elegant way of using a range using an if statement?

Question

I've got this very crude way of writing this IF statement .

for a in range (2,3000):
    if ( a % 1) == 0 and ( a % 2) == 0 and ( a % 3) == 0 and ( a % 4) == 0 and ( a % 5) == 0 and ( a % 6) == 0 and ( a % 7) == 0 and ( a % 8) == 0  and ( a % 9) == 0 and ( a % 10) == 0 :
    print a

I assume there is a much better way to write this, using for example a range function combined with the IF statement ?

Answer 1

For a more-or-less direct translation, how about

for a in range(2, 3000):
    if all(a % k == 0 for k in range(1,11)):
        print a

although of course a % 1 == 0 for all integers a, so that check is unnecessary.

Answer 2

What you need is the multiples of LCM(1, 2, 3, 4, 5, 6, 7, 8, 9, 10) which fall within your range. There's multiple ways of computing LCM (see. http://en.wikipedia.org/wiki/Least_common_multiple)

Since LCM(1, 2, 3, 4, 5, 6, 7, 8, 9, 10) = 2^3 * 3^2 * 5 * 7 = 2520, you can do something like

lcm = 2520
i = 2/lcm
j = 3000/lcm
for k in range(i, j)
  print (k + 1) * lcm