If n is a positive integer such that \((10−−√7)(10−−√7)2…(10−−√7)n > 999\), then the smallest value of n is
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:
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:
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.
The smallest integer value for \( n \) is: \( \boxed{6} \)