I'm in a group chat but it's the identity group

This is a polyhedron with cutaway faces, based on the snub dodecadodecahedron, which I posted here some time ago. In this version there is a single connected piece for each face, instead of multiple disconnected facets. I designed the model faces so that they avoid each other when assembled, with the holes making the interior of the model visible.

This is a given.

If the last digit is divisible by 2 (a.k.a. even), then so is the whole number.

If the sum of the digits is divisible by 3, then so is the whole number. The recursivity of this means that if the sum has multiple digits, you can add them up again until you get a single digit and see if it's 3, 6, or 9.

Like the rule for 2, but check if the last two digits are divisible by 4.

If it ends in 5 or 0.

If the rules for both 2 and 3 apply.

No shortcut. Alas.

Like the rules for 2 and 4, but check if the last three digits are divisible by 8. (Yes, this pattern keeps going for 16, 32, etc.)

Like the rule for 3, but the sum of the digits (or the sum of the sum of the digits, etc.) must be 9.

Suppose that p is such a prime. Then, by Bezout's lemma, there exist integers X, Y such that 10X + pY = 1 (and clearly any such X is not a multiple of p). Let N := 10a + b. Then p | N if and only if p | NX. But NX = 10aX + bX = (1-pY)a + bX = (a+bX) + p(aY). So p | N if and only if p | (a+bX).

This is a given.

If the last digit is divisible by 2 (a.k.a. even), then so is the whole number.

If the sum of the digits is divisible by 3, then so is the whole number. The recursivity of this means that if the sum has multiple digits, you can add them up again until you get a single digit and see if it's 3, 6, or 9.

Like the rule for 2, but check if the last two digits are divisible by 4.

If it ends in 5 or 0.

If the rules for both 2 and 3 apply.

No shortcut. Alas.

Like the rules for 2 and 4, but check if the last three digits are divisible by 8. (Yes, this pattern keeps going for 16, 32, etc.)

Like the rule for 3, but the sum of the digits (or the sum of the sum of the digits, etc.) must be 9.

OK, here we go!

Assume our number is a multiple of 7 - i.e. it's congruent to 0 mod 7. I'm using letters to represent the digits here, but there could be an arbitrarily large number of digits, hence the ellipsis.

Each digit is multiplied by a progressively higher power of 10.

Now 10 is congruent to 3 mod 7, so we can replace all the 10's with 3's.

A bit of rearranging gives us this way of expressing the last digit a in terms of the other digits.

When we take all the digits except the last one and move them down one place, we're dividing everything by 3.

Here's where 5 being the multiplicative inverse of 3 mod 7 comes in. Dividing by 3 is the same as multiplying by 5.

Speaking of multiplying by 5, let's do that to our alternative expression for a, and add it on.

Everything cancels out and we get 0.

So if our original number was a multiple of 7, doing this trick will also give us a multiple of 7.

"What about the converse?" you might ask.

If our original number isn't divisible by 7, we get a copy of the remainder in our expression for a. Then instead of cancelling down to 0, we'll be left with 5 times the remainder. This won't be 0 because 7 is prime, and so no non-zero numbers will multiply to make 0 mod 7.

This is a given.

If the last digit is divisible by 2 (a.k.a. even), then so is the whole number.

If the sum of the digits is divisible by 3, then so is the whole number. The recursivity of this means that if the sum has multiple digits, you can add them up again until you get a single digit and see if it's 3, 6, or 9.

Like the rule for 2, but check if the last two digits are divisible by 4.

If it ends in 5 or 0.

If the rules for both 2 and 3 apply.

No shortcut. Alas.

Like the rules for 2 and 4, but check if the last three digits are divisible by 8. (Yes, this pattern keeps going for 16, 32, etc.)

Like the rule for 3, but the sum of the digits (or the sum of the sum of the digits, etc.) must be 9.

This is a given.

If the last digit is divisible by 2 (a.k.a. even), then so is the whole number.

If the sum of the digits is divisible by 3, then so is the whole number. The recursivity of this means that if the sum has multiple digits, you can add them up again until you get a single digit and see if it's 3, 6, or 9.

Like the rule for 2, but check if the last two digits are divisible by 4.

If it ends in 5 or 0.

If the rules for both 2 and 3 apply.

No shortcut. Alas.

Like the rules for 2 and 4, but check if the last three digits are divisible by 8. (Yes, this pattern keeps going for 16, 32, etc.)

Like the rule for 3, but the sum of the digits (or the sum of the sum of the digits, etc.) must be 9.

This is a given.

If the last digit is divisible by 2 (a.k.a. even), then so is the whole number.

If the sum of the digits is divisible by 3, then so is the whole number. The recursivity of this means that if the sum has multiple digits, you can add them up again until you get a single digit and see if it's 3, 6, or 9.

Like the rule for 2, but check if the last two digits are divisible by 4.

If it ends in 5 or 0.

If the rules for both 2 and 3 apply.

No shortcut. Alas.

Like the rules for 2 and 4, but check if the last three digits are divisible by 8. (Yes, this pattern keeps going for 16, 32, etc.)

Like the rule for 3, but the sum of the digits (or the sum of the sum of the digits, etc.) must be 9.

"Just taking my pet 7 for a walk!"

Counting