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

Given that:
$(^7\sqrt{10})(^7\sqrt{10})^2).....(^7\sqrt{10})^n) > 999$

We can approximate 999 to 1000 since we are working with powers of 10, and 1000 can be expressed as $10^3$.
\\\((^7\sqrt{10})(^7\sqrt{10})^2).....$(^7\sqrt{10})^n)$ $>$ $10^3)$

Since the bases are identical in the product, the exponents will sum up to n.
Therefore,
$(10)^{\frac{1}{7}}.\frac{n(n+1)}{2} >10^3$
When you compare the bases on both sides of the inequality
$\frac{n(n+1)}{14}>3$
$n(n+1) > 42$
If n were 6, then $6 × 7 = 42$. This is a valid solution because we approximated 999 as 1000, so without the approximation, the right-hand side would be less than 42.
Therefore, 6 fulfills the requirement.