AMC12 2002 B

AMC12 2002 B · Q7

AMC12 2002 B · Q7. It mainly tests Linear equations, Word problems (algebra).

The product of three consecutive positive integers is $8$ times their sum. What is the sum of their squares?

三个连续正整数的乘积是它们和的 $8$ 倍。它们的平方和是多少？

(A) 50 50

(B) 77 77

(D) 149 149

(E) 194 194

Answer

Correct choice: (B)

正确答案：（B）

Solution

Let the three consecutive positive integers be $a-1$, $a$, and $a+1$. Since the mean is $a$, the sum of the integers is $3a$. So $8$ times the sum is just $24a$. With this, we now know that $a(a-1)(a+1)=24a\Rightarrow(a-1)(a+1)=24$. $24=4\times6$, so $a=5$. Hence, the sum of the squares is $4^2+5^2+6^2=\boxed{\mathrm{ (B)}\ 77}$.

设这三个连续正整数为 $a-1$、$a$、$a+1$。由于平均数为 $a$，它们的和为 $3a$，所以 $8$ 倍的和为 $24a$。因此 $a(a-1)(a+1)=24a\Rightarrow(a-1)(a+1)=24$。因为 $24=4\times6$，所以 $a=5$。于是平方和为 $4^2+5^2+6^2=\boxed{\mathrm{ (B)}\ 77}$。

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