To simplify the expression 167 using the factors of the base (which is 16 in this case), we need to express 167 in terms of the factors of 16, if possible.
Let's analyze the given factorizations of 16:
- \( 16 = 4 \cdot 2 \cdot 2 \)
- \( 16 = 8 \cdot 8 \)
- \( 16 = 2 \cdot 2 \cdot 2 \cdot 2 \)
- \( 16 = 4 \cdot 4 \)
However, since 167 does not have any apparent relationship with 16 (beyond the fact that they both are integers), we consider the factors of 16 to see how they can help.
But simplifying 167 using the factors of 16 is a bit misleading since 167 is not a multiple of 16, and there's no immediate simplification like dividing 167 by any of its factors.
If we are talking about modular simplifications, \( 16 \) itself is mostly irrelevant, and you can't directly reduce 167 using basic arithmetic relationships with the factors of 16.
None of the options provided will "simplify" the expression meaningfully, as 167 remains unchanged regardless of the factorization of 16. Therefore, out of the options provided, none are better suited to simplifying or reducing the expression 167.
If you need to ultimately reduce 167 modulo 16, we could compute \( 167 \mod 16 \) to find a remainder, but again, that's a different context altogether.
If you meant to ask which factorization might help express \( 167 \) in some fractional or divisibility sense, you might lean towards using the basic prime factorization of 16, which starts with \( 2 \), because both \( 16\) and \( 2\) are powers of \( 2\). This would be true if you were considering some relation to powers of \( 2 \) rather than simply "simplifying" 167 directly.
In conclusion, there's no factorization of 16 that directly simplifies 167; hence, this question might require a different interpretation or context.
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