Question 49108
The easist way is to do this is graphically with a calculator.  Plot y = | | |x - 2| - 4| - 5 | as one equation, and then graph y = 1 as the second equation, and then just count how many times they intersect.
But my guess is that you're not allowed to do this.
So you can try to solve it algebraically, but this involves many, many possibilities, and can be hard to keep track of.
Therefore, let's approach this graphically.  To do this, you need to know two basic items.  First, if you add or subtract a number to a graph, it shifts the graph up or down.  Second, the absolute value function makes all negative numbers positive but leaves positive numbers unchanged.  For a graph, that means that any piece of the graph below the x-axis gets flipped back up to the positive side of the graph.
So we just need to put those two pieces together to build your graph.
Let's start with a simple y = x graph.  You should know that this is a simple diagonal line through the origin (0, 0) with a slope of one (through points (-3, -3), (-2, -2), (-1, -1), (1, 1), (2, 2), (3, 3) etc)
Now, let's start building your graph.  We move to y = x - 2.  What does the - 2 do to the graph?  It shifts every point down two units.  So the origin (0, 0) moves to (0, -2), (1, 1) moves to (1, -1), (2, 2) moves to (2, 0).  OK so far?
Now, let's apply the first absolute value function.  Move to y = | x - 2 |.  This takes all of the diagonal line below the x-axis and flips it upwards.  So the points (2, 0), (3, 1) (4, 2) etc don't do anything, because they're y-values are already positive.  But on the other side of (2, 0), these points get flipped - like (1, -1) flips up to (1, 1) and (0, -2) flips up to (0, 2).  What we have is a V-shape with a vertex at (2, 0).  OK so far?
Now we just need to keep going.  Let's add the - 4 to the graph now.  Since we're subtracting 4 from everything, we shift the graph down 4 units.  So the vertex at (2, 0) gets shifted down to (2, -4), (3, 1) gets shifted down to (3, -3), and on the other side,  (1, 1) gets shifted down to (1, -3).  Got it?  We still have a V-shape, but the vertex now sits 4 units below the x-axis.
OK, keep adding to it.  Now, let's add the absolute value to this one.  Every point below the x-axis gets flipped up positive, while anything above the x-axis get's left alone.  Therefore, the vertex at (2, -4) gets flipped up to (2, 4), while (3, -3) gets flipped to (3, 3), and (1, -3) ends at (1, 3). The important points that don't change (lying the x-axis) (6, 0) stays at (6, 0), and (-2, 0) stays at (-2, 0).  What we have here is a weird shaped W, with the bottom of the W at (6, 0) and (-2, 0), and the peak of the middle of the W at (2, 4).  Got it?
OK, we're almost done.  Let's add the - 5 to it.  Again, this shifts the graph straight down 5 units.  So using the points above, the bottom of the W points: (-2, 0) shifts down to (-2, -5) and (6, 0) shifts down to (6, -5), while the peak in the middle of the W: (2, 4) shifts down to (2, -1). OK?
Last one!!  Take the absolute value of that W!  Everything below the x-axis flips up positive, while anything above stays the same.  So the very outside ends of the graph stay the same, while the W below the axis (between x = -7 and x = 11) gets flipped up into a M.  (-7, 0) stays (-7, 0), and (11, 0) stays (11, 0).  The lower peaks of the W at (-2, -5) and (6, -5) flip up to (-2, 5) and (6, 5), while the middle of the W at (2, -1) flips to (2, 1).
Our graph is an M with wings going up on each end!
OK, all we need to do now if find our how many places have a height of 1.  The answer is 5.  One on the left wing, one on the first leg of the M, one at the middle of the M, one on the last leg of the M, and the last one on the right wing.
This is probably not the best way to do it.  Definitely the easiest is to let a calculator do the work.  But if you thought this way was bad, don't even try to do this algebraically!  It really gets ugly. Trust me, I tried it.