(All my rings are commutative.)
Let $R$ denote a ring. Then given ideals $I,J$ and $K$ of $R$, from $J=IK$, we may deduce that $J \subseteq I$. In some rings, the converse doesn't necessarily hold. For example, let $R = \mathbb{R}[x,y]$. Then:
$$xR \subseteq xR+yR$$
but there is no ideal $K$ of $R$ satisfying $xR = (xR+yR)K$
However, suppose that every ideal of $R$ is principal. Then from $J \subseteq I,$ we can conclude the existence of $K$ satisfying $J=IK$. For example, in the integers, from $12\mathbb{Z} \subseteq 2\mathbb{Z}$, we may deduce that $12\mathbb{Z} = (2\mathbb{Z})K$ for some ideal $K$, namely $K = 6\mathbb{Z}$.
Question. Does this characterize rings in which every ideal is principal? Explicitly: suppose we're given a ring $R$ such that for all ideals $I,J$ of $R$, from $J \subseteq I$ we may deduce the existence of an ideal $K$ of $R$ such that $J=IK$. Does it follow that every ideal of $R$ is principal?