CAT Exam 2021 Slot 3 Solved Paper

$(7√10) (7√10)^2 ..... (7√10)^\n > 999$
$(7√10)^{1+2+... \n} > 999$
$10^{{1+2+... \n}/7 } > 999$
For minimum value of n,
$^{{1 + 2= ... +\n}/7} 3 $
$1 + 2 + ... + \n = 21$
We can see that if n = 6, 1 + 2 + 3 + ... + 6 = 21.