To express \( 10032 \) as a product of identical factors, we first need to factor \( 10032 \).
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Divisibility by 2: Since \( 10032 \) is an even number, we can divide it by \( 2 \): \[ 10032 \div 2 = 5016 \]
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Repeat dividing by 2: \[ 5016 \div 2 = 2508 \] \[ 2508 \div 2 = 1254 \] \[ 1254 \div 2 = 627 \]
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Next, we see \( 627 \) is not even, so we check next for \( 3 \): \[ 627 \div 3 = 209 \]
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Checking \( 209 \) for factors: It can be factored further. It isn't divisible by \( 3 \) or \( 5 \), but: \[ 209 \div 11 = 19 \]
Now, we have factored \( 10032 \) into prime factors: \[ 10032 = 2^4 \times 3 \times 11 \times 19 \]
Next, we need to express \( 10032 \) as a product of identical factors. First, we combine the terms:
To find identical factors, we can group these factors. We note:
- \( 2^4 = 16 \), so we could express \( 10032 \) as \((16 \times 3 \times 11 \times 19)\).
However, for a simpler uniform expression, we need to express the number wholly in terms of squared bases or a similar structure, which may not directly yield identical factors in a clean way.
To find any potential square roots:
- Finding the square root of \( 10032 \): \[ \sqrt{10032} \approx 100.16 \] Which does not yield whole identical factors clearly.
To simplify, we can check if any specific powers hold true. Given it's \( 16 \) would cube all factors into a uniform base, but retaining non-square values complicates.
Ultimately, we focus to express: \[ 10032 = (2^2 \times 3 \times 11 \times 19)^2 \text{ if taken with square base, otherwise, would not yield identical factors cleanly}. \]
So
Thus, the complete factorization summary of \( 10032 \): \[ \text{Product: } 10032 = 2^4 \cdot 3^1 \cdot 11^1 \cdot 19^1 \text{ resultantly has non-identical bases in multiples, i.e. } \]
Conclusively, \( 10032 = 10032^1 \) does indeed show potential restructuring but doesn't render identical factors simply.
Ultimately the product remains \( 10032 \) essentially unless refined under context whether other results hold under symbolic notation! For the clarity deliverance \( 2^4 \) can yield to potential multiples as identified forms under factorization to express multiple forms but with fractions through products.
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