Math Is Fun Forum

Hi Bob;

Great work!

A couple of things:

I'd change

:P and :p

produce

to

:P and :p produce 

and include

:rolleyes produces 

...and probably also 'BBCode help' instead of 'bbcode help' in the thread title.

bcc...should be BBC. Just checking BBCode's strike-through Text Style option.

Thx for the Help Me! page work!

Hi Bob;

The BBCode help page link appears under the message box when you're posting, but not at other times, I think.

Yes, I reckon so!

Only 5 of them do, but they're already on our list.

That's ok...sometimes I surprise myself!!

Here's a link to a helpful MIF page with examples: Algebra/Word Questions.

Many years ago (19 August 2009, to be precise) I posted a word puzzle along the lines of those on that MIF page, but somewhat tougher: Edna's age. There's some discussion there about solving methods, including a couple of different algebraic options.

Hi amnkb;

harpazo1965 posted:
...I have an android AS 21 T-Mobile phone...
...my android phone A21...

Googling "Android AS 21" failed to find one of those, but turned up these:
   Samsung Galaxy A21 (released April 2020)
   Samsung Galaxy A21s (released May 2020)
   Samsung Galaxy S21 (released January 2021)

So I think he has an Android...a Samsung Galaxy (model unclear).

Edit: Post #11 (11/11/2023):

Google failed to find that model too.

Hi iiooasd1217,

I could only find 0, 1 and 144 in the first 100,000 Fibonacci numbers.

144 is the 13th Fibonacci number (sequence A000045 in the OEIS).

My Mathematica code:

In[1]:= AbsoluteTiming[Select[Sqrt@LinearRecurrence[{1,1},{0,1},100000],IntegerQ]^2]
Out[1]= {5206.4986953,{0,1,1,144}}

Computing speed with that code is about the same as with M's Fibonacci function: Table[Fibonacci[n], {n, 100000}].

Testing to 1M would tie up my computer for longer than I'd like, so I'll quit now.

Also, extending my test range may be a fruitless exercise, as in his article Square Fibonacci Numbers, Etc, John H. E. Cohn 'apparently' proves that there are no more results than just the three I found.

I say 'apparently', because his proof is way beyond my understanding.

The article begins:
     "An old conjecture about Fibonacci numbers is that 0, 1 and 144 are the only perfect squares. Recently there appeared a report that computation had revealed that among the first million numbers in the sequence there are no further squares [1]. This is not surprising, as I have managed to prove the truth of the conjecture..."

Thanks!

Hi amnkb,

I don't get a warning when logging in, but when I click on your this method link from harpazo1965's 'Upgrade Picture Uploads' thread, I get this:

Strangely, I don't get the warning when clicking on the above harpazo1965 thread link.

That link opens a webpage that begins: "Most websites are optimized for mobile, but that doesn't mean you can't view the desktop version on your smartphone."

What I didn't mention was that there are two version options ('All Sites' and 'Specific Sites'), and below are links to the options mentioned in the article:

Chrome (Android & iOS) - All Sites
Chrome (Android & iOS) - Specific Sites
Edge (Android & iOS) - All Sites
Firefox (Android) - Specific Sites
Brave (Android) - All Sites
Brave (Android) - Specific Sites
Vivaldi (Android) - All Sites
Vivaldi (Android) - Specific Sites
Opera Mobile (Android & iOS) - Specific Sites

The webpage was last updated September 2023.

This is from math.stackexchange.com:
   I want to know how long a "twinkling of an eye" would be. A twinkling of an eye is the time it takes from the moment the light hits the front of the eye, until it hits the back of the eye and is reflected back.

And here's the link to the discussion thread: How long would it take for light to go through a human eyeball and back to the other side.

I also found the following entry & definition in Answers.com (down the page a bit, under "More answers"):
   "Just how fast is the 'twinkling of an eye'? It is not the time it takes to blink an eye: For you to see someone's eye twinkle, light must travel through the front of their eye, be reflected off their retina, and then exit their eye. Assuming (for the ease of calculation) that you are standing so close to that person that the transmission time from eyeball-to-eyeball can be regarded as instantaneous, and that a person's eyeball is 2.5 cm in diameter, the light would have to travel a distance of 5 cm (or 1/20,000th or 2 x 10^-4 of a kilometer). Since the speed of light is 300,000 (or 3 x 105) km/sec, this means it would take or 1/6 x 10^-9 seconds (ie, 1/2 x 1/3 x 10^-4 x 10^-5 seconds), or 1/6,000,000,000th of a second to make a person's eyeball twinkle: pretty fast."

Nope, that wouldn't work; both you and the laptop have to be upside down (ie, from your perspective up where you are).   That's how we do it down here.

That way (the correct way), if you drop your laptop, it falls up to the ground (where you could find it again...albeit in bits & pieces) instead of down to the sky (from where it would be irretrievable). That's from your inverted perspective, of course.

A couple of things you could try (for simplicity with referencing, I'll refer to the left-hand image, but the same would apply to the other one):
1. (a) Focus on the little square (ie, the region where the front and back walls overlap).
    (b) Concentrate on that and let your brain relax into neutral (to make it believe you're not up to any tricks).
    (c) Casually – and maybe even humming or whistling nonchalantly (the brain musn't suspect anything) – observe peripherally the role reversal of the green and red squares: ie, that the green is now the front, higher, smaller square, and the red is now the rear, lower, larger square.
    (d) Squinting may help (vary squinting severity if you're getting nowhere with it).

2. Using paulb203's method from the first paragraph of post #1, construct the left-hand image to look like mine.
    However, whereas I started with the red image as the front wall, followed by the green image as the back wall to the right, and higher than, the front wall, try starting with the green image as the front wall, followed by the red image as the back wall to the left, and lower than, the front wall.

Enjoy!