Kiana has two older twin brothers.  The product of their three ages is 128.  What is the sum of their three ages?
$\mathrm{(A)}\ 10\qquad \mathrm{(B)}\ 12\qquad \mathrm{(C)}\ 16\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 24$
The age of each person is a factor of $128 = 2^7$.  So the twins could be $2^0 = 1, 2^1 = 2, 2^2 = 4, 2^3 = 8$ years of age and, consequently Kiana could be $128$, $32$, $8$, or $2$ years old, respectively.  Because Kiana is younger than her brothers, she must be $2$ years old. So the sum of their ages is $2 + 8 + 8 = \boxed{18}$.  The answer is $\mathrm{(D)}$.