Understanding Trailing Zeros in Factorials

The number of trailing zeros in a factorial \(n!\) is determined by the number of times 10 is a factor in its prime factorization. Since \(10 = 2 \times 5\), we need to count the factors of 2 and 5 in \(n!\). In any factorial \(n!\), the number of factors of 2 is always greater than or equal to the number of factors of 5. Therefore, the number of trailing zeros is limited by the number of factors of 5.

The number of factors of a prime number \(p\) in \(n!\) is given by Legendre's formula:

Number of factors of \(p\) in \(n!\) = \(\sum_{k=1}^{\infty} \lfloor \frac{n}{p^k} \rfloor = \lfloor \frac{n}{p} \rfloor + \lfloor \frac{n}{p^2} \rfloor + \lfloor \frac{n}{p^3} \rfloor + \dots\)

To find the number of trailing zeros in \(n!\), we apply Legendre's formula for \(p=5\):

Number of trailing zeros in \(n!\) = \(\lfloor \frac{n}{5} \rfloor + \lfloor \frac{n}{25} \rfloor + \lfloor \frac{n}{125} \rfloor + \dots\)

We are given that \(n!\) has 17 zeros. We need to find an integer value of \(n\) such that the sum of the floor values in Legendre's formula for \(p=5\) equals 17. Let's examine the number of zeros for values of \(n\) around the expected range.

Calculating Zeros for Potential n Values

Let's calculate the number of zeros for values of \(n\) that might yield around 17 zeros, as well as checking the options provided.

• For \(n = 70\): Number of zeros = \(\lfloor \frac{70}{5} \rfloor + \lfloor \frac{70}{25} \rfloor + \lfloor \frac{70}{125} \rfloor = 14 + 2 + 0 = 16\).
• For \(n = 71\): Number of zeros = \(\lfloor \frac{71}{5} \rfloor + \lfloor \frac{71}{25} \rfloor + \lfloor \frac{71}{125} \rfloor = 14 + 2 + 0 = 16\).
• For \(n = 72\): Number of zeros = \(\lfloor \frac{72}{5} \rfloor + \lfloor \frac{72}{25} \rfloor + \lfloor \frac{72}{125} \rfloor = 14 + 2 + 0 = 16\).
• For \(n = 73\): Number of zeros = \(\lfloor \frac{73}{5} \rfloor + \lfloor \frac{73}{25} \rfloor + \lfloor \frac{73}{125} \rfloor = 14 + 2 + 0 = 16\).
• For \(n = 74\): Number of zeros = \(\lfloor \frac{74}{5} \rfloor + \lfloor \frac{74}{25} \rfloor + \lfloor \frac{74}{125} \rfloor = 14 + 2 + 0 = 16\).

For all integers \(n\) from 70 to 74, the number of trailing zeros in \(n!\) is 16.

• For \(n = 75\): Number of zeros = \(\lfloor \frac{75}{5} \rfloor + \lfloor \frac{75}{25} \rfloor + \lfloor \frac{75}{125} \rfloor = 15 + 3 + 0 = 18\).

When \(n\) becomes 75, the number of trailing zeros increases from 16 to 18. This jump occurs because 75 is a multiple of 5 (adding one factor of 5 from \(\lfloor 75/5 \rfloor\)) and a multiple of 25 (adding one factor of 5 from \(\lfloor 75/25 \rfloor\) compared to multiples of 5 less than 25). Specifically, 75 contributes \(\lfloor 75/5 \rfloor - \lfloor 74/5 \rfloor = 15-14 = 1\) from the \(\lfloor n/5 \rfloor\) term, and \(\lfloor 75/25 \rfloor - \lfloor 74/25 \rfloor = 3-2 = 1\) from the \(\lfloor n/25 \rfloor\) term. The total increase is \(1+1=2\).

Let's check the provided options:

• Option 1: \(n = 95\). Number of zeros = \(\lfloor \frac{95}{5} \rfloor + \lfloor \frac{95}{25} \rfloor + \lfloor \frac{95}{125} \rfloor = 19 + 3 + 0 = 22\).
• Option 2: \(n = 85\). Number of zeros = \(\lfloor \frac{85}{5} \rfloor + \lfloor \frac{85}{25} \rfloor + \lfloor \frac{85}{125} \rfloor = 17 + 3 + 0 = 20\).
• Option 3: \(n = 80\). Number of zeros = \(\lfloor \frac{80}{5} \rfloor + \lfloor \frac{80}{25} \rfloor + \lfloor \frac{80}{125} \rfloor = 16 + 3 + 0 = 19\).

Analysis of Number of Zeros Progression

We observed that the number of zeros for \(n \in [70, 74]\) is 16 and for \(n \in [75, 79]\) is 18. The number of trailing zeros in \(n!\) is a non-decreasing function of \(n\) and only increases when \(n\) is a multiple of 5. The number of zeros jumps from 16 to 18 between \(n=74\) and \(n=75\). This means there is no integer \(n\) for which \(n!\) has exactly 17 trailing zeros.

Conclusion on Finding n for 17 Zeros

Based on the calculations using Legendre's formula, we see that the number of trailing zeros does not take on every integer value. There is a jump from 16 zeros (for \(n\) from 70 to 74) to 18 zeros (for \(n\) from 75 to 79). Since 17 is between 16 and 18, there is no integer value of \(n\) such that \(n!\) has exactly 17 trailing zeros.

Therefore, the correct answer is that no such value of \(n\) exists.

The table confirms that the number of zeros jumps from 16 to 18, skipping 17.

Revision Table: Key Concepts

Additional Information: Properties of Factorials and Zeros

Factorials grow very rapidly. The number of trailing zeros in a factorial \(n!\) represents the power of 10 that divides \(n!\). Understanding how the number of prime factors, especially 5, accumulates in the product \(1 \times 2 \times \dots \times n\) is key to this problem. Multiples of 5 contribute one factor of 5 (\(5, 10, 15, \dots\)), multiples of 25 contribute an additional factor of 5 (\(25, 50, 75, \dots\)), multiples of 125 contribute yet another factor of 5 (\(125, \dots\)), and so on. Legendre's formula systematically counts these factors.

The number of zeros increases only when \(n\) reaches a new multiple of 5. When \(n\) reaches a multiple of 25, 125, etc., the number of zeros increases by more than one at that specific step, causing some integer values for the number of zeros to be skipped. This is exactly what happens when the number of zeros transitions from 16 to 18 between \(n=74\) and \(n=75\), skipping the value 17.