This problem involves finding values for the digits $a$, $b$, and $c$ in a 5-digit number $abca4$, ensuring that it is divisible by 11. We are given that $c > b$ and that the possible pairs of values for $b$ and $c$ are:

• $b = 0$, $c = 7$
• $b = 1$, $c = 8$
• $b = 2$, $c = 9$

To solve this problem, we need to use the divisibility rule for 11, which states:

$\text{The alternating sum of the digits must be divisible by 11.}$

For a 5-digit number $abca4$, the alternating sum is:

$a - b + c - a + 4$

Simplifying, we get:

$-b + c + 4$

For divisibility by 11, we require:

$-b + c + 4 \equiv 0 \pmod{11}$

This equation can be rearranged as:

$b - c \equiv 4 \pmod{11}$

Now, let's check the possible pairs of $b$ and $c$:

1. For $b = 0$ and $c = 7$: $0 - 7 + 4 = -3$ $-3 \mod 11 = 8$, so this pair does not work.

2. For $b = 1$ and $c = 8$: $1 - 8 + 4 = -3$ $-3 \mod 11 = 8$, so this pair does not work.

3. For $b = 2$ and $c = 9$: $2 - 9 + 4 = -3$ $-3 \mod 11 = 8$, so this pair does not work either.

None of the pairs satisfy the divisibility condition as written. It appears there may be a mistake in the original conditions or assumptions. Could you confirm the problem setup, or if any other conditions are given?