Combustion Waltz

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.

The basic principle is that if A is divisible by N, and B is also divisible by N, then A+B and A-B are divisible by N. So we want to split the number you're testing into a number we know is divisible by N, plus a small number that you can just look at.

2 and 5: the number is in base 10, which is 2*5. You can split the number into a number ending in zero, plus the last digit (9876 = 9870 + 6). The number ending in zero is always a multiple of 10, so it's a multiple of 2. So the whole number is a multiple of 2 if, and only if, the last digit is also a multiple of 2. (And the same for 5.)

4, 8, 16...: do the same, but split off the last n digits (9800 + 76, 9000 + 876). The number ending in n zeroes is a multiple of 10^n, so it's a multiple of 2^n. Then check the last n digits.

9: Much more interesting!

If you sum the digits of a two-digit number, what happens? You're subtracting some number of tens (whatever is in the tens place), and adding the same number of ones, which is the same as subtracting nines. If there's a third digit then you're subtracting hundreds and adding ones, which is subtracting a multiple of 99. And so on: each digit corresponds to a power of 10, so when you add it to the digit sum, you're subtracting 999... which is a multiple of 9. So the starting number is the sum of its digits, plus some multiple of 9. (This works because 9 is 10 - 1.)

And the sum of the digits has a sum of its digits, and so on until you get to a single digit. So the original number is that digit plus a big stack of nines, which is divisible by 9 only if the single digit is 9.

3: Same as 9, except at the end the single digit has to be divisible by 3, so it can be 3, 6, or 9.

Berzerk appreciation post. I remember encountering this game at University of Hartford's student center in 1981. This Stern Electronics 1980 maze arcade game stood out with its (crude but virtually unprecedented) voice synthesis, taunting players to "fight like a robot" and occasionally saying "coin detected in pocket" on the attract screen as you watched it.

One quirk of this game that always initially confuses me to this day when I play it is how you cannot shoot unless the joystick is out of the neutral position (moving), and how once you start shooting (fire button down) you stop moving.

Thus you cannot move and shoot, and the mechanic takes some getting used to. I have often wondered if this was less of a design choice and more of a limitation of game design experience by Stern Designer Alan McNeil, who left his previous company to work at Stern because he was turned down as a video game design lead due to lack of experience.

Whatever the case, it's very distinct and turned out to have an interesting exploit. If you keep holding the fire button down and move the joystick, you can stay in one spot, but fire in all directions at will.

https://www.kxan.com/news/texas/unblock-my-f-ing-anime-dps-arrests-houston-man-after-threatening-call-to-governor/amp/