Question

The number of zeros at the end of the product of ${222}^{111}\times {35}^{53}+(7!{)}^{6!}\times (10!{)}^{5!}+{42}^{42}\times {25}^{25}$ is

• A. 42
• B. 53
• C. 1055
• D. None of these

Answer 1

The correct option is A 42

The number of zeros at the end of ${222}^{111}\times {35}^{53}$ is 53.
The number of zeros at the end of $(7!{)}^{6!}\times (10!{)}^{5!}$ is 960.
The number of zeros at the end of ${42}^{42}\times {25}^{25}$ is 42.
Thus the number of zeros at the end of the whole expression is 42.