simpler way to find if a number is prime? random inquiry

[wooly_chicken]

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There are a few tricks to see if numbers are divisible by other numbers

all even numbers are out as stated earlier

numbers divisible by three you can add all digits in a number ie 213 is 2+1+3=6....if the sum is divisible by 3 then the full number is divisible by 3.

4 is even so it doest matter...but if the last two digits are divisible by 4, then the entire number is divisible by 4.

5 Obviously all numbers ending in 5 or 0 are div by 5

6 and 8 again both even

9 works the same as 3

That just leaves 7 to screw things up.

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Multiply the last digit of a number by five and add what you get to the remaining digits. What you get is divisible by 7 iff what you started with was. For example:

1547 is divisible by 7 iff 154+35=189 is divisible by 7 iff 18+5*9=63 is divisible by 7 iff 6+5*3=21 is divisible by 7 iff 2+5*1=7 is divisible by 7.

You can come up with similar tests for any number (the key idea for 7 is that 5 is the multiplicative inverse of 10 in $\Z/7\Z$), but they don't seem particularly useful to me.

EDIT: I suppose it would be easier to use -2 instead of 5.