AMC12 2002 B

AMC12 2002 B · Q13

AMC12 2002 B · Q13. It mainly tests Factoring.

The sum of $18$ consecutive positive integers is a perfect square. The smallest possible value of this sum is

18 个连续正整数的和是一个完全平方数。这个和的最小可能值是

(A) 169 169

(B) 225 225

(D) 361 361

(E) 441 441

Answer

Correct choice: (B)

正确答案：（B）

Solution

Let $a, a+1, \ldots, a + 17$ be the consecutive positive integers. Their sum, $18a + \frac{17(18)}{2} = 9(2a+17)$, is a perfect square. Since $9$ is a perfect square, it follows that $2a + 17$ is a perfect square. The smallest possible such perfect square is $25$ when $a = 4$, and the sum is $225 \Rightarrow \mathrm{(B)}$. Notice that all five choices given are perfect squares. Let $a$ be the smallest number, we have \[a+(a+1)+(a+2)+...+(a+17)=18a+\sum_{k=1}^{17}k=18a+153\] Subtract $153$ from each of the choices and then check its divisibility by $18$, we have $225$ as the smallest possible sum. $\mathrm {(B)}$

设这 18 个连续正整数为 $a, a+1, \ldots, a + 17$。它们的和为 $18a + \frac{17(18)}{2} = 9(2a+17)$，且是完全平方数。由于 $9$ 是完全平方数，因此 $2a + 17$ 也必须是完全平方数。满足条件的最小完全平方数是 $25$，此时 $a = 4$，和为 $225 \Rightarrow \mathrm{(B)}$。注意给出的五个选项都是完全平方数。令 $a$ 为最小的数，则 \[a+(a+1)+(a+2)+...+(a+17)=18a+\sum_{k=1}^{17}k=18a+153\] 将每个选项减去 $153$ 并检查其是否能被 $18$ 整除，可得 $225$ 是最小可能的和。$\mathrm {(B)}$

Topics

AL.006 Factoring