UPSC NDA 2 2019 Mathematics Solved Paper

Concept:The number of trailing zeros in $n!$ equals the highest power of $5$ dividing $n!$.It is given by the sum $\left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{25} \right\rfloor + \left\lfloor \frac{n}{125} \right\rfloor + \cdots$.Explanation:We need to find $n$ such that the total zeros in $n!$ is $17$.Check the given options by calculating the sum of floors.For $n = 80$: $\left\lfloor \frac{80}{5} \right\rfloor = 16$, $\left\lfloor \frac{80}{25} \right\rfloor = 3$, $\left\lfloor \frac{80}{125} \right\rfloor = 0$. Sum $= 16 + 3 = 19$.For $n = 85$: $\left\lfloor \frac{85}{5} \right\rfloor = 17$, $\left\lfloor \frac{85}{25} \right\rfloor = 3$, sum $= 17 + 3 = 20$.For $n = 95$: $\left\lfloor \frac{95}{5} \right\rfloor = 19$, $\left\lfloor \frac{95}{25} \right\rfloor = 3$, sum $= 19 + 3 = 22$.None of these values equals $17$.Also, $n$ can be less than $80$ (e.g., $75!$ gives $18$ zeros, $70!$ gives $16$ zeros), but the options only include $80, 85, 95$.Thus, no given $n$ yields exactly $17$ zeros.Answer:D. No such value of $n$ exists.