Question

Number of zeros at the end of the following expression $(5!{)}^{5!}+(10!{)}^{10!}+(50!{)}^{50!}+(100!{)}^{100!}$ is:

• A. 165
• B. 120
• C. 125
• D. None of these

Answer 1

The correct option is B 120

The number of zeros at the end of $(5!{)}^{5!}=120$
[$\because$ 5! = 120 and thus $(120{)}^{120}$] will give 120 zeros] and the number of zeros at the end of the $(10!{)}^{10!},(50!{)}^{50!}and(100!{)}^{100!}$ will be greater than 120.
Now since the number of zeros at the end of the whole expression will depend on the number which has least number of zeros at the end of the number among other given numbers.
So, the number of zeros at the end of the given expression is 120.