Fast way to check divisible by 2

What is the quickest way to check whether a number is divisible by 2

Someone started and deleted a thread. For those looking for the answer:

check and see if the lowest bit is set

if (number and &b1) > 0
     if it is its an odd number
     if it isnt its even & therefor divisible by 2
end if

dunno if this is quicker than mod but it should be

I tested, they are about the same.

if ( number mod 2 ) = 0 then
  // it's even
  // it's odd
end if