Question

Rewrite the expression as a product of identical factors to find the product for 10032
. Submit your answer in simplified form.

Answer 1

To rewrite 10032 as a product of identical factors, we first need to factor it.

1. **Start by dividing by 2** (since 10032 is even):
\[
10032 \div 2 = 5016
\]
2. **Divide by 2 again**:
\[
5016 \div 2 = 2508
\]
3. **Divide by 2 again**:
\[
2508 \div 2 = 1254
\]
4. **Divide by 2 again**:
\[
1254 \div 2 = 627
\]
5. **627 is not even, so try dividing by 3**:
\[
627 \div 3 = 209
\]
6. **209 is not divisible by 3, but let's check if it's a prime**. Checking 209:
- Not divisible by 2 (it's odd).
- Not divisible by 3 (sum of digits is 11).
- Not divisible by 5 (doesn't end in 0 or 5).
- Dividing 209 by 7 gives approximately 29.857 (not an integer).
- Dividing 209 by 11 gives exactly 19.

Thus, we can factor 10032 as:
\[
10032 = 2^4 \times 3 \times 11 \times 19
\]

This expression indicates that we cannot fully express 10032 as a product of identical factors such as \( n^k \) where \( n \) is an integer (except for partial components).

However, we can express it using powers of its prime factors. The factorization indicates that it has no duplicate groups other than 2 raised to a power.

In conclusion, the prime factorization of 10032 is:
\[
10032 = 2^4 \times 3^1 \times 11^1 \times 19^1
\]

There isn't a single group of identical factors as you might have been looking for with just multiples (like \( n^2, n^3, \) etc.), but the closest simplified product of its prime factors is:
\[
10032 = 2^4 \cdot 3^1 \cdot 11^1 \cdot 19^1
\]