Answer
a. \(6! = 2^4 \cdot 3^2 \cdot 5\).
b. \(20! = 2^{18} \cdot 3^8 \cdot 5^4 \cdot 7^2 \cdot 11 \cdot 13 \cdot 17 \cdot 19\).
c. \((20!)^2\) ends in **8 zeros**.
Work Step by Step
Recall that trailing zeros in a number come from factors of 10, and each factor of 10 is made by a pair of a 2 and a 5. In a factorial, there are usually more factors of 2 than 5, so the number of 5’s determines the number of trailing zeros. When we square the factorial, the exponents in its prime factorization double.
---
**a. Write 6! in Standard Factored Form**
First, compute 6!:
\[
6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720.
\]
Now, factor 720 into primes:
- \(6 = 2 \times 3\)
- \(5 = 5\)
- \(4 = 2^2\)
- \(3 = 3\)
- \(2 = 2\)
Collecting factors:
- Count of 2’s: from 6 (one 2), from 4 (two 2’s), from 2 (one 2) → \(1+2+1 = 4\).
- Count of 3’s: from 6 (one 3), from 3 (one 3) → \(1+1 = 2\).
- Count of 5’s: one from 5.
Thus, in standard factored form:
\[
6! = 2^4 \cdot 3^2 \cdot 5.
\]
---
**b. Write 20! in Standard Factored Form**
To write \(20!\) in standard factored form, we use the formula for counting the exponent of a prime \(p\) in \(n!\):
\[
\text{Exponent of } p = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \cdots
\]
For \(n = 20\):
- **For \(p = 2\):**
\[
\lfloor 20/2 \rfloor + \lfloor 20/4 \rfloor + \lfloor 20/8 \rfloor + \lfloor 20/16 \rfloor = 10 + 5 + 2 + 1 = 18.
\]
- **For \(p = 3\):**
\[
\lfloor 20/3 \rfloor + \lfloor 20/9 \rfloor = 6 + 2 = 8.
\]
- **For \(p = 5\):**
\[
\lfloor 20/5 \rfloor = 4.
\]
- **For \(p = 7\):**
\[
\lfloor 20/7 \rfloor = 2.
\]
- **For \(p = 11\), \(13\), \(17\), and \(19\):**
\[
\lfloor 20/11 \rfloor = 1,\quad \lfloor 20/13 \rfloor = 1,\quad \lfloor 20/17 \rfloor = 1,\quad \lfloor 20/19 \rfloor = 1.
\]
Thus, the prime factorization is:
\[
20! = 2^{18} \cdot 3^8 \cdot 5^4 \cdot 7^2 \cdot 11 \cdot 13 \cdot 17 \cdot 19.
\]
---
**c. Trailing Zeros in \((20!)^2\)**
The number of trailing zeros is determined by the number of factors of 10, which equals the minimum of the exponents of 2 and 5 in the prime factorization.
From part (b):
- In \(20!\), the exponent of 2 is 18 and of 5 is 4.
- Therefore, \(20!\) has \(\min(18, 4) = 4\) factors of 10, so it ends in 4 zeros.
When squaring \(20!\), the exponents double:
- Exponent of 2 becomes \(18 \times 2 = 36\).
- Exponent of 5 becomes \(4 \times 2 = 8\).
Thus, \((20!)^2\) has \(\min(36, 8) = 8\) factors of 10, meaning it ends in **8 trailing zeros**.
**Justification:**
Since the trailing zeros come solely from pairs of 2’s and 5’s and 2’s are in excess, the limiting factor is the number of 5’s. Doubling the exponents in \(20!\) gives 8 factors of 5, so there are 8 trailing zeros in \((20!)^2\).
