jmc

A) By the definition of factor, there must be some integer $n$ such that $60=b\cdot n.$ In addition, there must be some integer $m$ such that $b=a\cdot m.$ Substituting the second equation into the first yields $60=(a\cdot m)\cdot n=a\cdot (mn).$ Because $m$ and $n$ are integers, so is $mn.$ Thus, $a$ is a factor of $60.$ This statement is true.

B) By the definition of divisor, there must exist some integer $n$ such that $60=b\cdot n.$ However, because $n$ is an integer, this also means that $60$ is a multiple of $b.$ This statement is true.

C) $b$ and $c$ are both factors of 60, and $b<c.$ In many cases, this statement is true. However, there are counterexamples. For example, $c=30$ and $b=20.$ Both numbers are divisors of $60,$ but $20$ is not a factor of $30$ because there is no integer $n$ such that $30=20\cdot n.$ This statement is false.

D) If $a$ were to be $20,$ then the given inequality would be $20<b<c<60$ where $b$ and $c$ are factors of $60.$ Listing out the factors of $60,$ we see $1,$ $2,$ $3,$ $4,$ $5,$ $6,$ $10,$ $12,$ $15,$ $20,$ $30,$ $60.$ However, there is only one factor of $60$ that is between $20$ and $60,$ so it is impossible to choose a $b$ and $c$ that satisfy the conditions. Thus, this statement is true.

E) If $b$ is negative, then by the given inequality, so is $a$ because $a<b.$ We also know that $a$ is a divisor of $b.$ Thus, there exists an integer $n$ such that $b=a\cdot n.$ Dividing both sides by $a$ yields $n=\frac{b}{a}.$ Because both $a$ and $b$ are negative, $n$ must be positive. Recall that $\frac{x}{y}=\frac{-x}{-y}.$ Thus, the fraction $\frac{b}{a}$ where $a<b$ and both are negative is the same as the fraction $\frac{-b}{-a}$ where $-a>-b.$ However, because both the numerator and denominator are positive, and the denominator is greater than the numerator, it is impossible for this fraction to be an integer. But $n$ must be an integer, so this statement is false.

Thus, the false statements are $\boxed{\text{C,E}}.$