What Is The Highest Power Of 2 That Divides 533! ? The Highest Power Of 3? The Highest Power Of 6
Show That 95! Ends In 22 Zeros; With How Many Zeros Does 1001 End?
The Correct Answer and Explanation is :
To determine the highest powers of 2, 3, and 6 that divide (533!), and the number of trailing zeros in (95!) and (1001!), we will need to apply the formula for the highest power of a prime (p) that divides (n!):
[
\text{Highest power of } p \text{ dividing } n! = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor
]
1. Highest Power of 2 in (533!):
For powers of 2, we apply the formula for (p = 2):
[
\left\lfloor \frac{533}{2} \right\rfloor + \left\lfloor \frac{533}{4} \right\rfloor + \left\lfloor \frac{533}{8} \right\rfloor + \cdots
]
Summing these gives the highest power of 2 in (533!).
2. Highest Power of 3 in (533!):
For powers of 3, we use the same formula but with (p = 3):
[
\left\lfloor \frac{533}{3} \right\rfloor + \left\lfloor \frac{533}{9} \right\rfloor + \left\lfloor \frac{533}{27} \right\rfloor + \cdots
]
3. Highest Power of 6 in (533!):
Since (6 = 2 \times 3), the highest power of 6 that divides (533!) is the minimum of the highest powers of 2 and 3.
4. Number of Zeros in (95!):
The number of trailing zeros in (n!) is determined by the number of factors of 5 in (n!), because each factor of 5 pairs with a factor of 2 to form a trailing zero. The formula is:
[
\left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{25} \right\rfloor + \left\lfloor \frac{n}{125} \right\rfloor + \cdots
]
For (n = 95), we compute:
[
\left\lfloor \frac{95}{5} \right\rfloor + \left\lfloor \frac{95}{25} \right\rfloor + \left\lfloor \frac{95}{125} \right\rfloor = 19 + 3 + 0 = 22
]
Thus, (95!) ends in 22 zeros.
5. Number of Zeros in (1001!):
For (n = 1001), we compute the number of factors of 5 in (1001!):
[
\left\lfloor \frac{1001}{5} \right\rfloor + \left\lfloor \frac{1001}{25} \right\rfloor + \left\lfloor \frac{1001}{125} \right\rfloor + \left\lfloor \frac{1001}{625} \right\rfloor = 200 + 40 + 8 + 1 = 249
]
Thus, (1001!) ends in 249 zeros.
Conclusion:
- The highest power of 2 that divides (533!) is found by summing the floor divisions of (533) by powers of 2.
- The highest power of 3 that divides (533!) is found similarly.
- The highest power of 6 dividing (533!) is the minimum of the highest powers of 2 and 3.
- (95!) ends in 22 zeros, and (1001!) ends in 249 zeros.