ℝolliℵg M∀th Thr∑a∂

i don't think it's surprising or even particularly concerning that most parents can't do a GCSE maths question

mainstream secondary school math curricula are all horrible, just Cold War artifacts unsuited to either the exploration of math as a liberal art or to its practical applications in basically any career.

at least such was my experience in the aughts and I have no reason to believe anything has changed all that much

what's interesting is that people who are interested in "secondary math as gatekeeper for physical science education" and people who are interested in "secondary math as a liberal art" and people who are interested in "practical career applications of secondary math" and people who are interested in "abolish compulsory secondary math education" all agree with you, but each for different reasons

as long as they all line up behind my reform program I don't care what motivations they have

This is a fun one:

https://mindyourdecisions.com/blog/wp-content/uploads/2018/09/homework-in-china-how-tall-table-preview-600.png

No multiplication required.

table is 150, will show working on request but i couldn't remember what happens if you try to hide several paragraphs on ilx

also i found a less round-the-wrekin way to do it

B-b-but did you also calculate how much taller the cat is than the turtle?

anywhere between 1.15x and infinitely...

yes i know that too

tab + tur - cat = 130, tab - tur + cat = 170, just add them together to get tab + tab = 300

hence cat - tur = 20, however i do not believe we can calculate cat or tur in themselves despite knowing the diff between them

i assumed tur=0 to (marginally) simplify things

MIND YOUR DECISIONS ILXOR LEDGE

but both the creature heights disappear when you add the two, no need to assume anything!

i know. nevertheless, i like assuming.

I used the sin double angle formula.

No I didn’t but it is vaguely related

i do not believe we can calculate cat or tur in themselves despite knowing the diff between them

― mark s

yes, more unknowns than equations

even more unknowns: how many kids (out of all the kids in china) are doing this problem, how many do it correctly, do they do it with help, are they expected to get it right, why do we even care

Here's a question for y'all: what are the maximum heights of the cat and the turtle?

also, how is that table still standing with that cat sat on the edge like it is?

table is bolted to the floor, next question

maximum height of the cat is the height of the table, no? what's interesting is that as you make one animal taller, the other gets taller with it. that seem counterintuitive to me but that intuition is probably just based on my accumulated experience of (superficially) similar problems

the picture breaks if the cat gets bigger than the table but I'm not sure the maths does.

Right. Turtle will eventually break the table too.

have only just noticed that this is the average of the two values. is that always true?

yeah, trivial

yes bcz of yr post at 16.22 london time

https://www.youtube.com/watch?v=tS8oyl1gygs

Slight variation on the problem: instead of limiting ourselves to 2 measurements and 2 animals, imagine we could do a million measurements. For each measurement we select at random two animals from the total animal population of earth. We place one animal on the floor and one on the table (assume table is of infinite strength), and measure the distance between the top of the floor animal to the top of the table animal (which could be a positive or negative number). Given measurements M1 through M1000000, what is our best estimate of the actual height of the table?

51"

Bear in mind that if we select 2 million animals at random from total earth animal population, its likely that most or all will be bacteria and hence of negligible height.

bacteria aren't animals!

You're right. Please disregard my previous statement. It seems the most common animal will be an insect, whose height may or may not introduce significant error in our measurement, depending on the height of the table.

Come now Let us be crooked but never common.

Challenging myself to figure out why this works:

https://pbs.twimg.com/media/FmBgdAxWAAA_O_h?format=jpg&name=small

Happy to explain if you don't figure it out

Oh, I'll let it percolate in the back of my mind for at least a day or two before relenting and looking for hints. Interestingly this trick doesn't preserve the remainder of the number after division by seven in the general case. Only in the case of remainder zero. So its not a perfect modular algorithm.

Took me a few minutes of percolating but it makes perfect sense now.

Thought of two ways to do it. One is more obvious and clear, one is slightly fancier but more interesting. They both amount to the same thing anyway.

Can’t believe I never came across that before. I did spend time long ago thinking about why the decimal representation of one seventh is what it is though.

Not much percolating in my brain yet, except a vague idea that the equation 5 * 10 - 1 = 7 ^ 2 is somehow involved.

facts

as a physical sciences person i suck at number theory. here’s how i can prove it for numbers between 100 and 1000, no idea how to generalize to the result. pretty sure a slick method would use mod but as i suck at number theory idk how to do that

suppose you have a three digit integer with digits p, q and r which can be written 100p + 10q + r

using chika’s method we drop the last digit r and divide by 10, then add 5r. assuming this new number is divisible by 7, we can write

10p + q + 5r = 7m, where m is some rando integer

multiply both sides by 10 to get

100p + 10q + 50r = 70m

to reconstruct our original integer, we subtract 49r from both sides to get

100p + 10q + r = 70m - 49r

since 70 and 49 divide by 7 (and m and r are integers) we’ve proven 100p + 10q + r is divisible by 7

some distance still to go before i’m as slick as a 12 year old math whiz

actually maybe it’s not as hard as all that - with a four digit number i think you just get 700m on the other side, 7000m with five digits, and so on?

sadly i think you always get a -49r and never a 490 ;_;

er no, scratch that, i think you always have 70m, just bigger and bigger junk in front of the 50r to keep track of

One way:
Think instead of just peeling off the last digit, multiplying it by 50 and adding it back to the rest. This will just be ten times the number in the suggested trick and is divisible by seven if and only if that number is. If the original pieces were p and q mod 7, the new number mod 7 is p + 50q = p + 49q + q which is again p + q mod 7, so the operation does not change divisibility by 7

yeah see that’s slick, i understand it but i just can’t generate math like that