No. If you can’t multiply in your head, you are crippled in any quantitative reasoning. You have interjected too many steps in estimating, calculating, judging, etc.

This is not to say there aren’t useful, non-quantitative pursuits.

I don't understand what you're saying or objecting to.

Incase this is what you mean: remembering the multiplication table is the antithesis of multiplying in ones head.

> remembering the multiplication table is the antithesis of multiplying in ones head

But it's not! Because in order to be able to multiply numbers in your head efficiently you must remember the multiplication table.

No. By definition remembering an answer is not the same as doing.

Additionally my experience tells me otherwise.

I'm sorry but this seems patently wrong to me. There's only so much working stack space in your brain. If you're constantly having to think through multiple steps to multiply single digits then you're going to be at a serious disadvantage when you need to solve a bigger problem that involves more work than just single-digit multiplication.

No remembering the table is the ROM-table part. You need the algorithm part too.

The best is the 9's table, how all the digits add up to 9. My mom's an elementary teacher and there's always a few third graders who figure that out on their own and love it.

Additionally 9 * n = [n-1, 10-n] for n = 2-11; where n-1 is the digit in the 10's place and 10-n in the single place. This just an aesthetic curiosity. I know the pattern continues for larger n I've just never bothered to generalize it. Also never compared it to other bases.

It's not an aesthetic curiosity at all! 10 is equal to 1 mod 9... So suppose X is written in base 10 (a0 * 10^0 + a1 * 10^1 + a2 * 10^3 +...) and you want to find X mod 9. Then all of the 10^k's are just 1 (mod 9), so you just get the sum of the digits.

So if X is divisible by 9, then the sum of the digits (mod 9) is zero.

Same works for 3 (x is div by 3 iff the sum of the digits id divisible by 3). And 11 gets an /alternating/ sum of the digits, since 10 is -1 mod 11...

> [n-1, 10-n] for n = 2-11

I think the usual 10 * n - n makes more sense and is much easier to remember.

I was legit angry that I didn't learn about the divisibility checks for 3 and 11 until college...