Question:medium

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

Updated On: Jan 15, 2026
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Correct Answer: 6

Solution and Explanation

The provided inequality is:

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

This simplifies to:

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

The exponent is the sum of the first \( n \) natural numbers, which is:

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

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

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

The inequality now is:

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

Taking the natural logarithm of both sides:

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

Approximating the logarithms:

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

Solving for the exponent:

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

$$\frac{n(n+1)}{2} < -10.44$$

This result is a contradiction, as the left side of the inequality is positive, while the right side is negative.

Therefore, we test values of \( n \) to find the smallest integer that satisfies the original inequality.

  • Using \( x = 0.516 \)
  • For \( n = 6 \), the exponent is \( \frac{6 \cdot 7}{2} = 21 \)
  • \( x^{21} = (0.516)^{21} \approx 1010.2 \)
  • Since \( 1010.2 > 999 \), \( n=6 \) is a valid solution.

The smallest integer value for \( n \) is: \( \boxed{6} \)

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