CAT 1993 — QA Question 48

The number of zeros at the end will be same as the highest power of 10 in 100!.

Number of 2 in the product of all integers from 1 to 100 $=\frac{100}{2}+\frac{100}{4}+\frac{100}{8}+\frac{100}{16}+\frac{100}{32}+\frac{100}{64}$

= 50 + 25 +12 + 6 + 3 + 1= 97 and number of 5 in the product of all integers from 1 to 100 = $\frac{100}{5}+\frac{100}{25}$ = 20 + 4 = 24.

∴ 100! = 2⁹⁷ × 5²⁴ × ... 
⇒ 100! = 2⁷³ × (2 × 5²⁴) × ... 
⇒ 100! = 2⁷³ × (10²⁴) × ... 

∴ Highest power of 10 in 100! is 24.

Hence, option (b).