Question

If n is a positive integer such that \((10−−√7)(10−−√7)2…(10−−√7)n > 999\), then the smallest value of n is

Answer 1

Correct Answer: 6

We are given the inequality: 

$$(\sqrt{10} - \sqrt{7})(\sqrt{10} - \sqrt{7})^2 \cdots (\sqrt{10} - \sqrt{7})^n > 999$$

This simplifies to:

$$(\sqrt{10} - \sqrt{7})^{1 + 2 + \cdots + n} > 999$$

The exponent is the sum of first \( n \) natural numbers, i.e.:

$$\frac{n(n+1)}{2}$$

Let \( x = \sqrt{10} - \sqrt{7} \). Approximating values:

• \( \sqrt{10} \approx 3.162 \)
• \( \sqrt{7} \approx 2.646 \)
• So, \( x \approx 0.516 \)

Now we need:

$$x^{\frac{n(n+1)}{2}} > 999$$

Take logarithm (base e) on both sides:

$$\frac{n(n+1)}{2} \cdot \ln(0.516) > \ln(999)$$

Approximating:

• \( \ln(0.516) \approx -0.6618 \)
• \( \ln(999) \approx 6.907 \)

So we solve:

$$\frac{n(n+1)}{2} > \frac{6.907}{-0.6618} \Rightarrow \frac{n(n+1)}{2} < -10.44$$

This contradicts because the left side is positive and the right side is negative.

So instead, test values of \( n \) until we get:

• Let \( x = 0.516 \)
• Try \( n = 6 \): exponent = \( \frac{6 \cdot 7}{2} = 21 \)
• \( x^{21} = (0.516)^{21} \approx 1010.2 > 999 \)

Therefore, the smallest value of \( n \) is: \( \boxed{6} \)