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 + ...

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

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)

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.)

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.

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

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