For example, for the 47 times table, write down the grid with 7 in the top right:

| 7 4 1

| 8 5 2

| 9 6 3

Then prepend the five times table, knocked down by one for each line you've crossed:

\( \begin{pmatrix} 47 & 94 & 141 \\ 188 & 235 & 282 \\ 329 & 376 & 423 \end{pmatrix} \)

Boom!

I think these two grids are all you need to memorise. Suppose you're multiplying by \(10a+b\), [^0]. Rotate the grid so that \( b\) is in the top left. If \( b \lt 5\), write down the \(a\) times table in front, but bump it up by one every time you cross a vertical line. If \(b\gt5\), write down the \(a+1\) times table, but drop it by one each time you cross a vertical line.

[^0] Sigh, because mathematicians: \(a\) and \(b\) integers, \(0\le a\lt10\) and \(1\le b\le 9\), \(b\ne 5\). Honestly.

Show thread

Today while the youngest swims, I'm playing with this: http://507movements.com/mm_152.html

It took a while to convince myself it was really an ellipse, but Desmos to the rescue: https://www.desmos.com/calculator/9hsixaciwt

I've made one of those daily word game things! But it's not just Wordle in a hat, it's a totally different puzzle and I'm really pleased with how it came out.

Another proof

@icecolbeveridge

along the lines of the √2 proof:

Suppose p/q is a reduced rational solution. Then clearing denominators, 𝑎𝑝²+𝑏𝑝𝑞+𝑐𝑞²=0. If exactly one of 𝑝,𝑞 are even, then this is two even terms and an odd one. If neither of them is even, then it's three odd terms. So they're both even, contradicting "reduced" fraction.

#ProofInAToot

My proof

@icecolbeveridge (hope my first tray at a content warning works). Here's mine

Any odd square is 1 mod 8, but \(b^2-4ac\) is 5 mod 8.

My proof

If \( ax^2 + bx + c =0 \) has rational solutions, it can be written as \((px+q)(rx+s)\), with \(p\),\(q\),\(r\) and \(s\) integer.

\( a= pr\) and is odd, so \(p\) and \(r\) are both odd.

Similarly, \( c = qs\) and is odd, so \(q\) and \(s\) are both odd.

\( b \) is odd, but must equal \( ps+qr\), which is even. Contradiction.

Something I've wondered more than once on birdsite:

\( 1729 = 12^3 + 1^3\), so it's a multiple of 13.

\(1729 = 10^3 + 9^3 \), so it's a multiple of 19.

It's also a multiple of 7. Where does that come from? (i.e., is there a way to find *all* of the factors of a taxicab number given the cubes that sum to it, or similar?)

- Website
- colinbeveridge.co.uk

- Location
- Dorset, UK

- Latest book
- https://amzn.to/2MUuEbs

- Pronouns
- He/his

A mathematician with nothing to prove.

Joined May 2017