The problem asks for the digits $a$, $b$, and $c$ such that the 5-digit number $abc a 4$ is divisible by 11, given the condition that $c > b$.

To approach this, we apply the divisibility rule for 11. The rule states that a number is divisible by 11 if the difference between the sum of its digits in odd positions and the sum of its digits in even positions is a multiple of 11 (including 0).

For the 5-digit number $abc a 4$, the digits are arranged as follows:

• The odd-position digits are $a$, $c$, and $4$.
• The even-position digits are $b$ and $a$.

Let's calculate the alternating sum of these digits:

$S = a - b + c - a + 4 = c - b + 4$

For the number to be divisible by 11, this sum $S$ must be divisible by 11:

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

Simplifying the equation:

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

Since $-4 \equiv 7 \pmod{11}$, we have:

$c - b \equiv 7 \pmod{11}$

This means that $c = b + 7$, where $c > b$.

Now, we look for pairs of digits $b$ and $c$ such that $c = b + 7$ and both $b$ and $c$ are digits (i.e., between 0 and 9). The valid pairs are:

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

Thus, the possible pairs of $b$ and $c$ are:

$(0, 7), (1, 8), (2, 9)$

Do you need further clarification or assistance with other parts of the problem?