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gmax137 said:
Drove to Vegas yesterday from home near Reno. Saw a few snowflakes while in Red Rock Canyon this afternoon, that's a first.
My daughter who lives in Golden, Colorado told me it's been in the 30s and windy.
 
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gmax137 said:
Drove to Vegas yesterday from home near Reno. Saw a few snowflakes while in Red Rock Canyon this afternoon, that's a first.
I live a few miles south of Red Rock Canyon. Saw snow flurries outside several times this Winter. This year the snow melted once on the ground but a few inches accumulated late Feb 2019.

IMG_20190221_074102106.jpg
 
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How a Fake Mountain Range Slowed Down Arctic Exploration

The 19th-century naval officer John Ross had a knack for seeing land where it wasn’t.
https://www.atlasobscura.com/articles/john-ross-arctic-mountain-fake
On August 31st, 1818, around 3 p.m., the Arctic explorer John Ross was called away from his dinner and onto the deck of the ship he commanded, the Isabella. Ross and his crew were moored in Baffin Bay, just south of Greenland, seeking a way through to the Arctic sea beyond. All day, they had been waiting for the fog to clear, so they could take a look around and try to find it.

Ross stepped out onto the deck and began scanning the horizon: ice, more ice, and, in between, an imposing set of peaks. “I distinctly saw the land, round the bottom of the bay, forming a connected chain of mountains with those which extended along the north and south sides,” he wrote soon after. There was, he concluded, no way through.

Some are born great; some achieve greatness; some have greatness thrust upon them. And some narrowly miss greatness, kept from it by a pesky propensity to imagine land where there is none.

I'm wondering if he saw an iceberg, which is a transient phenomenon. It eventually moves and/or melts.
 
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Astronuc said:
How a Fake Mountain Range Slowed Down Arctic Exploration

The 19th-century naval officer John Ross had a knack for seeing land where it wasn’t.
https://www.atlasobscura.com/articles/john-ross-arctic-mountain-fakeI'm wondering if he saw an iceberg, which is a transient phenomenon. It eventually moves and/or melts.
... the Arctic explorer John Ross was called away from his dinner and onto the deck of the ship he commanded, the Isabella. ...
"Isabella was the imaginary friend of Clara Sutter, one of the children aboard the USS Enterprise-D."
See: https://memory-alpha.fandom.com/wiki/Isabella
 
Did you ever recognize that ordinary people get convenient finger positions while state guests and presidents have to use outfield positions with a stair?
 
fresh_42 said:
Did you ever recognize that ordinary people get convenient finger positions while state guests and presidents have to use outfield positions with a stair?
Can you elucidate?
 
Wow, overheard high schoolers trying to solve an exam problem ( that had been marked as wrong ; exam had been returned to them), that did not seem too easy nor obvious: Worker A starts on the 5th of April and travels every 3rd day . Worker B starts on the 2nd of April and travels every 4th day. During what days , if any, in April will they both be traveling? This was,from what I gathered, a timed test. I could only tell it had to see with the LCM of 3 and 4. Yes, we need to find positive Integer solutions to 5+3k =2+4j. I fail to see a quick solution to it.
 
WWGD said:
Yes, we need to find positive Integer solutions to 5+3k =2+4j. I fail to see a quick solution to it.
That's ##3k=4j.## Where is the problem?
 
fresh_42 said:
That's ##3k=4j.## Where is the problem?

Not
 
WWGD said:
Yes, we need to find positive Integer solutions to 5+3k =2+4j.
RHS is even so 8+6k suffices for LHS. Thus the LHS can only be 14, 20 or 26, and the RHS isn't divisible by four. So the 14th and the 26th, surely?
 
Not quite , k=4, j=3 is not a solution:
##5+12 \neq 2+12 ##
 
dlgoff said:
That IS chilly for where you're at.
Santiago de Chilly?
 
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\begin{align*}
5+3k &=2+4j \Longleftrightarrow 3+3k=4j \Longleftrightarrow 3(k+1)=4j \Longleftrightarrow 3k'=4j\\[6pt]
4\,|\,k'&\wedge 3\,|\,j \Longrightarrow k'=4n \wedge j=3m \Longrightarrow 12n=12m \Longrightarrow n=m\\[6pt]
5+3\cdot(4n-1)&=2+4\cdot 3n
\end{align*}
 
I never said they were intrinsically difficult. I said that, for 14-15 year olds with limited Math skills, on time pressure, it's not an easy problem.
 
WWGD said:
I never said they were intrinsically difficult. I said that, for 14-15 year olds with limited Math skills, on time pressure, it's not an easy problem.
Guess I misunderstood you.
 
fresh_42 said:
Guess I misunderstood you.
Neinert problemachkeit.
 
WWGD said:
Not quite , k=4, j=3 is not a solution:
##5+12 \neq 2+12 ##
2 + 3 + 3k = 4j + 2
3 + 3k = 4j
3(k+1)=4j
 
WWGD said:
Neinert problemachkeit.
PF's first law: the easier the question the more posts, especially if it is answered in post #2.
 
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Jarvis323 said:
2 + 3 + 3k = 4j + 2
3 + 3k = 4j
3(k+1)=4j
But the point is this is for a timed test for high-school kids, i.e., 14-15 year olds.
 
I would draw a 7x4 calendar grid and put As and Bs in the traveling days, probably take less than a minute.
 
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gmax137 said:
I would draw a 7x4 calendar grid and put As and Bs in the traveling days, probably take less than a minute.
Way back in a competition we got a programming example around the same difficulty.
I just did something like you suggested and provided results in a minute or so. Then I just left (the whole idiotic competition too, by the way.)
The winner took arond half an hour and wrote some recursive self-calling whatever mess which provided the same results.

Even that time I had a strong expectations for textbook examples to have at least some sense.
 
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If the question asked for results April a year from now, then maybe the mathematical approach would be faster. Maybe...
 
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gmax137 said:
If the question asked for results April a year from now, then maybe the mathematical approach would be faster. Maybe...

You'd have to consider issues of leap years, etc. And you'd have to consider number of days each month if you wanted to know beyond April dates.
 
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WWGD said:
Wow, overheard high schoolers trying to solve an exam problem ( that had been marked as wrong ; exam had been returned to them), that did not seem too easy nor obvious: Worker A starts on the 5th of April and travels every 3rd day . Worker B starts on the 2nd of April and travels every 4th day. During what days , if any, in April will they both be traveling? This was,from what I gathered, a timed test. I could only tell it had to see with the LCM of 3 and 4. Yes, we need to find positive Integer solutions to 5+3k =2+4j. I fail to see a quick solution to it.

gmax137 said:
I would draw a 7x4 calendar grid and put As and Bs in the traveling days, probably take less than a minute.
I suppose we can call this a matrix geometrical solution as opposed to algebraic?
Using @gmax137 's method:

Drew 7x4 grid on paper with pencil.
Labeled first column 1, 8, 15, 22, 29 for indexing. (not actually necessary)
Offset first A to square 5 and B to square 2,
Counted 3 squares from A and marked an "A" until end of grid.
Likewise for worker B counting by 4.

A and B meet on squares 14 and 26 meaning workers A and B both travel on 14 and 26 April.

Entire exercise including checking my work and reciting the mnemonic "Thirty days have November., April, ...", required <2 minutes. Less time than reading the posts. :cool:

Reminds me of using "Sieve of Eratosthenes" to find primes among first 100 natural numbers.
 
Klystron said:
Reminds me of using "Sieve of Eratosthenes"
Exactly my thought!
 
Too easy. Do it using Roman numerals now.
 
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