AMC10 2024 A

AMC10 2024 A · Q23

AMC10 2024 A · Q23. It mainly tests Systems of equations, Number theory misc.

Integers $a$, $b$, and $c$ satisfy $ab + c = 100$, $bc + a = 87$, and $ca + b = 60$. What is $ab + bc + ca$?

整数 $a$、$b$ 和 $c$ 满足 $ab + c = 100$，$bc + a = 87$，$ca + b = 60$。求 $ab + bc + ca$？

(A) 212 212

(B) 247 247

(D) 276 276

(E) 284 \qquad 284 \qquad

Answer

Correct choice: (D)

正确答案：（D）

Solution

Subtracting the first two equations yields $(a-c)(b-1)=13$. Notice that both factors are integers, so $b-1$ could equal one of $13,1,-1,-13$ and $b=14,2,0,-12$. We consider each case separately: For $b=0$, from the second equation, we see that $a=87$. Then $87c=60$, which is not possible as $c$ is an integer, so this case is invalid. For $b=2$, we have $2c+a=87$ and $ca=58$, which by experimentation on the factors of $58$ has no solution, so this is also invalid. For $b=14$, we have $14c+a=87$ and $ca=46$, which by experimentation on the factors of $46$ has no solution, so this is also invalid. Thus, we must have $b=-12$, so $a=12c+87$ and $ca=72$. Thus $c(12c+87)=72$, so $c(4c+29)=24$. We can simply trial and error this to find that $c=-8$ so then $a=-9$. The answer is then $(-9)(-12)+(-12)(-8)+(-8)(-9)=108+96+72=\boxed{\textbf{(D) }276}$.

前两式相减得 $(a-c)(b-1)=13$。因式为整数，故 $b-1=13,1,-1,-13$，$b=14,2,0,-12$。 $b=0$：$a=87$，$87c=60$，非整数。 $b=2$：$2c+a=87$，$ca=58$，$58$ 因数无解。 $b=14$：$14c+a=87$，$ca=46$，无解。 $b=-12$：$a=12c+87$，$ca=72$，$c(12c+87)=72$，$c(4c+29)=24$。试得 $c=-8$，$a=-9$。则 $ab+bc+ca=(-9)(-12)+(-12)(-8)+(-8)(-9)=108+96+72=\boxed{\textbf{(D) }276}$。

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