GingerAle

Once Andy stops rowing, his relative velocity and direction are the same as the hat. Rowing away from the hat for five minutes means it took five minutes for him to return to it, for a total of ten minutes. In that ten-minute length of time, the hat traveled one kilometer. 

Current flow: 1 km/10 min = 100 meters per minute

= 1.67 meters per second.

GA

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.

GA

Typical Mr. BB BS!

An intuited solution with no explanations.

Followed by an explanation of fluff and blarney— borderline useless!

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.

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

Solution: 

\(\text {With six dice }\\ \small \text {There are } \dbinom{6}{2} \text { ways to chooses two dice for the “ones” }\\ \small \text{There are then} \dbinom{4}{2} \text { ways to choose two dice for the “twos” }\\ \small \text{There are } 6^2 = 36 \text{ ways to roll two dice. There is one way to roll two dice for two 1s (snake eyes) (1/36). }\\ \small \text{There is one way to roll two dice for two 2s (1/36).} \\ \small \text{Of the remaining two dice, there are 4 of 6 ways to roll each die without a 1 or 2 (2/3). }\\ \dbinom{6}{2} * (1/36) * \dbinom{4}{2} * (1/36) * (2/3) * (2/3) = 5/162 \\ \)

We really do get it right if we monkey around with it enough.

GA

Alles Gute zum Geburtstag, Heureka

Solution:

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

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

..

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

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

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

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

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

-------------------------------------------------------------------..

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

-----

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

GA

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

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

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:

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

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

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

The producers of this recording label never individually identify the soloist or choral singers.  I happen to know who the soloist is though: It’s Hectictar.

.