Originally Posted by

**Jhevon** 1) will five examples be enough?

24, 36, 72, 144, 216

2) Let x be a multiple of 3. then 3|x. (3|x means "3 divides x," that is 3 goes into x and leaves no remainder.)

since 3|x, x = 3m for some integer m.

similarly, if x is also a multiple of 4, then 4|x. so x = 4n for some integer n

=> 3m = 4n

=> 3m/4 = n, where n is some integer divisible by 3 and 4.

Now suppose 6|n, then n = 6k for some integer k. that is 3m/4 = 6k.

so 3m = 24k = 6(4k).

since 4k is an integer, and 3m = 6(4k), 6|3m. but 3m is x, and so 6|x, and therefore, x is a multiple of 6 as well.

there's something i dont like about this proof, i'll think about it and get back to you