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Comment by Jensson | Hacker News Reader

Symbolic manipulation and intuitive understanding doesn't seem to be very correlated to me. I've met people who are very good at manipulating symbols, but they aren't very good at understanding less formal things like strategy games well, they tend to invent way too rigid rules for themselves and lose since they are bad at adapting.

So maybe you are very smart with symbols, but less so with intuitive understanding? There is nothing wrong with that, makes it easier to decide what to work on, focus on your strengths and let others cover for your weaknesses.

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My memory is kinda shit.

So when I was in the gifted program, they got real concerned about how I didn't know my times tables and was super slow at a few of them. This was the 1970s, and before I knew binary, so 7s and 8s caused me issues. So I'd figure out 8x8 by going from 6x6=36 (real easy to memorize) and then adding +6+6 to mentally fill out the 8x6 block and then adding +8+8 to fill it out to a 8x8 block (I'm visual/geometrical).

I was the first kid in school to pass the AHSME to get invited to take the AIME though. Being before the internet, and being stuck on an island in Alaska, didn't get me enough exposure to higher math through my own self-direction to get anywhere on the AIME. If you don't live anywhere near a University library and/or don't know you can use it, that'll set you behind (at least back then, these days there's YouTube and sci-hub and friends).

I still think I would have hit a wall anyway with Math, even with perfect exposure, because I'm visual and higher Math seems to require being very good with symbols and memory as well.

I suspect lots of people still underestimate me because my memory is ass, and we associate memory with intelligence so much (e.g. Jeopardy).

I used to believe that I was good at math because my memory was shit. So I had to actually understand everything, because I couldn't rely on memorizing it.

But some random facts were easy to memorize, for example that the chessboard has 64 fields. I had problems with 6×9 and 7×8 though, always confused about which one was 54 and which one was 56.

Oh! So that’s why my teachers made us memorize row by row: For me 54 and 56 are in an entirely different category (resp the 6 and 8 categories), didn’t even realize they landed in the same dozen when learning my tables!

And I don’t have a good memory either.

I can't tell you what 6x9 is. But I can tell you what 9x6 is. I always turn that one around. My mind immediately jumps to a visual 60 coz 10x6 is super easy of course and subtracting 6 from it is easily 54.

Similarly 8x7 I can't tell you but 7x8=56 in my brain feels like a little "rhyme" I just need to repeat and I have the answer.

That's also how I remember (somewhat) arbitrary passwords. If it "flows" well almost like a rhyme and can be typed fluently I'll remember. Actual arbitrary ones don't work as well.

Yeah, after a while I also learned that 9x6 = 10x6 - 6 = 54.

And then I just remembered that 7x8 is "that other difficult number", because by that time I already remembered that 54 and 56 are the two most difficult numbers in the multiplication table. :)

Btw, same here, 9x6 and 7x8 feels much more natural than the other way round.

So, fun fact from calculus (though you can easily prove this with basic algebra as I do below):

- You want to compare two products: in this case 6x9 and 7x8.

- And in each product, if you add the two numbers together, you get the same result. In this case, 6+9 = 7+8.

Then the product will be larger for the pair of numbers that are closer together. So 7x8 > 6x9. That might help you remember which is 56 and which is 54.

You typically see this in a word problem where you are given a fixed amount of fence and you have to enclose the largest rectangular area. The answer is to use a square area (two sides being equal). If the problem has constraints that prevent the sides from being equal, then you pick the length and width to be as close to each other as possible.

In case you want to transfer the geometric intuition to an algebraic proof: If the sum of the two side lengths is 2m, then the two side lengths can be written as (m+n) and (m-n) for some positive n. If you multiply the two, you get (m+n)(m-n) = m²-n². To maximize the product, you need n to be as close to 0 as possible - i.e. for both sides to be as close to each other as possible.

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