If $a$, $b$, and $c$ are three prime numbers such that $abc = 23(a + b + c)$, then the maximum possible value of $a + b - c$ is

Since $abc = 23(a + b + c)$ and $23$ is prime, $23$ must divide one of $a$, $b$, $c$. As these are primes, one of them equals $23$.

To maximise $a + b - c$, take $c$ small, so set $a = 23$ (not $c$).

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With $a = 23$:

$23 \cdot bc = 23(23 + b + c)$

$bc = 23 + b + c$

$bc - b - c = 23$

$(b - 1)(c - 1) = 24$

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Factor pairs of $24$ give possible $(b - 1, c - 1)$:

$(1, 24)$ gives $(b, c) = (2, 25)$. $25$ is not prime. Reject.

$(2, 12)$ gives $(3, 13)$. Both prime.

$(3, 8)$ gives $(4, 9)$. Neither prime. Reject.

$(4, 6)$ gives $(5, 7)$. Both prime.

Swapped pairs also valid since $b, c$ are interchangeable.

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Valid prime triples (with $a = 23$): $\{23, 3, 13\}$ and $\{23, 5, 7\}$.

For maximum $a + b - c$, pick $c$ smallest among the two non-$23$ primes:

Triple $\{23, 3, 13\}$: take $b = 13$, $c = 3$, so $a + b - c = 23 + 13 - 3 = 33$

Triple $\{23, 5, 7\}$: take $b = 7$, $c = 5$, so $a + b - c = 23 + 7 - 5 = 25$

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Maximum value $= 33$

Answer $= 33$