GingerAle

After clicking on this link I heard a gobbling and clucking noise; I thought Mr. BB gave the forum a turkey for Thanksgiving. However on closer inspection, it’s easy to tell it’s just a chicken with a thyroid condition, stuffed with Blarney Dressing.   ^Yum!

\(\begin{array}{rcll} n &\equiv& {\color{red}3331} \pmod {{\color{green}231}} \\ n &\equiv& {\color{red}1361} \pmod {{\color{green}1247}} \\ \text{Set } m &=& 231\cdot 1247 = 288057\\ \\ \end{array}\)

\(\begin{array}{rcll} n &=& {\color{red}3331} \cdot {\color{green}1247} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ {\color{green}1247}^{\varphi({\color{green}231})-1} \pmod {{\color{green}231}} ] }_{=\text{modulo inverse 1247 mod 231} } }_{=1247^{230-1} \mod {231} }}_{=1247^{229} \mod {231}}}_{=113} + {\color{red}1361} \cdot {\color{green}231} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ {\color{green}231}^{\varphi({\color{green}1247})-1} \pmod {{\color{green}1247}} ] }_{=\text{modulo inverse 231 mod 1247} } }_{=231^{1246-1} \mod {1247} }}_{=231^{1245} \mod {1247}}}_{=637}\\\\ n &=& {\color{red}3331} \cdot {\color{green}1247} \cdot [ 113] + {\color{red}1361} \cdot {\color{green}231} \cdot [637] \\ n &=& 469374541 + 200267067 \\ n &=& 669641608 \\\\ && n\pmod {m}\\ &=& 669641608 \pmod {288057} \\ &=& 197140 \\\\ n &=& 197140 + k\cdot 288057 \qquad k \in Z\\\\ \mathbf{n_{min}} & \mathbf{=}& \mathbf{197140 } \end{array}\)

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\(\small \text {Related formulas and principles compliments of Leonhard Euler }\scriptsize \text {(totient function),} \\ \small \text {Euclid of Alexandria }\scriptsize \text {(Extended Euclidean algorithm), and Brilliant Chinese mathematicians “Chinese Remainder Theorem” } \\ \small \text {LaTex layout and coding adapted from Heureka’s mathematical solution and Latex presentation:}\\ \)

That bridge needs a Troll. Aside from collecting fares, he can have an occasional snack of puréed dumb dumb.  

These lil’imps often give us big chimps a bad reputation.  They never pay for their fares, but they expect on time, first-class service, and they have fits when they don’t get it.  The little monkeys are much smarter than the bridge running fool is, though. They are much less likely to become monkey mincemeat.

Although this subtly phrased question appears to lack sufficient information for a solution. It does in fact have one. 

Part of the information is contained in this phrase.

if it is possible for the jogger to just escape being hit by the train by running at full speed to either end of the bridge

Here the use of the word “if” is not so much a conditional clause, as it is a presented condition or supposition.  Rephrasing makes this more clear:

"Suppose the jogger can just barely escape being hit by the train by running to either end of the bridge at his greatest speed."

This tells us the runner will meet the train at end of the bridge and just miss being hit, whether he runs toward the train or away from the train. Along with this information, there are two pieces of numerical data: the train’s 30kph speed and the man’s relative position of 3/4 on the bridge. 

Using this information, set up an equation where the train and the man meet at either end of the bridge.   

\(\dfrac{(30 + F) }{(30-F)} = \dfrac{3}{1} \tiny \text { F is the speed the Fool can run.}\\ (30+F) =3(30-F)\\ 30+F = 90-3F\\ 4F = 60\\ \,\;F= \dfrac{60}{4}\\ \,\\ F = 15kph\\ \)

Answer  (A). 

Use these equations to solve:

\(x+y=74 \tiny \text { Total heads; x = number of humans and y = number of horses} \\ 2x+4y=196 \tiny \text{ Total legs: each human has two and each horse has 4}\\ \text { Solve for x and y}\\ \)

To find the solution for this, calculate the weighted average:

 (10×55+20×56+70×57) / (10+20+70) = 56.6 amu

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You may have missed your calling, JB. You should have been a philosopher. I’m adding your philosophical thought to my toolbox of aphorisms.

GOOD JOB!! Mr. BB. You sourced your quote, though Google isn’t the source. The quote is from: https://en.wikipedia.org/wiki/Knot_(unit)  

So . . .  no banana for you.

Also, your post is knot really an answer to his question.  From the context, his question should be read as “what is the linear value of one nautical mile at the equator?” 

Johannes von Gumpach makes the same error, using “knot” when he means “nautical mile” on page 253 of his book The True Figure and Dimensions of the Earth (1862) 

I use a quote from Gumpach’s book to partly answer the question

The explanation however will appear as simple, when as it is remembered that the nautical mile is an angular, rather than a linear measure, being one of 360x60=21,600 parts of the Earth's equatorial circumference whatever be the true linear value of that circumference. Hence considered as a linear measure it has as yet no definite value and its correctness depends absolutely on the correct linear measurement of an equatorial degree. If therefore the circumference of the Earth is taken too great by 166 or 167 miles, the nautical mile being one of its equal parts, and the subdivisions of the nautical mile or knots of the log-line__by which the distance sailed by a vessel is actually measured—are  likewise taken too great; and consequently, the linear distance sailed by a vessel when reduced to angular distance, is reduced by means of too  great a unit of measure; whence the number of nautical miles sailed both by computation and by the log-line, falls short of the true number.

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This book is in the public domain. Here’s a link for a high-quality PDF image copy of an original book in the NY Public Library archive collection.  The next time I’m there, I may “check it out.”

LOL!! Somehow, I am not surprised. In few more weeks, your mind will interpret it as \Left parenthesis (Log Natural close parenthesis). Followed by “Wait, that isn’t right ... there’s no argument.”  

During Spring break, when I was studying physics, I decided to spray my sunbathing friends with a garden hose, and I saw an equation like this:

\(V_f = V_0 \cdot sin( \theta) \Bigg( \dfrac {x}{V_0 \cdot cos(\theta)}\Bigg) + \dfrac {1}{2} \Bigg(\dfrac {x} {V_0 \cdot cos(\theta)}\Bigg)^{2}\)

But when they got up to chase me, I just ran like heII, and calculated my “escape velocity” later.