UPSC NDA I 2022 Mathematics Solved Paper

Concept:The highest power of a prime $p$ in $n!$ is given by $\left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \cdots$.Explanation:We need the maximum $n$ such that $5^n$ divides $(30! + 35!)$.Factor out $30!$: $(30! + 35!) = 30! (1 + 35 \times 34 \times 33 \times 32 \times 31)$.Now, $35 \times 34 \times 33 \times 32 \times 31 = 7 \times 5 \times 17 \times 2 \times 3 \times 11 \times 2^5 \times 31$.Thus the term in parentheses becomes $1 + (7 \times 5 \times 17 \times 2 \times 3 \times 11 \times 2^5 \times 31)$.This expression contains exactly one factor of $5$, so the total power of $5$ in the product comes entirely from $30!$.Compute the power of $5$ in $30!$: $\left\lfloor \frac{30}{5} \right\rfloor + \left\lfloor \frac{30}{25} \right\rfloor + \left\lfloor \frac{30}{125} \right\rfloor = 6 + 1 + 0 = 7$.Hence, $30!$ contributes $7$ fives, and the other factor contributes zero additional fives.Therefore, the maximum $n$ is $7$.Answer:Option C: 7