the perimeter are equal to the number of tiles on the inside?



2

The following are the dimensions of the inner rectangle:

3x10
4x6
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
ThatDonGuy
ThatDonGuy
ThatDonGuy
  • Threads: 122
  • Posts: 6738
Joined: Jun 22, 2011
Thanked by
rsactuaryGialmere
November 27th, 2020 at 12:26:01 PM permalink

Let H and W be the dimensions, with H >= W
There are a total of HW squares, of which (H-2)(W-2) are interior
The number of interior = the number of exterior when HW = 2 (H-2)(W-2) = 2HW - 4H - 4W + 8
This is HW = 4H + 4W - 8
H (W - 4) = 4W - 8
H = (4W - 8) / (W - 4)
H = (4W - 16 + 8) / (W - 4)
H = 4 + 8 / (W - 4)
Since H is an integer, (W - 4) is a factor of 8
The only factors of 8 are 1, 2, 4, and 8, with resulting H values of 12, 8, 6, and 5, but the latter two have W > H
The only H values where H > W are 12 and 8; the corresponding W values are 5 and 6

12 x 5 and 8 x 6 are the only solutions

Gialmere
Gialmere
Gialmere
  • Threads: 45
  • Posts: 3047
Joined: Nov 26, 2018
Thanked by
miplet
November 27th, 2020 at 8:45:18 PM permalink
Quote: Wizard

2

The following are the dimensions of the inner rectangle:

3x10
4x6


Quote: ThatDonGuy


Let H and W be the dimensions, with H >= W
There are a total of HW squares, of which (H-2)(W-2) are interior
The number of interior = the number of exterior when HW = 2 (H-2)(W-2) = 2HW - 4H - 4W + 8
This is HW = 4H + 4W - 8
H (W - 4) = 4W - 8
H = (4W - 8) / (W - 4)
H = (4W - 16 + 8) / (W - 4)
H = 4 + 8 / (W - 4)
Since H is an integer, (W - 4) is a factor of 8
The only factors of 8 are 1, 2, 4, and 8, with resulting H values of 12, 8, 6, and 5, but the latter two have W > H
The only H values where H > W are 12 and 8; the corresponding W values are 5 and 6

12 x 5 and 8 x 6 are the only solutions


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

The Wi-Fi at my house went out this Thanksgiving.

There are 148 tiles in my bathroom.
Have you tried 22 tonight? I said 22.
Gialmere
Gialmere
Gialmere
  • Threads: 45
  • Posts: 3047
Joined: Nov 26, 2018
November 30th, 2020 at 8:35:38 AM permalink
I have some difficult puzzles this week so even easy Monday isn't all that easy...



You go to visit your pool playing friend and shoot a few games. He's a very superstitious player and had a special 16-ball made to replace his table's 8-ball.

He also insists on racking the balls in a certain way. He arranges them so that each ball is the difference of the two balls above it. He begins to show you how to do it when his phone rings and he leaves the room.

Can you figure out how to finish racking the balls?

Have you tried 22 tonight? I said 22.
ThatDonGuy
ThatDonGuy
ThatDonGuy
  • Threads: 122
  • Posts: 6738
Joined: Jun 22, 2011
Thanked by
Gialmere
November 30th, 2020 at 10:29:05 AM permalink
Quote: Gialmere

Can you figure out how to finish racking the balls?



It's not so much hard as exhaustive, as it's a bit of a brute force problem, but the solution is:
5, 14, 16, 3, 15
9, 2, 13, 12
7, 11, 1
4, 10
6

Gialmere
Gialmere
Gialmere
  • Threads: 45
  • Posts: 3047
Joined: Nov 26, 2018
November 30th, 2020 at 5:08:12 PM permalink
Quote: ThatDonGuy



It's not so much hard as exhaustive, as it's a bit of a brute force problem, but the solution is:
5, 14, 16, 3, 15
9, 2, 13, 12
7, 11, 1
4, 10
6


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

Have you tried 22 tonight? I said 22.
Gialmere
Gialmere
Gialmere
  • Threads: 45
  • Posts: 3047
Joined: Nov 26, 2018
December 1st, 2020 at 9:04:07 AM permalink
It's toughie Tuesday...



A 5 by 5 square lattice is formed by drilling holes in a piece of wood. Three pegs are placed in this lattice at random.

What is the probability that the three pegs will form a triangle?
Have you tried 22 tonight? I said 22.
charliepatrick
charliepatrick
charliepatrick
  • Threads: 39
  • Posts: 3017
Joined: Jun 17, 2011
December 1st, 2020 at 9:31:11 AM permalink
Total perms =25*24*23/6 = 2208
For a triangle to have zero area the holes must be in the same line, diagonal or horizontal/vertical.
Diagonal 3-length lines - 4 of them, 1 perm = 4 combos
Diagonal 4-length lines - 4 of them, 4 perms = 16 combos
Diagonal 5-length lines - 2 of them, 10 perms = 20 combos
Vertical/Horizontal 5-length lines - 10 of them, 10 perms = 100 combos
Total = 140 combos.
So P=140/2208.
Postscript: I know this is wrong (I was on a zoom call at the time!). I agree with the comments made later in the thread.
Last edited by: charliepatrick on Dec 1, 2020
USpapergames
USpapergames
USpapergames
  • Threads: 18
  • Posts: 807
Joined: Jun 23, 2020
December 1st, 2020 at 10:21:09 AM permalink
Nice work! Is anyone else having a difficult time making sense of this form?
Math is the only true form of knowledge
ThatDonGuy
ThatDonGuy
ThatDonGuy
  • Threads: 122
  • Posts: 6738
Joined: Jun 22, 2011
December 1st, 2020 at 10:32:19 AM permalink
Quote: charliepatrick

Total perms =25*24*23/6 = 2208
For a triangle to have zero area the holes must be in the same line, diagonal or horizontal/vertical.
Diagonal 3-length lines - 4 of them, 1 perm = 4 combos
Diagonal 4-length lines - 4 of them, 4 perms = 16 combos
Diagonal 5-length lines - 2 of them, 10 perms = 20 combos
Vertical/Horizontal 5-length lines - 10 of them, 10 perms = 100 combos
Total = 140 combos.
So P=140/2208.



...having three in the same "line" that are knight's moves from each other?
i.e. if you label the rows 1-5 from top to bottom and the columns A-E from left to right, then if the pegs are in A1, C2, and E3, that's not a triangle?
There appear to be 12 of these:
A1-C2-E3, A2-C3-E4, A3-C4-E5
A3-C2-E1, A4-C3-E2, A5-C4-E3
A1-B3-C5, B1-C3-D5, C1-D3-E5
A5-B3-C1, B5-C3-D1, C5-D3-E1

USpapergames
USpapergames
USpapergames
  • Threads: 18
  • Posts: 807
Joined: Jun 23, 2020
December 1st, 2020 at 10:47:16 AM permalink
Quote: ThatDonGuy

Quote: charliepatrick

Total perms =25*24*23/6 = 2208
For a triangle to have zero area the holes must be in the same line, diagonal or horizontal/vertical.
Diagonal 3-length lines - 4 of them, 1 perm = 4 combos
Diagonal 4-length lines - 4 of them, 4 perms = 16 combos
Diagonal 5-length lines - 2 of them, 10 perms = 20 combos
Vertical/Horizontal 5-length lines - 10 of them, 10 perms = 100 combos
Total = 140 combos.
So P=140/2208.



...having three in the same "line" that are knight's moves from each other?
i.e. if you label the rows 1-5 from top to bottom and the columns A-E from left to right, then if the pegs are in A1, C2, and E3, that's not a triangle?
There appear to be 12 of these:
A1-C2-E3, A2-C3-E4, A3-C4-E5
A3-C2-E1, A4-C3-E2, A5-C4-E3
A1-B3-C5, B1-C3-D5, C1-D3-E5
A5-B3-C1, B5-C3-D1, C5-D3-E1



Nice, can't you go a step beyond the knight's move & do a 1 diagonal and 3 spaces up movement or 2 diagonal & 2 spaces up movement to get a triangle as well?

Wait looks like Charlie is only counting the permutations of non-triangle outcomes. So he is accounting for all other outcomes to give a triangle as a result.
Math is the only true form of knowledge
USpapergames
USpapergames
USpapergames
  • Threads: 18
  • Posts: 807
Joined: Jun 23, 2020
December 1st, 2020 at 11:09:11 AM permalink
Ok, so I have the actual answer! Looks like you guys need help with combination analysis.

Total combinations = 25!/(25-3)!3! = 2,300
Total permutations = 25*24*23 = 13,800

For a triangle to have zero area the holes must be in the same line, diagonal or horizontal/vertical. It's ironic that Charlie knows this but still got the wrong answer.

Vertical / Horizontal - 10 combos per 10 lines = 100 combos.
Diagonal - 10 combos per 2 (5 hole) lines + 4 combos per 4 (4 hole) lines + 1 combo per 4 (3 hole) lines = 40 combos.

So P = 140/2,300

Last edited by: USpapergames on Dec 1, 2020
Math is the only true form of knowledge
unJon
unJon
unJon
  • Threads: 16
  • Posts: 4807
Joined: Jul 1, 2018
December 1st, 2020 at 11:26:14 AM permalink
Quote: USpapergames

Ok, so I have the actual answer! Looks like you guys need help with combination analysis.

Total combinations = 25!/(25-3!)3! = 2,300
Total permutations = 25*24*23 = 13,800

For a triangle to have zero area the holes must be in the same line, diagonal or horizontal/vertical. It's ironic that Charlie knows this but still got the wrong answer.


Vertical / Horizontal - 10 combos per 10 lines = 100 combos.
Diagonal - 10 combos per 2 (5 hole) lines + 4 combos per 4 (4 hole) lines + 1 combo per 4 (3 hole) lines = 40 combos.

So P = 140/2,300



You missed the same extra diagonals that Donguy pointed out above. For example, A1-C2-E3
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
ChesterDog
ChesterDog
ChesterDog
  • Threads: 9
  • Posts: 1730
Joined: Jul 26, 2010
Thanked by
Gialmere
December 1st, 2020 at 11:29:27 AM permalink
Quote: Gialmere

It's toughie Tuesday...



A 5 by 5 square lattice is formed by drilling holes in a piece of wood. Three pegs are placed in this lattice at random.

What is the probability that the three pegs will form a triangle?




I counted 152 combinations that are colinear out of a total of 2300 leaving 2148 combinations to make triangles.

Probability of a triangle = 2148 / 2300 = 537 / 575 or approximately 93.39%

USpapergames
USpapergames
USpapergames
  • Threads: 18
  • Posts: 807
Joined: Jun 23, 2020
December 1st, 2020 at 11:31:42 AM permalink
Quote: unJon

Quote: USpapergames

Ok, so I have the actual answer! Looks like you guys need help with combination analysis.

Total combinations = 25!/(25-3!)3! = 2,300
Total permutations = 25*24*23 = 13,800

For a triangle to have zero area the holes must be in the same line, diagonal or horizontal/vertical. It's ironic that Charlie knows this but still got the wrong answer.


Vertical / Horizontal - 10 combos per 10 lines = 100 combos.
Diagonal - 10 combos per 2 (5 hole) lines + 4 combos per 4 (4 hole) lines + 1 combo per 4 (3 hole) lines = 40 combos.

So P = 140/2,300



You missed the same extra diagonals that Donguy pointed out above. For example, A1-C2-E3



It's funny because I thought so also but I believe Donguy is wrong since I am only counting outcomes that don't result in a triangle outcome. So all the knight moves are definitely counted in terms of successful triangle outcomes.
Math is the only true form of knowledge
USpapergames
USpapergames
USpapergames
  • Threads: 18
  • Posts: 807
Joined: Jun 23, 2020
December 1st, 2020 at 11:34:41 AM permalink
Quote: ChesterDog

Quote: Gialmere

It's toughie Tuesday...



A 5 by 5 square lattice is formed by drilling holes in a piece of wood. Three pegs are placed in this lattice at random.

What is the probability that the three pegs will form a triangle?




I counted 152 combinations that are colinear out of a total of 2300 leaving 2148 combinations to make triangles.

Probability of a triangle = 2148 / 2300 = 537 / 575 or approximately 93.39%



Please share your work, this can't be right. Your saying I missed 12 outcomes that don't result in a triangle, and that's just hard for me to believe because from what I have seen on this form, I am the Combo King.
Math is the only true form of knowledge
ChesterDog
ChesterDog
ChesterDog
  • Threads: 9
  • Posts: 1730
Joined: Jul 26, 2010
December 1st, 2020 at 11:47:10 AM permalink
Quote: USpapergames

...
Please share your work, this can't be right. Your saying I missed 12 outcomes that don't result in a triangle, and that's just hard for me to believe because from what I have seen on this form, I am the Combo King.




I found 12 combinations congruent to A1-B3-C5.
USpapergames
USpapergames
USpapergames
  • Threads: 18
  • Posts: 807
Joined: Jun 23, 2020
December 1st, 2020 at 11:58:04 AM permalink
Here are the remaining outcomes with line shapes.


2 Vertical/Horizantal | 1 Diagonal = 6
1 Vertical/Horizantal | 2 Diagonal = 6
Last edited by: USpapergames on Dec 1, 2020
Math is the only true form of knowledge
USpapergames
USpapergames
USpapergames
  • Threads: 18
  • Posts: 807
Joined: Jun 23, 2020
December 1st, 2020 at 11:58:10 AM permalink
Quote: ChesterDog


I found 12 combinations congruent to A1-B3-C5.



Your right, I completely misunderstood the point Donguy was making. The entire time I thought he was attempting to count all the triangle outcomes and was forgetting to include all the triangles from knight moves lol. So exclusion method doesn't work unless you are aware of all possible straight line combinations. Ok so then there are 12 that we missed :(

It's possible I was too narrow-minded after agreeing with charliepatrick but I think the real culprit was being under the impression the question would be easy. Coming up with the answer in less than 2 minutes felt right to me. No need to overthink a simple question lol. It's times like this that I realize my arrogance can be a huge weakness if I were to ever underestimate a problem that could cost me a fortune.
Last edited by: USpapergames on Dec 1, 2020
Math is the only true form of knowledge
charliepatrick
charliepatrick
charliepatrick
  • Threads: 39
  • Posts: 3017
Joined: Jun 17, 2011
Thanked by
Gialmere
December 1st, 2020 at 12:01:13 PM permalink
Quote: charliepatrick

...this is wrong...

Total perms =25*24*23/6 = 2300
For a triangle to have zero area the holes must be in the same line, diagonal or horizontal/vertical.

Diagonal 3-length lines - 4 of them, 1 perm = 4 combos
Diagonal 4-length lines - 4 of them, 4 perms = 16 combos
Diagonal 5-length lines - 2 of them, 10 perms = 20 combos
Vertical/Horizontal 5-length lines - 10 of them, 10 perms = 100 combos
Knight moves (each one only have 1 perm - e.g. A1 C2 E3) = 12 combos
(i) West to East - going up/North - three
(ii) West to East - going down/South - three
(iii) North to South - going right/East - three
(iv) North to South - going left/West - three

Total = 152 combos.

So P=152/2300.
USpapergames
USpapergames
USpapergames
  • Threads: 18
  • Posts: 807
Joined: Jun 23, 2020
December 1st, 2020 at 2:20:17 PM permalink
Now here is an easy math puzzle ;) How far can you walk into a forest?

f/2 or half way
Math is the only true form of knowledge
charliepatrick
charliepatrick
charliepatrick
  • Threads: 39
  • Posts: 3017
Joined: Jun 17, 2011
December 1st, 2020 at 2:27:49 PM permalink
Quote: USpapergames

Now here is an easy math puzzle ;) How far can you walk into a forest?

Assuming there's a clear boundary, then once you've fully crossed it you're inside. So not very far.

Of course if you didn't see it you might walk straight into a tree - gosh that hurt! And if you had come from a long way away, then it might be miles.

btw your "toes" will have come "far" to make it to "a forest" (anag).
USpapergames
USpapergames
USpapergames
  • Threads: 18
  • Posts: 807
Joined: Jun 23, 2020
December 1st, 2020 at 2:31:42 PM permalink
Oh Charlie you make me laugh so hard! You do bring up some great points as usual. We could reframe this question so that it's a shopping mall & not a forest, yet the answer will still be the same ;)

The question isn't about how the traveler would know when he has reached the middle but rather how far is it possible for the traveler to be within the forest.
Math is the only true form of knowledge
Ace2
Ace2
Ace2
  • Threads: 32
  • Posts: 2706
Joined: Oct 2, 2017
December 1st, 2020 at 8:09:50 PM permalink
Quote: USpapergames

Ok, so I have the actual answer! Looks like you guys need help with combination analysis.

You make me laugh long time
It’s all about making that GTA
Gialmere
Gialmere
Gialmere
  • Threads: 45
  • Posts: 3047
Joined: Nov 26, 2018
December 1st, 2020 at 8:47:28 PM permalink
Quote: ChesterDog


I counted 152 combinations that are colinear out of a total of 2300 leaving 2148 combinations to make triangles.

Probability of a triangle = 2148 / 2300 = 537 / 575 or approximately 93.39%


Quote: charliepatrick

Total perms =25*24*23/6 = 2300
For a triangle to have zero area the holes must be in the same line, diagonal or horizontal/vertical.

Diagonal 3-length lines - 4 of them, 1 perm = 4 combos
Diagonal 4-length lines - 4 of them, 4 perms = 16 combos
Diagonal 5-length lines - 2 of them, 10 perms = 20 combos
Vertical/Horizontal 5-length lines - 10 of them, 10 perms = 100 combos
Knight moves (each one only have 1 perm - e.g. A1 C2 E3) = 12 combos
(i) West to East - going up/North - three
(ii) West to East - going down/South - three
(iii) North to South - going right/East - three
(iv) North to South - going left/West - three

Total = 152 combos.

So P=152/2300.


Correct!

(With an assist from ThatDonGuy)


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

Have you tried 22 tonight? I said 22.
USpapergames
USpapergames
USpapergames
  • Threads: 18
  • Posts: 807
Joined: Jun 23, 2020
December 2nd, 2020 at 9:00:29 AM permalink
Quote: Ace2

You make me laugh long time



Glad to know my confusion was amusing ;) And I was assuming all Charlie did wrong was calculating the combinations wrong and title it as permutations. Even assumed Donguy was attempting to solve the problem by calculating all the possible triangle combos before realizing he was just finding more exclusion outcomes (or more straight lines).

Question, what type of math puzzles come to mind when they are easy? To me, the definition of an easy math puzzle is a puzzle that requires no hidden information to solve, where all parts of the puzzle are straightforward & obvious. The puzzle could even take a very long time to solve as long as the average high school student could complete the puzzle with no help.

P.S. It's not like my combinatorial analysis was wrong either ;) Just didn't account for the extra straight-line outcomes which weren't obvious to me within the 2 minutes it took for me to complete my analysis lol. Don't forget I'm the only person to realize Charlie's permutation value was wrong & to come up with the correct combinations & permutation values for the problem.
Math is the only true form of knowledge
Gialmere
Gialmere
Gialmere
  • Threads: 45
  • Posts: 3047
Joined: Nov 26, 2018
December 2nd, 2020 at 9:04:27 AM permalink


The digits 0-9 are written down on ten 3x5 cards. The cards are then separated into five groups of two such that each group makes a prime number. For example...



How many different sets of five primes can you form this way?
Have you tried 22 tonight? I said 22.
charliepatrick
charliepatrick
charliepatrick
  • Threads: 39
  • Posts: 3017
Joined: Jun 17, 2011
December 2nd, 2020 at 9:49:50 AM permalink
Quote: Gialmere

...How many different sets of five primes can you form this way?...

Logic said you didn't need to actually work out which 10-digits numbers were primes (obviously the ones ending in 5 can't be), but as the sum of the digits is a multiple of 9 (or 3) then none can be. Nice question! (btw it's almost a shame the total of the odd and even placed numbers aren't the same as that would make it divisible by 11 as well and give another chance to spot it.)
Edit: I've misread the question assuming you had to use all the cards to make a 10-digit number.
Last edited by: charliepatrick on Dec 2, 2020
CrystalMath
CrystalMath
CrystalMath
  • Threads: 8
  • Posts: 1911
Joined: May 10, 2011
December 2nd, 2020 at 10:09:37 AM permalink
Quote: Gialmere


How many different sets of five primes can you form this way?




Each card must contain an odd and an even number, with the even preceding the odd.
05 is the only prime you can create with the number 5, so we only need to look at the ways to place the numbers 2,4,6,8 with 1,3,7,9.

We have the following possible primes:
41, 61
23, 43, 83
47, 67
29, 89

Because 41/61 choice will determine what number we use for 47/67, then we can never use the number 43.

This gets us down to:
41, 61
23, 83
47, 67
29, 89

So, it looks to me like there are only 4 choices:
05, 41, 67, 23, 89
05, 41, 67, 83, 29
05, 61, 47, 23, 89
05, 61, 47, 83, 29
I heart Crystal Math.
unJon
unJon
unJon
  • Threads: 16
  • Posts: 4807
Joined: Jul 1, 2018
December 2nd, 2020 at 11:02:45 AM permalink
Quote: CrystalMath

Quote: Gialmere


How many different sets of five primes can you form this way?




Each card must contain an odd and an even number, with the even preceding the odd.
05 is the only prime you can create with the number 5, so we only need to look at the ways to place the numbers 2,4,6,8 with 1,3,7,9.

We have the following possible primes:
41, 61
23, 43, 83
47, 67
29, 89

Because 41/61 choice will determine what number we use for 47/67, then we can never use the number 43.

This gets us down to:
41, 61
23, 83
47, 67
29, 89

So, it looks to me like there are only 4 choices:
05, 41, 67, 23, 89
05, 41, 67, 83, 29
05, 61, 47, 23, 89
05, 61, 47, 83, 29



Your first assumption is incorrect:

You can make a card with 02 which would then let you make a 53 card or a 59 card.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
ThatDonGuy
ThatDonGuy
ThatDonGuy
  • Threads: 122
  • Posts: 6738
Joined: Jun 22, 2011
Thanked by
Gialmere
December 2nd, 2020 at 12:37:14 PM permalink
Quote: Gialmere

How many different sets of five primes can you form this way?



0 cannot be the second digit as no primes end in 0, so there are four two-digit and one one-digit primes (i.e. 2, 3, 5, or 7).
The only primes < 100 that contain a 5 are 5, 53, and 59
4, 6, and 8 have to be the first digit of a prime as otherwise it would be an even number > 2
If 2 is not the one-digit prime, it must also be the first digit of a prime

The two-digit primes that begin with 2 are 23 and 29
The two-digit primes that begin with 4 are 41, 43, and 47
The two-digit primes that begin with 6 are 61 and 67
The two-digit primes that begin with 8 are 83 and 89

If 5 is one of the primes, then the other four begin with 2, 4, 6, 8 and end with 1, 3, 7, 9
The possibilities are:
05, 23, 41, 67, 89
05, 23, 47, 61, 89
05, 29, 41, 67, 83
05, 29, 47, 61, 83

If 53 is one of the primes, then 2 is another one, and the other three begin with 4, 6, 8 and end with 1, 7, 9
The possibilities are:
02, 53, 41, 67, 89
02, 52, 47, 61, 89

If 59 is one of the primes, then 2 is another one, and the other three begin with 4, 6, 8 and end with 1, 3, 7
The possibilities are:
02, 59, 41, 67, 83
02, 59, 47, 61, 83

There are a total of 8 sets of five primes

CrystalMath
CrystalMath
CrystalMath
  • Threads: 8
  • Posts: 1911
Joined: May 10, 2011
December 2nd, 2020 at 4:51:41 PM permalink
Quote: unJon

Quote: CrystalMath

Quote: Gialmere


How many different sets of five primes can you form this way?



Your first assumption is incorrect:

You can make a card with 02 which would then let you make a 53 card or a 59 card.




d'oh
I heart Crystal Math.
Gialmere
Gialmere
Gialmere
  • Threads: 45
  • Posts: 3047
Joined: Nov 26, 2018
December 2nd, 2020 at 5:50:23 PM permalink
Quote: ThatDonGuy



0 cannot be the second digit as no primes end in 0, so there are four two-digit and one one-digit primes (i.e. 2, 3, 5, or 7).
The only primes < 100 that contain a 5 are 5, 53, and 59
4, 6, and 8 have to be the first digit of a prime as otherwise it would be an even number > 2
If 2 is not the one-digit prime, it must also be the first digit of a prime

The two-digit primes that begin with 2 are 23 and 29
The two-digit primes that begin with 4 are 41, 43, and 47
The two-digit primes that begin with 6 are 61 and 67
The two-digit primes that begin with 8 are 83 and 89

If 5 is one of the primes, then the other four begin with 2, 4, 6, 8 and end with 1, 3, 7, 9
The possibilities are:
05, 23, 41, 67, 89
05, 23, 47, 61, 89
05, 29, 41, 67, 83
05, 29, 47, 61, 83

If 53 is one of the primes, then 2 is another one, and the other three begin with 4, 6, 8 and end with 1, 7, 9
The possibilities are:
02, 53, 41, 67, 89
02, 52, 47, 61, 89

If 59 is one of the primes, then 2 is another one, and the other three begin with 4, 6, 8 and end with 1, 3, 7
The possibilities are:
02, 59, 41, 67, 83
02, 59, 47, 61, 83

There are a total of 8 sets of five primes


Correct!

(Although, I spy a composite number typo in there.)
--------------------------------------

Have you tried 22 tonight? I said 22.
Gialmere
Gialmere
Gialmere
  • Threads: 45
  • Posts: 3047
Joined: Nov 26, 2018
December 3rd, 2020 at 8:05:26 AM permalink


A fair die is rolled ten times.

What is the probability that the sequence of rolls is non-decreasing?

That is, each roll is greater than or equal to the previous roll.
Have you tried 22 tonight? I said 22.
ThatDonGuy
ThatDonGuy
ThatDonGuy
  • Threads: 122
  • Posts: 6738
Joined: Jun 22, 2011
Thanked by
Gialmere
December 3rd, 2020 at 8:41:48 AM permalink
Quote: Gialmere

What is the probability that the sequence of rolls is non-decreasing?


This looks like a Markov chain problem - suitable more for computing than math.

Let P(a,b) be the probability of success after a rolls where the most recent roll is b
P(10,n) = 1 for all n
P(a,n) = 1/6 (P(a+1,n) + P(a+1,n+1) + ... + P(a+1,6)) for all n, a <= n <= 6
The solution is P(0,1)

Rolls123456
915 / 62 / 31 / 21 / 31 / 6
87 / 125 / 125 / 181 / 61 / 121 / 36
77 / 2735 / 2165 / 545 / 1081 / 541 / 216
67 / 7235 / 64835 / 1,2965 / 4325 / 1,2961 / 1,296
57 / 2167 / 4327 / 9727 / 2,5921 / 1,2961 / 7,776
477 / 7,77635 / 7,7767 / 3,8887 / 11,6647 / 46,6561 / 46,656
311 / 3,88855 / 46,6565 / 11,6641 / 7,7761 / 34,9921 / 279,936
2143 / 186,62455 / 186,62455 / 559,8725 / 186,6241 / 186,6241 / 1,679,616
11,001 / 5,038,848715 / 10,077,69655 / 2,519,42455 / 10,077,6965 / 5,038,8481 / 10,077,696

Result = 1,001 / 20,155,392 = about 1 / 20,135

Gialmere
Gialmere
Gialmere
  • Threads: 45
  • Posts: 3047
Joined: Nov 26, 2018
December 3rd, 2020 at 4:57:37 PM permalink
Quote: ThatDonGuy


This looks like a Markov chain problem - suitable more for computing than math.


Let P(a,b) be the probability of success after a rolls where the most recent roll is b
P(10,n) = 1 for all n
P(a,n) = 1/6 (P(a+1,n) + P(a+1,n+1) + ... + P(a+1,6)) for all n, a <= n <= 6
The solution is P(0,1)

Rolls123456
915 / 62 / 31 / 21 / 31 / 6
87 / 125 / 125 / 181 / 61 / 121 / 36
77 / 2735 / 2165 / 545 / 1081 / 541 / 216
67 / 7235 / 64835 / 1,2965 / 4325 / 1,2961 / 1,296
57 / 2167 / 4327 / 9727 / 2,5921 / 1,2961 / 7,776
477 / 7,77635 / 7,7767 / 3,8887 / 11,6647 / 46,6561 / 46,656
311 / 3,88855 / 46,6565 / 11,6641 / 7,7761 / 34,9921 / 279,936
2143 / 186,62455 / 186,62455 / 559,8725 / 186,6241 / 186,6241 / 1,679,616
11,001 / 5,038,848715 / 10,077,69655 / 2,519,42455 / 10,077,6965 / 5,038,8481 / 10,077,696

Result = 1,001 / 20,155,392 = about 1 / 20,135



Correct!

Very good.
-------------------------------

Have you tried 22 tonight? I said 22.
Ace2
Ace2
Ace2
  • Threads: 32
  • Posts: 2706
Joined: Oct 2, 2017
Thanked by
Gialmere
December 3rd, 2020 at 6:47:24 PM permalink
Quote: Gialmere

A fair die is rolled ten times.

What is the probability that the sequence of rolls is non-decreasing?

That is, each roll is greater than or equal to the previous roll.

There's another way to calculate this. If we only count permutations that have at least one of each side 1-6, then the answer would be combin(9,5) = 126 (out of 6^10). This is because there are 126 ways to insert 5 dividers between 10 values so that 6 bins are created. This, for example, would represent 1123334556: xx/x/xxx/x/xx/x

However, a bin can contain zero values, so we add 1 for each of the 6 bins and "overcount" each side by 1. But not exactly since the value we are overcounting is a zero. x/xxx/x/xxxxxx/xxx/xx represents 2244444556, for instance.

Therefore the answer becomes combin(15,5) = 3,003 (out of 6^10). This is because there are 3,003 ways to insert 5 dividers between 16 values so that 6 bins are created.
It’s all about making that GTA
Gialmere
Gialmere
Gialmere
  • Threads: 45
  • Posts: 3047
Joined: Nov 26, 2018
December 4th, 2020 at 8:00:05 AM permalink


Oh no! You're driving your car through town when you see you're going to be stuck behind a big Budget Rental Truck at the next traffic light.

A Budget Rental Truck is 12 feet 8 inches tall (from the road to the top of the Van). The cab of the truck is 8 feet long and the van is 24 feet long (see diagram below).

If the front of the truck is 8 feet from the stoplight suspended 18 feet above the road, how far behind the truck must your car be so that you can still see the stoplight?

Assume your eyes are 4 feet above the road.

Have you tried 22 tonight? I said 22.
ThatDonGuy
ThatDonGuy
ThatDonGuy
  • Threads: 122
  • Posts: 6738
Joined: Jun 22, 2011
Thanked by
Gialmere
December 4th, 2020 at 8:26:19 AM permalink

The top of the truck is 5 1/3 feet below the stoplight, and 40 feet to the right of the stoplight
Let X be the distance you are behind the truck
Your eyes are 14 feet below the stoplight, and (40 + X) feet to the right of the stoplight
The minimum distance where you can see the stoplight is where the two height/length ratios are equal
40 / (5 1/3) = (40 + X) / 14
15 / 2 = (40 + X) / 14
210 = 80 + 2X
X = 65 feet

chevy
chevy
chevy
  • Threads: 3
  • Posts: 151
Joined: Apr 15, 2011
Thanked by
Gialmere
December 4th, 2020 at 8:46:28 AM permalink

ray from light from signal to top right corner of rear of truck's van continues on to your eye.

similar triangles give (in feet)

x/(8+8+24) =((12+2/3) - 4) / (18 - (12+2/3))

x= (40 ft) * (104 in) / (64 in) =

x= 65 ft


Though this assumes your eyes are located at the front of your car. If we assume a car length of maybe 12 ft and driver's seat is say 4 ft from the front bumper....

x = 61 ft


EDIT: i.e. the car's front bumper is at 61 ft from rear of truck....your eyes are still at 65ft
charliepatrick
charliepatrick
charliepatrick
  • Threads: 39
  • Posts: 3017
Joined: Jun 17, 2011
December 4th, 2020 at 8:57:18 AM permalink
I had mnisread it as 12'6" rather than 12'8"! Nevertheless I agree that if you were doing 25mph and kept to the 2-second rule you would be able to see the lights. Therefore you wouldn't be too near to the truck! However that wouldn't apply further back - now that's a mathematical puzzle depending on both of your speeds.
Gialmere
Gialmere
Gialmere
  • Threads: 45
  • Posts: 3047
Joined: Nov 26, 2018
December 4th, 2020 at 5:47:16 PM permalink
Quote: ThatDonGuy


The top of the truck is 5 1/3 feet below the stoplight, and 40 feet to the right of the stoplight
Let X be the distance you are behind the truck
Your eyes are 14 feet below the stoplight, and (40 + X) feet to the right of the stoplight
The minimum distance where you can see the stoplight is where the two height/length ratios are equal
40 / (5 1/3) = (40 + X) / 14
15 / 2 = (40 + X) / 14
210 = 80 + 2X
X = 65 feet


Quote: chevy


ray from light from signal to top right corner of rear of truck's van continues on to your eye.

similar triangles give (in feet)

x/(8+8+24) =((12+2/3) - 4) / (18 - (12+2/3))

x= (40 ft) * (104 in) / (64 in) =

x= 65 ft


Though this assumes your eyes are located at the front of your car. If we assume a car length of maybe 12 ft and driver's seat is say 4 ft from the front bumper....

x = 61 ft


EDIT: i.e. the car's front bumper is at 61 ft from rear of truck....your eyes are still at 65ft


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


A man was waiting at a red light when an alligator and a parrot approached his car window.

The man rolled down his window to see what they wanted.

The alligator said "Excuse me sir, we have reason to believe you've taken hallucinogenics."
Have you tried 22 tonight? I said 22.
Gialmere
Gialmere
Gialmere
  • Threads: 45
  • Posts: 3047
Joined: Nov 26, 2018
December 7th, 2020 at 8:00:33 AM permalink
It's easy Monday. Let's dance...



The ratio of boys to girls at a high school dance is 9:10.

An extra 17 boys arrive and the ratio changes to 8:7.

How many girls are there at the dance?
Have you tried 22 tonight? I said 22.
Wizard
Administrator
Wizard
Wizard
  • Threads: 1520
  • Posts: 27118
Joined: Oct 14, 2009
Thanked by
Gialmere
December 7th, 2020 at 8:16:35 AM permalink
Quote: Gialmere

How many girls are there at the dance?



70
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
chevy
chevy
chevy
  • Threads: 3
  • Posts: 151
Joined: Apr 15, 2011
Thanked by
Gialmere
December 7th, 2020 at 8:17:28 AM permalink


B=# of Boys, G=# of Girls)

Originally : B = (9/10) G

Later : B+17 = (8/7) G

Solving

(9/10) G +17 = (8/7) G

G=70 girls (and B= 63 boys)

Gialmere
Gialmere
Gialmere
  • Threads: 45
  • Posts: 3047
Joined: Nov 26, 2018
December 7th, 2020 at 4:22:07 PM permalink
Quote: Wizard

70


Quote: chevy



B=# of Boys, G=# of Girls)

Originally : B = (9/10) G

Later : B+17 = (8/7) G

Solving

(9/10) G +17 = (8/7) G

G=70 girls (and B= 63 boys)


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

Trust me ... You can dance.

Love,

Alcohol
Have you tried 22 tonight? I said 22.
Wizard
Administrator
Wizard
Wizard
  • Threads: 1520
  • Posts: 27118
Joined: Oct 14, 2009
December 8th, 2020 at 8:01:18 AM permalink
It's Toughy Tuesday and I am ready!
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
Gialmere
Gialmere
Gialmere
  • Threads: 45
  • Posts: 3047
Joined: Nov 26, 2018
December 8th, 2020 at 8:08:22 AM permalink
It's toughie Tuesday. You're hexed...



A hexagon with consecutive sides of lengths 2, 2, 7, 7, 11, and 11 is inscribed in a circle.

What is the radius of the circle?
Have you tried 22 tonight? I said 22.
ChesterDog
ChesterDog
ChesterDog
  • Threads: 9
  • Posts: 1730
Joined: Jul 26, 2010
Thanked by
Gialmere
December 8th, 2020 at 9:15:09 AM permalink
Quote: Gialmere


......
What is the radius of the circle?




Let r = the radius of the circle.
Let A = the central angle subtending a side of length 2.
Let B = the central angle subtending a side of length 7.
Let C = the central angle subtending a side of length 11.

Apply the "law of cosines" for each angle:
cos A = ( r2 + r2 - 22 ) / (2 r2 ) = 1 - 22 / ( 2 r2 )
cos B = ( r2 + r2 - 72 ) / (2 r2 ) = 1 - 72 / ( 2 r2 )
cos C = ( r2 + r2 - 112 ) / (2 r2 ) = 1 - 112 / ( 2 r2 )

Using "trial and error" to find r such that 2A + 2B + 2C = 360 degrees yields r = 7.
RSS Feed
  • Jump to: