2024 AMC 12A Problem 17

Subtracting the second equation from the first gives $(a-c)(b-1)=13.$ Hence $$\begin{gathered} (a-c,b-1)\in\{(1,13),(13,1),\\ (-1,-13),(-13,-1)\}. \end{gathered}$$ The corresponding values of $b$ are $14,2,-12,0.$ Substituting $a=c+(a-c)$ into $ab+c=100$ eliminates the first two cases because they would require $15c=86$ or $3c=74.$ The case $b=-12$ gives $c=-8$ and $a=-9,$ which satisfies all three equations. The case $b=0$ gives $(a,c)=(87,100),$ which fails $ca+b=60.$ Thus the unique integer solution is $(-9,-12,-8),$ and $ab+bc+ca$ $=108+96+72=276.$ Thus, the correct answer is D.