UPSC CDS 2 2021 Math Paper

Concept:The number of trailing zeros equals the power of 10, which is the minimum of the exponents of 2 and 5 in the prime factorization. Since powers of 5 are rarer, we count the total exponent of 5.Explanation:The product is $\prod_{k=1}^{50} k^{51-k}$.We need the total exponent of 5 in this product.For each $k$ divisible by 5, contribution = $(51-k) \times v_5(k)$ where $v_5(k)$ is the exponent of 5 in $k$.Multiples of 5 up to 50: $5,10,15,20,25,30,35,40,45,50$.$v_5(5)=1$, weight $46$ → $46$$v_5(10)=1$, weight $41$ → $41$$v_5(15)=1$, weight $36$ → $36$$v_5(20)=1$, weight $31$ → $31$$v_5(25)=2$, weight $26$ → $52$$v_5(30)=1$, weight $21$ → $21$$v_5(35)=1$, weight $16$ → $16$$v_5(40)=1$, weight $11$ → $11$$v_5(45)=1$, weight $6$ → $6$$v_5(50)=2$, weight $1$ → $2$Sum: $46+41+36+31+52+21+16+11+6+2 = 262$.The exponent of 2 in the product is much larger than $262$ (since many even numbers contribute multiple powers of 2).Thus the number of trailing zeros = $\min(\text{exponent of 2}, \text{exponent of 5}) = 262$.Answer:262 (Option A).