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Ma(ya) + (p)atter(n) + (sylph)rena

Ma(ya) + (p)atter(n) + (sylph)rena

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.

Break it down into pairs of two digits and add them up (digit sum but in base 100). If this is divisible by 11, so is your original number.

For example: 121 -> 01; 21 -> 22

120350274

74;02;35;20;01

=132-> 32;01,-> 33 Done. Indeed, it's 11 times 10940934.

This one happened to be fairly easy to add, so I couldn't demonstrate further tricks. Maybe I'll remove the 0s and replace them with a less nice number.

197427274

74;72;42;97;01

This is much uglier to add, but it's okay! Here's where the other trick actually comes from.Since we're only checking divisiblity by 11, we can mod out the 11 whenever it gets inconvenient. This modding it easy as well. Subtract the first digit from the second, since our base is one off from the modulus in that direction.

-3,-5,-2,-2,1

-11

Done.