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.
#THIS POST IS WRONG 7 DOES HAVE A RULE FOR HOW TO TELL IF THE NUMBER IS DIVISIBLE#it's just. not a short cut#you're better off dividing it and seeing if you get a whole number than doing the method
but what IS the rule @anqelbean ???
Okay, so, basically, you remove the least important digit of the number (which is the first digit from the right), you double it, and then subtract it from the remaining number. If the number you get is still large, you do it again.
For example, 294
29 4
4×2=8
29-8=21
21 is divisible by 7
Second example, 1946
194 6
6×2=12
194-12=182
182 is still large, so let do it again
18 2
2×2=4
18-4=14
14 is divisible by 7
