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The person I talked to who actually plays Chutes-&-Ladders (BTW he is four years old ) said that you roll the dice "until I win."
That removes a little bit of the randomness from the game.
A google search turned up an interesting article written by Dr. David L. Morgan (http://www.math.niu.edu/~rusin/uses-math/games/chutes/) then states that the math average is 2.35 spaces per turn or 42.5 turns per game. He includes a C source code program to simulate the game. After a run of 10000 (ten thou) game the average number of turns per game was 35.7534
Actually that is more than I need to know about the game, if I'm going to lose when I play the game.
I will ask someone who actually plays the game.
Welcome emadstats
emadstats -> e mad stats or stats dame
Are statistics a casual or serious interest?
The original problem is as follows.
Suppose I have several members in the room.
I want to give to each of them different number.So member Q[1] has some number, member Q[2] different number, etc.
Then I need to come up with public parameter P that I will announce in the room.
Everybody is able to see that number, but only valid members Q[1], Q[2], Q[3] will be able to recover passwordQ[1] mod P = Z (which is a secret number)
Q[2] mod P = Z
Q[3] mod P = ZIn case I want to exclude Q[3] member from the list, algorithm should be able to come up with different public parameter, lets call it R that helps remaining members generate new secret number
Q[1] mod R = W (this is new secret)
Q[2] mod R = W
Obviously W != Z. since we cannot use old secret number.
numbers(prime?) ,1111,1719,2643,4177,4379,4797,5431,7817,9741,3789,4513,8741,1917
ID,Q[], &residues
A,103872994,49,700,451,3535,3114,3553,5119,698,4711,1348,1786,3691,349
B,157971872,1004,929,2405,1909,3826,1865,[b]375[/b],5936,2075,884,3333,4520,1487
C,105879513,102,1146,933,[b]917[/b],4051,129,2168,6065,4584,3486,20,8521,1686
D,149575546,505,199,247,1353,2043,289,[b]375[/b],5068,2491,982,1187,8295,1621
E,190090806,928,348,960,3890,2795,87,[b]375[/b],4817,4932,465,3246,279,1086
F,146230050,230,1596,789,1634,2103,3099,[b]375[/b],5248,7899,1173,4337,1861,1290
G,131537022,177,861,198,3292,620,3282,3633,363,4299,1887,1124,2454,150
H,108196757,911,1178,266,4103,425,422,[b]375[/b],1660,3470,1862,2095,659,1277
I,156049460,622,359,1454,[b]917[/b],3795,3050,537,6506,8381,3284,3459,5128,1826
J,159203095,128,1348,1990,[b]917[/b],171,259,4192,2073,5932,682,2507,3262,79
K,157757853,297,66,2469,[b]917[/b],4378,3711,3596,2976,2358,2838,1425,285,255
L,197629034,1021,761,1352,2633,385,2228,375,7457,3626,2372,251,3765,1670You specifiy a number for the entire group.
In this sample only two primes will produce the desired keys.
The secret keys are 917 and 375.
Is this what you are attempting?
This generation of number is very easy to do.
Trying different primes will require a lot of effort and my be effective enough to preclude being broken by that method.
HOWEVER, it is easily broken, NOT by trying different primes.
The Q[] numbers are only a few thousand times larger than the modulus.
Thus this is easily cracked by differences of the Q[] and residues.
If possible create a set of Q[] values, that you want to use and show them here.
In the following:
a1 means 1 across; a3 means 3 across; etc.
d1 means 1 down; d2 means 2 down; etc.
a1 is defined in terms of d4.
a1 [LHS] points to or refers to d4 [RHS]
or [in my notation]
a1 -> d4
d4 -> a3
a3 -> a5
a5 -> d2
d2 -> a5 & a6
a5 already refers to d2 -- self ref
a6 -> a3
a3 -> a5
a5 already refers to d2 -- self ref
This enters a self referencing loop.
That implys that a5 or d2 can be isolated.
Therefore, substituting, expanding and simplifying
an exact number is determined.
If you need further assistance...
I'd guess those are clue numbers, not leading digits.
OK.
abcde--
f---g--
h-ijklm
n-o-p-q
rstuv-w
--x---y
--záíóúUsing the clues above, fill in the numbers for the letters.
Simple enough.
But I did forget to mention leading zeros are allowed
eg 00123 = 123
Based on the clues it is IMPOSSIBLE to have a leading zero.
You have already specified the first digit of all the numbers.
I do not understand what you are trying to explain.
You can export the excel spreadsheet to a ".txt" or better to a ".csv" file and then include it
in your message as plain text.
This is what I have done below.
There are infinite amount of tables (Table 1, Table 2, Table 3, , Table 100, Table 101, etc), with ten rows (row 0,1,2, .9) each.
Inside of each tables, there are numbers from 1 to 92, 93 to 184, 185 to 276, and 277 to 284, which lying on their certain rows.
Here I gave the example tables that have been filled in for 40 tables.
By finding the patterns/ formulas, my friend asked me to extend the tables to fill in the blank tables 41,42,43,etc as given beneath of Table 40.
Just like SUDOKU, in each tables there will be no same numbers vertically, horizontally and diagonally. If these tables are using permutations from an ideal table that you can see beneath Table 43 (supposed the ideal table was right), then how to find the formulas of its permutations?
Can somebody help me about this?
These tables are truncated on the right (so that that fit on one line of this page).
Question: How is Table 1 supposed to differ from table 2.
Infinite number of tables?
It looks as if you only have 920 numbers with which to work ( 10 rows * 92 = 920)
Table 1
0 1 2 3 4 5 6 7 8
1 93 94 95 96 97 98 99 100
2 185 186 187 188 189 190 191 192
3 277 278 279 280 281 282 283 284
4 369 370 371 372 373 374 375 376
5 461 462 463 464 465 466 467 468
6 553 554 555 556 557 558 559 560
7 645 646 647 648 649 650 651 652
8 737 738 739 740 741 742 743 744
9 829 830 831 832 833 834 835 836No numbers are repeated horizontall, vertically nor diagonally.
How is this to be constructed differenly from Table 1
Table 2
0 1 2 3 4 5 6 7 8
1 93 94 95 96 97 98 99 100
2 185 186 187 188 189 190 191 192
3 277 278 279 280 281 282 283 284
4 369 370 371 372 373 374 375 376
5 461 462 463 464 465 466 467 468
6 553 554 555 556 557 558 559 560
7 645 646 647 648 649 650 651 652
8 737 738 739 740 741 742 743 744
9 829 830 831 832 833 834 835 836How is this to be form that is different from Table 1 and Table 2?
Table 3
0 1 2 3 4 5 6 7 8
1 93 94 95 96 97 98 99 100
2 185 186 187 188 189 190 191 192
3 277 278 279 280 281 282 283 284
4 369 370 371 372 373 374 375 376
5 461 462 463 464 465 466 467 468
6 553 554 555 556 557 558 559 560
7 645 646 647 648 649 650 651 652
8 737 738 739 740 741 742 743 744
9 829 830 831 832 833 834 835 836Tables 4 thorough Table 16384
How are they to be generated?
Give a example or more detailed clues.
Suppose:
Your are a teacher.
You have two students.
Each student brings you a paper with some "art work" on it.
Paper 1 shows a railroad track, and looks as if it were a picture.
It is not a tracing. You know it is the result of the students own work.
You know that because you saw the students doing it.
Paper 2 shows a free hand drawn rectangle with diagonals connecting
the corners.
The question.
How do you grade the papers? Which student should receive a promotion.
Suppose:
You have 2 employees.
You are charged with getting getting life saving food and water to a village
devastated by a natural diaster.
Your instructions to the employees. Each is to load a truck with food & water
and get it to those who need it.
Hours later:
Employee 1 has truck 1 about 1/4 filled, nicely stacked, neatly arranged, and
looking very good.
Employee 2 has truck 2 filled, but it is stuffed, packed and obvious a few ruined
containers.
Which employee is going to mean the most to you?
Which one would you recommend for a promotion.
It is obvious that employee 1 enjoys his work and loves what he does.
Employee 2 just gets the job done.
Who would you want on your team.?
Across
6. (3 across x 2)-2618
3xxx4 * 2 = ????8 - 2618 = ????0 <--last digit is a zero
Down
4. (3 across x 7)+257
3xxx4 * 7 = ????8 + 257 = ????5 <--last digit is a five
Seems inconsistent.
The Chinese Remaind Theorem (CRT) seems partially helpful.
The CRT typically has different moduli.
Could you give a specific example.
Exactly what numbers do you know, and what is unknown.
It looks as if all values of Q[] are known.
It appears as if you know what Z is and you are looking for P and P'
If that is the case:
subtract Z from each of the Q[], then perform a gcd operation on the group.
gcd( Q[1]-Z, Q[2]-Z, Q[3]-Z, Q[4]-Z, Q[5]-Z ) == P
I'm not sure that I understand the question.
Please elaborate with a couple small examples.
Could you elaborate and give a small example of what you want to happen.
If your example were change:
Q[1] mod P = Z
Q[2] mod P = Z
This would work.
The quadratic formula can be used to answer all three.
If you know it, then no problem.
If you don't understand it ....
Side Note: Question 3.
The unit_price relative to the quantity_sold is rarely linear.
Go in the other direction: at $0 cost you have a limited distribution of 420 students.
It is not the people of Alabama, the United States, Britian, France, Iran, Iraq, or any other place. It is persons who cannot control or direct anger in a more meaningful manner.
Sometimes death by violence is the only choice.
Most often, there are many other ways to reach a solution to a problem.
Re-arranging you prejudices doesn't make you a more rational person, but sometimes it allows insight into your specific bias. (And for that, I apologize to the original author - William James)
why is there 7 & seven alone??
Because of the other 4137 problems, no one is interested
in paying any money to get the answers.
Actually, there are more than seven that some groups
are willing to pay for someone to provide viable answers.
Lot of problems have only very small monetary rewards.
Take a lot at the $621 question.
What would your chances be if Monty Hall opened the door showing the winning prize?
Would that change your probability of winning?
To: charismath
Please accept my apology. I'm sorry that my post offended you.
It was therefore, to you, a lame attempt to repharse "things are not always what they seem."
Generally, there is more to a story than what is observed.
As you read more posts in more forums, you will be able to discern the difference between
good comments and bad ones, brilliant ones and stupid ones, sarcastic and mean, etc.
Again, it was not intended to annoy.
Either
the Dodecahedron (12 Faces, 20 Vertices , 30 Edges )
or
the Icosahedron (20 Faces , 12 Vertices, 30 Edges )
For the icosahedron this is my method of choice:
As you increase the size of the "sphere" the distance
from the center of the sphere to a vertex will become
larger than the distance from the center of the sphere
to the midpoint of an edge. Obvious. If will remain in
a constant proportion. However, when that distance
exceeds a preset number of pixels, replace each
equilateral triangle with 4 equilateral triangles, and
then reset the vertices to equal distances.
The decision depends on the resolution of the screen.
The calculations are done only when the difference
between the arc of the sphere and the edge (chord)
are noticeable.
I thought this was a brilliant idea.
Then I discovered it was a mistake.
I really like of idea of ME being able to tag a site as unfit, so
that I do not click on it again. There are too many sites that
re-direct the [BACK] button back into their site. It then becomes
necessary to abort the browzer and restart.
I would like to put those web sites in a nuisance folder so that if
a few weeks later, I clicked on a site, the system would remind me
that this site was previously flagged by me as off limits.
Testing IQ is not illegal.
What will put the company/employer in a cauldron of trouble is asking questions that are not directly related to the job for which the applicant is applying.
Even if the tester believes the question has merit, it can cause problems.
If you are looking to hire a person to bail water, and then ask questions about quantum physics, you might be forced to hire the person who has no arms.
It should be obvious, WE NEED MORE LAWYERS!
Why was it necessary to use coffee?
Was it coarse sand or fine sand?
Was it a standard silicon glass jar or
an expensive hand crafted specimen?
There are a lot of unanswered questions
about this lesson.
What brand of golf...
Need some (a lot of) help with Chinese Remainder Theorem
How can the Chinese Remainder Theorem be used to solve this:
Sorry for the poor form above.
Obviously, LaTex not fully understood, but I have downloaded the LaTeX tutorial.
Background
(not necessary, but you might be interested in how the above was found.
A middle school student (7th grade)
wanted to know why the Perrin, Lucas and
Fibonacci sequences do not have negative
numbers. Since they are defined as such
they do not. The negative values only occur
when moving in the negative direction.
However, other sequences with different
initial values can and do have negative
values on the positive side of zero.
The formula for the Perrin Sequence:
P(n+0) + P(n+1) = P(n+3)
with the intial values of: P(0)= 3, P(1)= 0, P(2)= 2
The Perrin Sequence also satisfies the equation:
P(n+0) + P(n+1) + P(n+2) + P(n+3) + P(n+4) = P(n+8) {1}
A + B + C + D + E = I
If we assign random values to five consecutive terms
in the squence, and using only equation {1} assign
values to all other terms in a quasi Perrin sequence,
we should have enough data to compute the intermediate terms:
P(n+5), P(n+6), P(n+7)
F G H
Our sequence defined:
P(-3) P(-2) P(-1) P(0) P(1) P(2) P(3) P(4) P(5) P(6) P(7) P(8) P(9)
X Y Z A B C D E F G H I J
X,Y,Z are unknown
A,B,C,D,E are assigned random numeric values
it follows that
F = X+Y+Z+A+B
G = Y+Z+A+B+C
H = Z+A+B+C+D
I = A+B+C+D+E
J = B+C+D+E+F
By substituting, combining, reducing we derived these equations:
X + 2Y + 3Z == v1 mod 11
2X + 3Y + 4Z == v2 mod 13
6X + 10Y + 13Z == v3 mod 17
9X + 16Y + 22Z == v4 mod 19
This is where we are -- enveloped in a fog.
The Chinese Remainder Theorem is easy enough to use on single
unknown values:
x == a mod b
x == c mod d
But with three unknowns we are lost.
An internet search only shows methods to solve for a single
unknown value.
Any one have an idea as to how we need to proceed so that
we may determine the smallest positive value for X?
Y and Z would then be defined in relation to X.
Any help, hints, pointers, references, or thoughts would
be appreciated.
Thanks.
You don't want to prove theorems that are false.
In my opinion -- GUESTS should not be able to post.
If anyone wants/needs to post, register first.
That will not stop the stupidity -- but it might slow it.
I like this forum and I would like for it to stay available.
Sometimes there are some BRILLIANT posts.
Thanks.
Memorization and repetition.
If it is of value you will get emotionally involved and it will stick forever.
It is important to know pi out to several decimal digits, but it is practically
useless to know pi to several hundred or more decimal digits. What would you do
or how could you use such information?
It is much more important to know where to get the information.
Knowing a lot of facts is okay, but knowing how to use information to achieve something
is everything.