Official Solution:The expression \(\frac{6^5 - 6^3}{(7^4 + 7^6)^{-1}}\) is NOT divisible by which of the following?A. 10
B. 16
C. 27
D. 99
E. 125
In order to find all possible factors of the expression above, we need to break the expression down into its prime factors.
First, though, we should deal with the negative exponent in the denominator. Negative exponents indicate the reciprocal of the positive version of the exponent. For example, \(x^{-2} = \frac{1}{x^2}\).
The expression \((7^4 + 7^6)^{-1}\) is thus the same as \(\frac{1}{7^4 + 7^6}\). So our original expression can be rewritten as \(\frac{6^5 - 6^3}{\frac{1}{7^4 + 7^6}}\).
Dividing by a fraction is the same as multiplying by its reciprocal. Thus,
\(\frac{6^5 - 6^3}{\frac{1}{7^4 + 7^6}} = (6^5 - 6^3)*(7^4 + 7^6)\)
The expression \(6^5 - 6^3\) can be factored as follows:
\(6^5 - 6^3 =\)
\(6^3(6^2 - 1) =\)
\(6^3(35) =\)
\(6^3(5)(7) =\)
\((2^3)(3^3)(5)(7)\)
The expression \(7^4 + 7^6\) can be factored as follows:
\(7^4 + 7^6 =\)
\(7^4(1 + 7^2) =\)
\(7^4(50) =\)
\(7^4(2)(25) =\)
\((2)(5^2)(7^4)\)
Thus it must be true that
\((6^5 - 6^3)*(7^4 + 7^6) =\)
\((2^3)(3^3)(5)(7)(2)(5^2)(7^4) =\)
\((2^4)(3^3)(5^3)(7^5)\)
The question asks which of the choices is not a factor of this expression.
\(10 = 2 \times 5\). The original expression contains both 2 and 5 as factors. Therefore 10 must be one of its factors. Eliminate A.
\(16 = 2 \times 2 \times 2 \times 2 = 2^4\). The original expression contains \(2^4\). Therefore 16 must be one of its factors. Eliminate B.
\(27 = 3 \times 3 \times 3 = 3^3\). The original expression contains \(3^3\). Therefore 27 must be one of its factors. Eliminate C.
\(99 = 3 \times 3 \times 11 = 3^2 \times 11\). The original expression contains \(3^2\) but not 11. Therefore 99 CANNOT be one of its factors.
CORRECT.
\(125 = 5 \times 5 \times 5 = 5^3\). The original expression contains \(5^3\). Therefore 125 must be one of its factors.
Answer: D.
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