[Jazza]

most of you know the shortcuts for figuring out if a whole number is divisible by:

2 - if it's even
3 - if the sum of the digits is divisble by 3
4 - if the last two digits are divisible by 4
5 - if the last digit is 0 or 5
6 - if it's divisible by 2 and 3 (duh)
8 - if the last 3 digits are divisible by 8
9 - if the sum of the digits is divisible by 9

but no one ever told me one for 7, and i thought that was bullshit, so i found it myself:

write the number in question backwards

under this number write the number 132645132645132645...

now multiply each digit with the number below it, then add them all together

if this number is divisible by 7, then the original is

so, lets say you wanted to know if 27351223 was divisible by 7, well...

2*1+7*3+3*2+5*6+1*4+2*5+2*1+3*3=84, and 84 is divisible by 7, so 27351223 is divisible by 7

also not that you can't figure out if your new number is divisible by 7, repeat the procss:

4*1+8*3=28 which is divisible by 7, so 84 is too

this methods kinda blows though, it's not as easy as the other ones, which is probably why they don't teach it