here's an algebraic solution:
http://www.manhattangmat.com/forums/if- ... t1163.html
if you'd rather conceptualize it (which is always a good idea for number-line problems like this one), you can think of it this way:
r is 3 times as far away from 0 as is m,
but we don't know in which direction.
that's the big thing.
since 12 is halfway between m and r, imagine m and r both starting out at 12, and 'sliding' equally in opposite directions, with r moving to the right and m moving to the left. (you can't slide r to the left and m to the right, because, if you do so, then r will be closer to 0 than is m.)
when the numbers have 'slid' a certain distance - specifically, 6 units each, so that m = 6 and r = 18 - they'll arrive at a point where the distance between m and 0 is 1/3 of the distance between r and 0. that's the first point that satisfies both criteria.
now keep sliding the points away from 12.
eventually, m will pass through 0 itself, and will come out on the negative side. if you keep sliding, you'll reach
another point at which the distance from 0 to m is 1/3 of the distance from 0 to r, only this time m is negative. (specifically, this will happen when m = -12 and r = 36.)
good stuff