CAT 2021 Slot 3 · Quant · Question 20

Each factor is a power of {10}^{\frac{1}{7}}, so multiplying them adds the exponents:

{\left({10}^{\frac{1}{7}}\right)}^{1+2+\cdots +n}={10}^{\frac{n\left( n+1 \right)}{14}}.

We need {10}^{\frac{n\left( n+1 \right)}{14}}>999. Since 999<1000={10}^{3}, it is enough to have \frac{n\left( n+1 \right)}{14}\ge 3, that is n\left( n+1 \right)\ge 42.

At n=6: 6\times 7=42, so the product is exactly {10}^{3}=1000, which is indeed greater than 999.

At n=5: 5\times 6=30, giving {10}^{\frac{30}{14}}={10}^{2.14\ldots}\approx 139, which is far below 999.

So the smallest value is n=6.