2024 AMC 10A Problem 23

Subtract the second equation from the first:
$$ab + c - bc - a = 13 \quad\Longrightarrow\quad (a - c)(b - 1) = 13.$$
Since $13$ is prime, $b - 1 \in \{1, 13, -1, -13\}$, i.e. $b \in \{2, 14, 0, -12\}$ with $a - c = 13, 1, -13, -1$ respectively. Test each in $ab + c = 100$:

• $b = 2$, $c = a - 13$: $2a + a - 13 = 100$, so $3a = 113$. Not an integer.
• $b = 14$, $c = a - 1$: $14a + a - 1 = 100$, so $15a = 101$. Not an integer.
• $b = 0$, $c = a + 13$: $c = 100$, $a = 87$. Then $ca + b = 8700 \ne 60$. Rejected.
• $b = -12$, $c = a + 1$: $-12a + a + 1 = 100$, so $a = -9$ and $c = -8$. Check: $bc + a = 96 - 9 = 87$ and $ca + b = 72 - 12 = 60$. Valid.

So $(a, b, c) = (-9, -12, -8)$ and
$$ab + bc + ca = 108 + 96 + 72 = 276.$$

The answer is $\boxed{\textbf{(D)}\ 276}$.

Symmetric-looking systems with products become tractable by subtracting equations, because $ab + c - (bc + a)$ factors as $(a - c)(b - 1)$. Choosing the pair whose difference is a prime ($13$) keeps the casework to four branches, and each branch reduces to a linear equation. Always verify a candidate against all three original equations; the $b = 0$ branch satisfies two of them.

Adding the three equations gives $(ab + bc + ca) + (a + b + c) = 247$. Once $(a,b,c) = (-9,-12,-8)$ is known, $a + b + c = -29$ and the requested sum is $247 + 29 = 276$; the value $247$ itself is choice (B), a deliberate trap.

• Adding the three equations to get ab + bc + ca + a + b + c = 247 and answering 247, forgetting to subtract a + b + c = −29.
• Assuming $a, b, c$ are positive and abandoning the search after the positive branches fail.