AMC 8 1986 Problem 7

AMC8 1986 年第 7 题

表达式与运算★★☆☆☆

How many whole numbers are between

\sqrt8

and

\sqrt80

\sqrt{8}

和

\sqrt{80}

之间有多少个整数？

正确答案 · Correct Answer

解析 · Solution

Solution 1

Note that

2 < \sqrt{8} < 3

and

8 < \sqrt{80} < 9

. Therefore, only the integers from 3 to 8 inclusive are between

\sqrt{8}

and

\sqrt{80}

, for a total of 6 integers.

Linear Approximation using Calculus

While this may not be the best, most efficient way to solve this problem, it can prove to be quite useful in similar problems. Using Calculus, we can use tangent-line approximation to approximate these square roots. First, we understand that these roots are of the form

f(x) = \sqrt{x}.

Given the values to approximate, 8, and 80, we find the closest square number to them, namely, 9, and 81.

Now, we take the derivative of

f(x)

to get

f'(x) = \dfrac{1}{2\sqrt{x}}.

We can approximate

f(x)

, to get

f(x) \approxeq f(x) + f`(x) \cdot \Delta{x},

where

x

is the perfect square closest to the the input value (in this case 8), and

\Delta{x}

is the difference between

x

and the input value.

The two cases are as follows:

𝑖 𝑛 𝑝 𝑢 𝑡 𝑥 𝑣 𝑎 𝑙 𝑢 𝑒 Δ 𝑥

8 9 8 − 9 = − 1

80 81 80 − 81 = − 1

Now, we can plug in our values into the equation, getting:

𝑓 ( 8 ) ≊ 𝑓 ( 9 ) + 𝑓 ' ( 9 ) ⋅ − 1 ⟹ 9 + 1 2 9 ⋅ − 1 ⟹ 3 − 1 6

𝑓 ( 80 ) ≊ 𝑓 ( 81 ) + 𝑓 ' ( 81 ) ⋅ − 1 ⟹ 81 + 1 2 81 ⋅ − 1 ⟹ 9 − 1 18

Thus, we see that our answer lies in the amount of whole numbers between

3 - \dfrac{1}{6},

and

9 - \dfrac{1}{18},

and as we are counting whole numbers, we can round each expression to narrow our interval to $[3,8].$ Therefore, we see that our answer is

6\text{(B)}.

Note: this solution is definitely not the fastest solution, and in a competition, you should probably use intuition to see the closest numbers in the interval of those square roots, and solve the problem from there.

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