The number of real solutions of the equation
\[|x-2|+|x-3|=1\]
is
$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 2\quad \text{(D) } 3\quad \text{(E) more than } 3$
For $2\le x\le 3$, the left-hand side is $(x-2)-(x-3)$ which is $1$ for all $x$ in the interval. $\fbox{E}$