AMC10 2023 B

AMC10 2023 B · Q9

AMC10 2023 B · Q9. It mainly tests Quadratic equations, Arithmetic misc.

The numbers $16$ and $25$ are a pair of consecutive positive squares whose difference is $9$. How many pairs of consecutive positive perfect squares have a difference of less than or equal to $2023$?

数字 $16$ 和 $25$ 是一对相邻正平方数，它们的差是 $9$。有多少对相邻正完全平方数的差小于或等于 $2023$？

(A) 674 674

(B) 1011 1011

(D) 2019 2019

(E) 2017 2017

Answer

Correct choice: (B)

正确答案：（B）

Solution

Let $x$ be the square root of the smaller of the two perfect squares. Then, $(x+1)^2 - x^2 =x^2+2x+1-x^2 = 2x+1 \le 2023$. Thus, $x \le 1011$. So there are $\boxed{\text{(B)}1011}$ numbers that satisfy the equation. A very similar solution offered by ~darrenn.cp and ~DarkPheonix has been combined with Solution 1.

设 $x$ 是两个完全平方数中较小的那个的平方根。那么，$(x+1)^2 - x^2 =x^2+2x+1-x^2 = 2x+1 \le 2023$。因此，$x \le 1011$。所以有 $\boxed{\text{(B)}1011}$ 个数满足该方程。 ~darrenn.cp 和 ~DarkPheonix 提供的一个非常相似的解法已与解法 1 合并。

Topics