AMC10 2002 B

AMC10 2002 B · Q11

AMC10 2002 B · Q11. It mainly tests Quadratic 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

(B) Let $n-1$, $n$, and $n+1$ denote the three integers. Then $$(n-1)n(n+1)=8(3n).$$ Since $n\neq 0$, we have $n^2-1=24$. It follows that $n^2=25$ and $n=5$. Thus, $$(n-1)^2+n^2+(n+1)^2=16+25+36=77.$$

（B）令 $n-1$、$n$ 和 $n+1$ 表示这三个整数。则 $$(n-1)n(n+1)=8(3n)。$$ 由于 $n\neq 0$，有 $n^2-1=24$。因此 $n^2=25$ 且 $n=5$。于是 $$(n-1)^2+n^2+(n+1)^2=16+25+36=77。$$

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