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Zeroes in a Factorial - Number Theory Video Lecture | Quantitative Aptitude for SSC CGL
FAQs on Zeroes in a Factorial - Number Theory Video Lecture - Quantitative Aptitude for SSC CGL
| 1. How do you calculate the number of trailing zeroes in a factorial? |  |
Ans.To calculate the number of trailing zeroes in a factorial, you can use the formula:
\[ \text{Number of trailing zeroes} = \left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{25} \right\rfloor + \left\lfloor \frac{n}{125} \right\rfloor + \ldots \]
This formula sums the integer division of \( n \) by powers of 5 until the result is zero. Each term counts how many times 5 is a factor in the numbers from 1 to \( n \).
| 2. Why do trailing zeroes in a factorial come from the factor 10? | |
Ans.Trailing zeroes in a factorial come from the factor 10, which is the product of 2 and 5. In factorials, there are generally more factors of 2 than factors of 5, so the number of trailing zeroes is determined by the number of times 5 is a factor in the numbers leading up to \( n \).
| 3. What is the maximum number of trailing zeroes in \( 100! \)? | |
Ans.To find the maximum number of trailing zeroes in \( 100! \), we apply the formula:
\[ \left\lfloor \frac{100}{5} \right\rfloor + \left\lfloor \frac{100}{25} \right\rfloor = 20 + 4 = 24 \]
Therefore, \( 100! \) has 24 trailing zeroes.
| 4. Does the number of trailing zeroes in a factorial increase with larger \( n \)? | |
Ans.Yes, the number of trailing zeroes in a factorial generally increases with larger \( n \). This is because as \( n \) increases, there are more multiples of 5, which contribute additional factors of 5 to the factorial, thus increasing the number of trailing zeroes.
| 5. Can you provide an example of calculating trailing zeroes in \( 50! \)? | |
Ans.To calculate the number of trailing zeroes in \( 50! \), we use the formula:
\[ \left\lfloor \frac{50}{5} \right\rfloor + \left\lfloor \frac{50}{25} \right\rfloor = 10 + 2 = 12 \]
Thus, \( 50! \) has 12 trailing zeroes.
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