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}} .$