Powers & Roots of Numbers - GMAT Quantitative Reasoning

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Question

and are integers. Is $A ^{B}$ positive, negative, or zero?

Statement 1: is negative.

Statement 2: is odd.

Answer

A negative integer to an even power is positive:

Example: $\left ( -5\right )^{4} =625$

A negative integer to an odd power is negative:

Example: $\left ( -5\right )^{3} =-125$

A positive integer to an odd power is positive:

Example: $\left 5^{3} =125$

So, as seen in the first two statements, knowing only that base is negative is insufficient to detemine the sign of $A^{B}$ ; as seen in the last two statements, knowing only that exponent is odd is also insufficient. But by the middle statement, knowing both facts tells us $A^{B}$ is negative.

The answer is that both statements together are sufficient to answer the question, but neither statement alone is sufficient.

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Question

Let $A\ \textup{and }B$ be positive integers. Is $\sqrt{A^{3} B}$ an integer?

Statement 1: is a perfect square.

Statement 2: is an even integer.

Answer

We examine two examples of situations in which both statements hold.

Example 1:

Then $\sqrt{A^{3} B} =\sqrt{2^{3} \cdot 4} = \sqrt{32}$

32 is not a perfect square, so $\sqrt{A^{3}B}$ is not an integer.

Example 2:

Then $\sqrt{A^{3} B} =\sqrt{4^{3} \cdot 4} = \sqrt{256} = 16$ , making $\sqrt{A^{3}B}$ an integer.

In both cases, both statements hold, but in only one, $\sqrt{A^{3}B}$ is an integer. This makes the two statements together insufficient.

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Question

Imagine an integer such that the units digit of is greater than 5. What is the units digit of ?

(1) The units digit of is the same as the units digit of $y^{2}$ .

(2) The units digit of is the same as the units digit of $y^{3}$ .

Answer

(1) The only single-digit integer greater than 5 whose unit digit of its square term is equal to itself is 6. This statement is sufficient.

(2) There are two single-digit integers where the unit digit of the cubed term is equal to the integer itself: 6 and 9. This statement is insufficient.

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Question

is a real number. Is positive, negative, or zero?

Statement 1:

Statement 2: $\frac{1}{2}x^{3} -1> 0$

Answer

If , then , and $x > \frac{5}{2} > 0$ , so must be positive.

If $\frac{1}{2}x^{3} -1> 0$ , then $\frac{1}{2}x^{3} > 1$ , $x^{3} > 2$ . and $x> \sqrt[3]{2}$ , so again, must be positive. Either statement is enough to answer the question in the affirmative.

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Question

Simplify this expression as much as possible:

$\sqrt[3]{16} + \sqrt[3]{128} - \sqrt[3]{54}$

Answer

$\sqrt[3]{16} + \sqrt[3]{128} - \sqrt[3]{54}$

$= \sqrt[3]{8} \cdot \sqrt[3]{2} + \sqrt[3]{64} \cdot \sqrt[3]{2} - \sqrt[3]{27} \cdot \sqrt[3]{2}$

$= 2 \sqrt[3]{2} +4 \sqrt[3]{2} -3 \sqrt[3]{2}$

$=\left ( 2+4-3 \right ) \sqrt[3]{2} = 3 \sqrt[3]{2}$

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Question

What is the value of twelve raised to the fourth power?

Answer

"Twelve raised to the fourth power" is 124. If you can translate the words into their mathematical counterpart, you're done, because the actual calculation should be done by your calculator. It will tell you that $12\cdot 12\cdot 12\cdot 12=20,736$ . There is not enough time on the test for you to try to do this by hand.

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Question

Calculate the fifth root of :

(1) The square root of is .

(2) The tenth root of is .

Answer

Using Statement (1):

$m^{\frac{1}{5}}= (m^{\frac{1}{2}})^{\frac{2}{5}}=(3125)^{\frac{2}{5}}=25$

Statement (1) ALONE is SUFFICIENT.

Using Statement (2):

$m^{\frac{1}{5}}= (m^{\frac{1}{10}})^{2}=(5)^{2}=25$

Statement (2) ALONE is SUFFICIENT.

Therefore EACH Statement ALONE is sufficient.

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Question

is a positive real number. True or false: $\sqrt{N}$ is a rational number.

Statement 1: is irrational.

Statement 2: $\sqrt[4]{N}$ is irrational.

Answer

If $\sqrt{N}$ is rational, then, since the product of two rational numbers is rational, $\sqrt{N} \cdot \sqrt{N} = N$ is rational. If Statement 1 alone is assumed, then, since is irrational, $\sqrt{N}$ must be irrational.

Assume Statement 2 alone, and note that

$\left (\sqrt[4]{N} \right )^{2}= \left (N ^{\frac{1}{4}} \right )^{2}= N ^{\frac{1}{4} \cdot 2} = N ^{\frac{1}{2} } = \sqrt{N}$

In other words, $\sqrt[4]{N}$ is the square root of $\sqrt{N}$ . Since both rational and irrational numbers have irrational square roots, $\sqrt[4]{N}$ being irrational does not prove or disprove that $\sqrt{N}$ is rational.

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Question

is a positive real number. True or false: $\sqrt{N}$ is a rational number.

Statement 1: $\sqrt[3]{N}$ is a rational number.

Statement 2: $\sqrt[4]{N}$ is a rational number.

Answer

Statement 1 alone provides insufficient information. is a number with a rational cube root, , and a rational square root, . is a number with a rational cube root, , but an irrational square root.

Now assume Statement 2 alone.

$\left (\sqrt[4]{N} \right )^{2}= \left (N ^{\frac{1}{4}} \right )^{2}= N ^{\frac{1}{4} \cdot 2} = N ^{\frac{1}{2} } = \sqrt{N}$

In other words, $\sqrt{N}$ is the square of $\sqrt[4]{N}$ . The rational numbers are closed under multiplication, so if $\sqrt[4]{N}$ is rational, $\left (\sqrt[4]{N} \right )^{2}= \sqrt{N}$ is rational.

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Question

is a positive real number. True or false: is a rational number.

Statement 1: $N ^{3}$ is a rational number.

Statement 2: $N ^{4}$ is a rational number.

Answer

Statement 1 alone is not enough to determine whether is rational or not; $\sqrt[3]{2}$ and both have rational cubes, but only is rational. By a similar argument, Statement 2 alone is insufficient.

Assume both statements are true. $N = \frac{N^{4}}{N^{3}}$ , the quotient of two rational numbers, which must itself be rational.

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Question

is a positive real number. True or false: is a rational number.

Statement 1: $N ^{3}$ is an irrational number.

Statement 2: $N ^{4}$ is an irrational number.

Answer

An integer power of a rational number, being a product of rational numbers, must itself be rational. Either statement alone asserts that such a power is irrational, so conversely, either statement alone proves irrational.

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Question

. True or false: is rational.

Statement 1: $\left (N + 1 \right )^{2}$ is rational.

Statement 2: $\left (N - 1 \right )^{2}$ is rational.

Answer

Statement 1 alone is not enough to prove is or is not rational. Examples:

If , then $\left (N+ 1 \right )^{2} = \left (2+ 1 \right )^{2} = 3^{2} = 9$

If $N = \sqrt{2} - 1$ , then $\left (N+ 1 \right )^{2} = \left ( \sqrt{2}-1 + 1 \right )^{2} =( \sqrt{2})^{2} = 2$

In both cases, $\left (N + 1 \right )^{2}$ is rational, but in one case, is rational and in the other, is irrational.

A similar argument demonstrates Statement 2 to be insufficient.

Assume both statements are true. $\left (N + 1 \right )^{2}$ and $\left (N - 1 \right )^{2}$ are rational, so their difference is as well:

$\left (N + 1 \right )^{2} - \left (N - 1 \right )^{2}$

$=\left ( N ^{2}+ 2N + 1 \right ) - \left ( N ^{2}- 2N + 1 \right )$

$= N ^{2}- N ^{2}+ 2N +2N + 1 -1$

is rational, so by closure under division, $4N \div 4 = N$ is rational.

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Question

. True or false: is rational.

Statement 1: $\sqrt{N+1}$ is irrational.

Statement 2: $\sqrt{N-1}$ is rational.

Answer

Statement 1 alone is not enough to prove rational or irrational. Examples:

If , then $\sqrt{N+1} = \sqrt{4+1} = \sqrt{5}$

If $N = \pi -1$ , then $\sqrt{N+1} = \sqrt{(\pi -1)+1} = \sqrt{\pi}$

In both cases, $\sqrt{N+1}$ is irrational, but in only one case, is rational.

Assume Statement 2 alone. $\sqrt{N-1}$ is rational, so, by closure of the rational numbers under multiplication,

$\sqrt{N-1} \cdot \sqrt{N-1} = N - 1$ is rational. The rationals are closed under addition, so the sum

is rational.

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Question

Simplify: $\frac{5}{\sqrt{4a}}$

Answer

When we are faced with a radical in the denominator of a fraction, the first step is to multiply the top and bottom of the fraction by the numerator:

$\frac{5}{\sqrt{4a}} \cdot \frac{\sqrt{4a}}{\sqrt{4a}} = \frac{5\sqrt{4a}}{4a}$

We can then reduce the fraction to:

$\frac{5\sqrt{4a}}{4a} = \frac{5\cdot 2\sqrt{a}}{4a} = \frac{5\sqrt{a}}{2a}$

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