The problem is below.
Let $I$ be a monomial ideal and let $J= (x_{1}, \dots, x_{r}).$ Show that $I:J \neq I$ if there exist integers $a_{i} >0$ such that $x_{i}^{a_{i}} \in G(I)$ for $i=1, \dots, r$.
For $r=1$, case is trivial; since any $I = (x_{1}^{a_{1}})$ for some $a_{1}>0$, we can find $x_{1}^{a_{1}-1} \not\in I$ such that $x_{1}^{a_{1}-1}(x_{1}) \subseteq I$.
However, I don't know how to prove it for $r>1$. Could you give me any hint?
For general $r$, suppose such $a_{i}$ exists and $I:J=I$. Note also that these $a_{i}$ are minimal, i.e., there is no integer $b$ less than $a_{i}$ such that $x_{i}^{b} \in G(I)$. (Otherwise, $G(I)$ should contain $x_{i}^{b}$, contradicting minimality.) Then let $x^{A_{1}}:=\prod_{j=1}^{r}x_{j}^{a_{j}-1}$. Clearly $x^{A_{1}} \in I:J$. Hence $G(I)$ must contain $x^{A_{2}}$ which divides $x_{A_{1}}$. Also, this $x^{A_{2}}$ should not be $x_{i}^{b}$ form, otherwise it contradicts minimality of $a_{i}$. Then, we can also conclude that $x^{A_{2}-1}:= \prod_{j=1}^{r}x_{j}^{(A_{2})_{j}-1} \in I:J$. Thus there are two cases; the one is that there exists another such $x^{A_{3}} \in G(I)$ which divides $x^{A_{2}}$ and do not have a form $x_{i}^{b}$. However, this contradicts the minimality of $G(I)$. Or, $x^{A_{2}}$ is degree 1 polynomial, so no such $x^{A_{3}}$ exists. However, this is also not the case because it shouldn't be a form of $x_{i}^{b}$. Hence $I:J=I$ leads to the contradiction.