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The number of zeros at the end of 60! is :
Answer & Solution
Correct Answer:
Option
B
Clearly, highest power of 2 is much higher as compared to that of 5 in 60!,
So, Required number of zeros
= Highest power of 5
= $$ \left[ {\frac{{60}}{5}} \right] + \left[ {\frac{{60}}{{{5^2}}}} \right]$$
= 12 + 2
= 14
So, Required number of zeros
= Highest power of 5
= $$ \left[ {\frac{{60}}{5}} \right] + \left[ {\frac{{60}}{{{5^2}}}} \right]$$
= 12 + 2
= 14
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