The number of zeros at the end of $60!$ is

### Concept & Formula To find the number of trailing zeros in a standard factorial ($n!$), we must count the number of times the prime factor $5$ appears in its prime factorization. We ignore the factor $2$ because in any sequential $1$ to $n$ product, $2$'s are always vastly more abundant than $5$'s. Use successive division by $5$ to evaluate the total occurrences: $$ \text{Trailing Zeros} = \left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{25} \right\rfloor + \left\lfloor \frac{n}{125} \right\rfloor + \dots $$ ### Step-by-Step Solution * **Given:** The expression is $60!$ (60 factorial). * **Calculation / Deduction:** Apply successive division to the number $60$: * Divide $60$ by $5$: $$ \lfloor 60 / 5 \rfloor = 12 $$ This accounts for numbers like $5$, $10$, $15$, $20$, etc., that contribute at least one $5$. * Divide the previous quotient ($12$) by $5$: $$ \lfloor 12 / 5 \rfloor = 2 $$ This accounts for the numbers that contain a *second* factor of $5$ (specifically $25$ and $50$). * Since $2$ is less than $5$, the division process terminates here. Sum the integer quotients: $$ \text{Total zeros} = 12 + 2 = 14 $$ ### Exam Strategy & Shortcut **Mental Ladder:** For numbers under $100$, do this purely in your head. "$60$ divided by $5$ is $12$." "$12$ divided by $5$ is $2$." "$12 + 2 = 14$." Do not write anything down for basic factorial trailing zero queries. ### Common Pitfall A recurring mistake for beginners is simply dividing by $5$ once and assuming the answer is $12$ (Option a). It is crucial to remember that higher powers of $5$ (like $25$) hide extra $5$'s that must be captured by continuing the division chain on the quotient. ### Final Answer **Therefore, the correct answer is 14.**