### Instruction: Let $a$ be a factor of $b,$ and let $b$ and $c$ be divisors of $60$ such that $a<b<c<60.$ Which of the following statements is/are false? List the letters in alphabetical order with commas separating the letters.

$\bullet$ A.) $a$ must be a divisor of $60.$

$\bullet$ B.) $60$ must be a multiple of $b.$

$\bullet$ C.) $b$ must be a factor of $c.$

$\bullet$ D.) $a$ cannot be $20.$

$\bullet$ E.) It is possible for $b$ to be negative.### Response: I need to find the false statements, and list them in alphabetical order.
I can start by checking each statement one by one, and see if I can find a counterexample or a proof.
A.) $a$ must be a divisor of $60.$
This seems true, because if $a$ is a factor of $b,$ then $a$ must divide $b$ evenly.
And if $b$ is a divisor of $60,$ then $b$ must divide $60$ evenly.
So, by transitivity, $a$ must divide $60$ evenly.
I don't see any way to make $a$ not a divisor of $60,$ without violating the given conditions.
So I think statement A is true.
B.) $60$ must be a multiple of $b.$
This also seems true, because if $b$ is a divisor of $60,$ then by definition, $60$ is a multiple of $b.$
I don't see any way to make $60$ not a multiple of $b,$ without violating the given conditions.
So I think statement B is true.
C.) $b$ must be a factor of $c.$
This seems false, because I can think of a counterexample.
For example, suppose $a=2,$ $b=6,$ and $c=15.$
Then $a$ is a factor of $b,$ and $b$ and $c$ are divisors of $60,$ and $a<b<c<60.$
But $b$ is not a factor of $c,$ because $c$ is not a multiple of $b.$
So statement C is false.
D.) $a$ cannot be $20.$
This seems true, because if $a=20,$ then $b$ and $c$ would have to be divisors of $60$ that are larger than $20.$
But the only divisors of $60$ that are larger than $20$ are $30$ and $60.$
So $b$ and $c$ would have to be either $30$ or $60,$ which would violate the condition that $b<c<60.$
So I don't see any way to make $a=20,$ without violating the given conditions.
So I think statement D is true.
E.) It is possible for $b$ to be negative.
This seems false, because if $b$ were negative, then $c$ would have to be negative as well, since $b<c.$
But then $b$ and $c$ would not be divisors of $60,$ because divisors are usually assumed to be positive integers.
So I don't see any way to make $b$ negative, without violating the given conditions.
So I think statement E is false.
Therefore, the only false statements are C and E.
I need to list them in alphabetical order, so the answer is C, E.