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Comment by lupire | Hacker News Reader

The author is talented, but these are mundane exercises, not problems. They are all trivial mechanical computations best done by a calculator, that do nothing to promote understanding and insight.

A lot of work went into this, to generate trivial exercises to train rote memorization, like flashcards.

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True, perhaps this would be better called an exercise generator. However, when studying this in school, the teachers always called them problems, hence why the title was chosen. The project is aims to help students K-12 and possibly undergrads whose curriculum is aimed to get them comfortable with solving these computations without aid of a calculator, or at least without programs that solve it for them. This is not supposed to be the primary means of teaching, but rather a tool that teachers can use to provide infinite exercises for their students to practice on, with the benefit of students not being able to copy answers directly out of a textbook without understanding how it works.

I have a degree in math and I actually enjoy these types of problems. Keeping sharp at mental arithmetic is really fun and it prompts the brain to come up with shortcuts that can lead to more interesting math. Like, is 51 prime? (No, 5+1 = 6). It's not really rote memorization but getting better at mental arithmetic, which is actually meditative in a way.

Why do you sum the digits? Goldbach?

It's divisible by 3 , because 6

To say this in more detail, it’s observed that any number with digits adding up to multiples of 3 is divisible by 3

https://math.stackexchange.com/questions/341202/how-to-prove...

https://everything2.com/title/m_turner%2527s+postulate

For any integer base B greater than 2, the multiples of the number B-1, when represented in base B will always have the sum of the digits be a multiple of B-1. If B-1 is a square number (4,9,16,...) then the square root of B-1 will also have this property.

(the following node has a proof)

> If B-1 is a square number (4,9,16,...) then the square root of B-1 will also have this property.

In fact, if d is any divisor of B-1 (including a trivial divisor), then d will have this property. (The author of the proof on the page you linked to is aware of this, but the author of the original conjecture isn't. For example, the reply notes that the digital root test for divisibility in hexadecimal works for 3 or 5, which are divisors of 16-1=15, regardless of the fact that 15 isn't a square. It also works for 1 or 15, which are also divisors of 15.)

It is because a number is base ten is like this:

a + 10b + 100c + 1000d + ....

But 10 leaves a remainder of 1 on division by 3 so this number divided by 3 is the same as

a + b + c + d + ...

Almost every physicist would disagree :-)

While mathematicians hate this kind of stuff, it is the bread and butter of physicists. While it is true most of them will not compute the cross product of two concrete vectors by hand, most will compute them symbolically, and the skills to do both are the same.

yet, he's right.

these are "exercises", not "problems" they don't involve creative thinking. look up the difference on any text about "problem solving"

Curiously, ChatGPT falls on its face with about half of these that I checked. It was confidently wrong about them too, but tried its logic again when I corrected it. YMMV.

So if your calculator is a huge neural net, maybe the problems aren't trivial!

Try ask ChatGPT what 12931239 + 712380 is. It can only solve 30 + 46 because it actually memorized/'saw' the latter, but that's nothing new.

But Wolfram Alpha will do them all easily.

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