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Zeroes in a Factorial - Number Theory Video Lecture | Quantitative Aptitude for SSC CGL

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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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