GingerAle

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UsernameGingerAle
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Questions 2
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 #9
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0

My criticism isn’t because of his intuition. It’s because of his lack of an explanation or the  fluff and blarney he uses in its stead.  His explanation of reasoning does not extend to similar problems nor does it really explain the solution method for this exact problem.

 

All mathamations develop evolving intuition in solving problems. This usually develops as the students learn the algorithms and rote formulas for solutions. For most, mathematics starts with “Ours is not to know the reason why, it’s to invert the divisor and multiply.”

 

While Mr. BB appears to have an intuition to solve this and other related modulo problems, his explanations never convey any substance for reason, nor foundation for a rote formula that is useful to anyone.

 

A case in point. You have a degree in mathematics and decades of practiced skill. You have a level intuition that far exceeds the vast majority of college students and most graduates. Yet, Mr. BB’s explanation didn’t trigger much, if any, understanding for the nature of the problem or its solution. If this doesn’t float any part of your banana boat, then no one has a prayer to the banana goddess of hope for any understanding.   

 

Though his comments may be true, they are only true by tautology. They are not very useful for this particular problem, and they are useless for any related problem.  No one learns anything from Mr. BB’s Blarney, except how to become a Blarney Master.  He is one too. . . . One of the best I’ve seen.indecision

 

 

 

GA

GingerAle Feb 7, 2018
 #6
avatar+902 
+2

Solve this by setting up a system of modular equations.

 

\(\begin{array}{rcll} n &\equiv& {\color{red}0} \pmod {{\color{green}8}} \\ n &\equiv& {\color{red}1} \pmod {{\color{green}9}} \\ n &\equiv& {\color{red}2} \pmod {{\color{green}10}} \\ \text{Set } m &=& 8\cdot 9\cdot 10 = 720 \\ \end{array}\)

 

The first product zero— included as a formality.

 Eurler totients calculated from non-prime numbers.

 

 

\(\small{ \begin{array}{l} n = {\color{red}0} \cdot {\color{green}9\cdot 10} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ { (\color{green}9 \cdot 10) }^{\varphi({\color{green}8}) -1 } \pmod {{\color{green}8}} ] }_{=\text{modulo inverse }(9\cdot 10) \mod 8 } }_{=(9\cdot 10)^{4-1} \mod {8}} }_{=(9\cdot 10)^{3} \mod {8}} }_{=(90\pmod{8})^{3} \mod {8}} }_{=(2)^{3} \mod {8}} }_{= 0} + {\color{red}1} \cdot {\color{green}8\cdot 10} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ { (\color{green}8\cdot 10) }^{\varphi({\color{green}9}) -1} \pmod {{\color{green}9}} ] }_{=\text{modulo inverse } (8\cdot 10) \mod {9}} }_{=(8\cdot 10)^{6-1} \mod {9}} }_{=(8\cdot 10)^{5} \mod {9}} }_{=(80\pmod{9})^{5} \mod {9}} }_{=(8)^{5} \mod {9}} }_{=8} + {\color{red}2} \cdot {\color{green}8\cdot 9} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ { (\color{green}8\cdot 9) }^{\varphi({\color{green}10}) -1 } \pmod {{\color{green}10}} ] }_{=\text{modulo inverse } (8\cdot 9) \mod 10 } }_{=(8\cdot 9)^{4-1} \mod { 10}} }_{=(8\cdot 9)^{3} \mod {10}} }_{=(72\pmod{10})^{3} \mod {10}} }_{=(2)^{3} \mod {10}} }_{=8}\\ \\ n = {\color{red}0} \cdot {\color{green}9\cdot 10} \cdot [0] + {\color{red}1} \cdot {\color{green}8\cdot 10} \cdot [8] + {\color{red}2} \cdot {\color{green}8\cdot 9} \cdot [8] \\ n = 0+ 640 + 1152 \\ n = 1792 \\\\ n \pmod {m}\\ = 1792 \pmod {720} \\ = 352 \\ \mathbf{n_{min}} \mathbf{=} \mathbf{352} \end{array} } \)

GingerAle Feb 6, 2018
 #6
avatar+902 
+3

This link has both the original Latin and English translation for Part 2 of Ars Conjectandi:  The Doctrine of Permutations and Combinations: Being an Essential and Fundamental Doctrine of Changes.  (1795). (Navigate to page 217 for the English translation of the page Heureka references.) Despite the archaic spelling, syntax, and mathematical notation, this text is very readable. This book includes several related essays and theories from notable mathematicians, contemporary to, and proceeding, Jakob Bernoulli.

 

Complete English translations of Ars Conjectandi are rare.  This is required reading for Lancelot Link’s school of Mathematics, Physics, Chemistry, and Trolling. I had to brush-up on my Latin.smiley

For anyone interested, here’s Part 4 of  Ars Conjectandi (English) Link

 

JB, will you sign my printed copies?laugh

 

 

 

GA

GingerAle Feb 2, 2018
 #6
avatar+902 
+2

Solution:

\(\text {The easy parts first: }\\ \text {There are }\dbinom{52}{5} = 2598960 \text{ ways to select 5 from 52 cards. } \\ \)

 

\(\text{Single suit probability – five cards of the same suit. }\\ \dbinom{4}{1}\dbinom{5}{13} = 5148\\ \rho(1) = \dfrac{5148}{2598960} = 0.19808 \% \\ ------------------- \)

..

\(\text{Four suit probability – five cards four suits. }\\ \text {A five-card hand will have at least two cards of the same suit. } \\ \text {Select two cards with matching suits. }\\ \text{There are }\underbrace {\dbinom{4}{1}}_ {\small \text { ways to choose suit }} * \underbrace {\dbinom{13}{2}}_ {\small \text { ways to choose 2 of 13 }} \\ \text{Then for the 3 remaining cards }\\ \text{ there are } \dbinom{13}{1}*\dbinom{13}{1}*\dbinom{13}{1} \text { ways to choose unique suit for each of remaining cards. } \\ \text {The product of these counts gives the total number of five-card hands with four unique suites. }\\ \dbinom{4}{1} * \dbinom{13}{2} * \dbinom{13}{1} * \dbinom{13}{1}* \dbinom{13}{1} = 685464 \\ \rho(4) = \dfrac {685464}{2598960} \text { 26.3746}\% \\ \)

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\(\text {Two suits. Five cards }\\ \text {There are } \dbinom{4}{2} \text { ways to chooses two suits. Twenty-six (26) cards represent these two suits. }\\ \text{There are } \dbinom{26}{5} \text { ways to choose five cards from the 26. }\\ \text{These selections include single suits, }\\ \text{so subtract the single-suit counts of three for the five-card selections. } -(3)\dbinom{4}{1} * \dbinom{13}{5}\\ \dbinom{4}{2} * \dbinom{26}{5} – (3)\dbinom{4}{1} * \dbinom{13}{5}= 379236\\ \rho(2) = \dfrac{379236}{2598960} = 14.5918\% \)

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\(\text{For three suits, subtract the counts for 1, 2, and 4 suits from } \dbinom{52}{5}\\ \dbinom{52}{5} - 5148 - 379236 - 685464 = 1529112\\ \rho(3) = \dfrac{1529112}{2598960} = 0.14.5918\% \\ \)

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\(\rho(1) = \dfrac{5148}{2598960}\\ \rho(2) = \dfrac{379236}{2598960}\\ \rho(3) = \dfrac{1529112}{2598960}\\ \rho(4) = \dfrac {685464}{2598960}\\ -----------------\\ \text{The probability of a five-card hand having at least three (3) unique suits is }\\ \dfrac{1529112}{2598960} + \dfrac {685464}{2598960} = \dfrac {2214576}{2598960} = \dfrac {507}{595}\\ \text{ }\\ \text{ }\\ \small \text { Source: Lancelot Link & Co. Solutions for AoPS probability questions. } \)

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The result matches Melody’s answer.  This proves we monkey around until we get it right. laugh

 

 

 

GA

GingerAle Jan 31, 2018
 #2
avatar+902 
+1

The following presentation is adapted from conversational dialogues with Lancelot Link for stochastic measurement errors in the physical sciences.  

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To calculate the error range for uncorrelated and random errors, use Gauss’s equations for normal distribution of errors. Using the Gauss equations, gives a 68% confidence interval the measurement error is within this range.

Here are some basic derivatives from the Gauss equations:

 

\(\small \text{If Q is a combination of sums and/or differences, then the sigma (error) of Q is equal to } \\ \sigma Q = \sqrt{(\sigma a)^2 + (\sigma b)^2 . . . + (\sigma x)^2 . . . } \\ \text{If } Q = b^2 \text{ then } \sigma Q = 2(\sigma (b)) \text{ where } \sigma (b) \text{ is the error of (b)}\\ \text{If } Q = \sqrt{(b)} \text{ then }\sigma Q = \frac{1}{2} (\sigma (b)) \text{ where } \sigma (b) \text{ is the error of (b)}\\ \text{Note how the exponent becomes the multiplier of the error.}\\ \small \text {For this equation, the easiest method is to calculate each error as a }\\ \small \text {decimal and then use the sum of the decimals to calculate the error range. }\\ \text {Uncertainties (In decimal) for a, b, and c. }\\ a_1 = \frac{0.1}{1.3} = 0.0769\\ b_1 = (2)\frac{0.2}{2.8} = 0.1428\\ c_1 = (0.5)\frac{0.1}{0.8} = 0.0625\\ Q= \frac{1.3 \times (2.8)^2}{\sqrt {0.8}} = 11.4 \pm ((0.0769 + 0.1428+ 0.0625) * (11.4))\\ Q= 11.4 \pm 3.2 \\ \small \text { note: pay attention to significant figures when recording the final error range. }\\ \\\)

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Relating to this question and others you’ve posted on here , another important thing you need to do is Stop being a lazy dolt!

I would have answered your chemistry questions, except I was too lazy. I caught this laziness from you—it’s more contagious than dumbness.indecision

 

 

GA

GingerAle Jan 6, 2018