5001 being divisible by 3 doesnt feel right
Shortcuts to determine if an integer is divisible by:
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.
If it ends in 0.
There are actually several rules for 7 (there's a bunch of rules for all of these, in fact). Here's a simple one:
"Subtract twice the last digit from the rest, check if it gives a multiple of 7. Repeat until it's a small enough multiple for you to recognise."
Example:
92911: 9291 - 2 = 9289
-> 9289: 928 - 18 = 910
-> 910: 91 - 0 = 91
-> 91: 9 - 2 = 7
Therefore 92911 is a multiple of 7 (13,273 x 7)
