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# The expression (6^5 - 6^3)/(7^4 + 7^6)^(-1) is NOT divisible by which

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Math Expert
Joined: 02 Sep 2009
Posts: 58381
The expression (6^5 - 6^3)/(7^4 + 7^6)^(-1) is NOT divisible by which  [#permalink]

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18 Nov 2014, 08:50
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Difficulty:

25% (medium)

Question Stats:

79% (02:05) correct 21% (02:22) wrong based on 252 sessions

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Tough and Tricky questions: Exponents.

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

Kudos for a correct solution.

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Joined: 21 Jul 2014
Posts: 119
Re: The expression (6^5 - 6^3)/(7^4 + 7^6)^(-1) is NOT divisible by which  [#permalink]

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18 Nov 2014, 09:44
2
2
Bunuel wrote:

Tough and Tricky questions: Exponents.

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

Kudos for a correct solution.

Here's how I attempted this problem, although there may be a faster way.

First, I converted $$\frac{6^5 - 6^3}{(7^4 + 7^6)^{-1}}$$ to $$\(6^5 - 6^3)(7^4 + 7^6)$$
Then, I factored to simplify the expression to $$\(6^3)(6^2-1)(7^4)(1+72)$$

This reduced to [m]$$6^3)(35)(7^4)(50) which I reduced to 6*36*35*49*49*50. Looking at the answer choices, 125 = 5^3, 27 = 3^3, 16 = 4^2 = 2^3, and 10 = 2*5. All of these factors can be found in the reduced expression. 99 cannot, so that should be the correct choice. Solution: D Senior Manager Joined: 17 Mar 2014 Posts: 429 Re: The expression (6^5 - 6^3)/(7^4 + 7^6)^(-1) is NOT divisible by which [#permalink] ### Show Tags 18 Nov 2014, 09:47 1 Simplify the expression. \(\frac{{6^5 - 6^3}}{{(7^4 + 7^6)^{-1}}}$$

$$\frac{{6^3*(6^2-1)}}{{1/(7^4+7^6)}}$$

$${6^3*(6-1)*(6+1}*(7^4 + 7^6)$$

$$6^3*(5)*(7)*7^4(1+7^2)$$

$$6^3*5*7^5*(50)$$

Prime factorize it

$$2^4*3^3*7^5*5^3$$

expression is divisible by 10, 16, 27 & 125.

expression is not divisible by 99 because it has not have multiple of 11

D
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Joined: 21 Jan 2014
Posts: 61
WE: General Management (Non-Profit and Government)
Re: The expression (6^5 - 6^3)/(7^4 + 7^6)^(-1) is NOT divisible by which  [#permalink]

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18 Nov 2014, 23:18
1
Given expression (6^5-6^3)/(7^4+7^6)^(-1)

This expression will further reduced to (6^3)(7^4)(6^2-1)(1+7^2)=(6^3)(7^4)35*50

This expression does not posses 11 or multiple of 11.
Hence,this shows that it will not be divisible by 99 which is a multiple of 11.

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Location: India
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Re: The expression (6^5 - 6^3)/(7^4 + 7^6)^(-1) is NOT divisible by which  [#permalink]

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18 Nov 2014, 23:44
1

$$\frac{(6^5 - 6^3)}{(7^4 + 7^6)^{-1}} = 6^3(6^2 - 1) * 7^4 (1+7^2) = 6^3 * 35 * 7^4 * 50 = 2^3 * 3^3 * 5 * 7 * 7^4 * 5^2 * 2$$

11 is not the factor here.

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Math Expert
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Posts: 58381
Re: The expression (6^5 - 6^3)/(7^4 + 7^6)^(-1) is NOT divisible by which  [#permalink]

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19 Nov 2014, 07:40
1
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.

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Math Expert
Joined: 02 Sep 2009
Posts: 58381
Re: The expression (6^5 - 6^3)/(7^4 + 7^6)^(-1) is NOT divisible by which  [#permalink]

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19 Nov 2014, 07:42
Bunuel wrote:
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.

Similar question to practice: baker-s-dozen-128782-20.html#p1057503
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Re: The expression (6^5 - 6^3)/(7^4 + 7^6)^(-1) is NOT divisible by which  [#permalink]

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01 Nov 2018, 23:39
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Re: The expression (6^5 - 6^3)/(7^4 + 7^6)^(-1) is NOT divisible by which   [#permalink] 01 Nov 2018, 23:39
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