Exponents - GMAT Quantitative

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Question

is a number not in the set $\left\{-1, 0, 1 \right\}$ .

Of the elements $\left \{ x^{4}, x^{5}, x^{6}\right \}$ , which is the greatest?

Statement 1: is a negative number.

Statement 2: $\left | x\right | > 1$

Tap to reveal answer

Answer

Statement 1 alone is inconclusive, as can be demonstrated by examining two negative values of other than .

Case 1: .

Then

$x^{4} = (-2)^{4} = 16$

$x^{5} = (-2)^{5} = -32$

$x^{6} = (-2)^{6} = 64$

$x^{6}$ is the greatest of these values.

Case 2: $x = -\frac{1}{2}$

Then

$x ^{4}=\left ( -\frac{1}{2} \right )^{4} = \frac{1}{16}$

$x ^{5}=\left ( -\frac{1}{2} \right )^{5} = -\frac{1}{32}$

$x ^{6}=\left ( -\frac{1}{2} \right )^{6} = \frac{1}{64}$

$x^{4}$ is the greatest of these values.

Now assume Statement 2 alone. Either or .

Case 1: .

Then $x \cdot x^{4} > 1\cdot x^{4}$ , so $x^{5} > x^{4}$ ; similarly, $x^{6} > x^{5}$ .

$x^{6}$ is the greatest of the three.

Case 2: .

Odd power $x^{5}$ is negative, and even powers $x^{4}$ and $x^{6}$ are positive, so one of the latter two is the greatest. Since , it follows that $x^{2} > 1$ . It then follows that $x^{2} \cdot x^{4} > 1 \cdot x^{4}$ , or $x^{6} > x^{4}$ .

Again, $x^{6}$ is the greatest of the three.

Statement 2 alone is sufficient, but not Statement 1.

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Question

is a number not in the set $\left\{-1, 0, 1 \right\}$ .

Of the elements $\left \{ x^{4}, x^{5}, x^{6}\right \}$ , which is the greatest?

Statement 1: is a negative number.

Statement 2: $\left | x\right | > 1$

Tap to reveal answer

Answer

Statement 1 alone is inconclusive, as can be demonstrated by examining two negative values of other than .

Case 1: .

Then

$x^{4} = (-2)^{4} = 16$

$x^{5} = (-2)^{5} = -32$

$x^{6} = (-2)^{6} = 64$

$x^{6}$ is the greatest of these values.

Case 2: $x = -\frac{1}{2}$

Then

$x ^{4}=\left ( -\frac{1}{2} \right )^{4} = \frac{1}{16}$

$x ^{5}=\left ( -\frac{1}{2} \right )^{5} = -\frac{1}{32}$

$x ^{6}=\left ( -\frac{1}{2} \right )^{6} = \frac{1}{64}$

$x^{4}$ is the greatest of these values.

Now assume Statement 2 alone. Either or .

Case 1: .

Then $x \cdot x^{4} > 1\cdot x^{4}$ , so $x^{5} > x^{4}$ ; similarly, $x^{6} > x^{5}$ .

$x^{6}$ is the greatest of the three.

Case 2: .

Odd power $x^{5}$ is negative, and even powers $x^{4}$ and $x^{6}$ are positive, so one of the latter two is the greatest. Since , it follows that $x^{2} > 1$ . It then follows that $x^{2} \cdot x^{4} > 1 \cdot x^{4}$ , or $x^{6} > x^{4}$ .

Again, $x^{6}$ is the greatest of the three.

Statement 2 alone is sufficient, but not Statement 1.

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Question

is a number not in the set $\left\{-1, 0, 1 \right\}$ .

Of the elements $\left \{ x^{4}, x^{5}, x^{6}\right \}$ , which is the greatest?

Statement 1: is a negative number.

Statement 2: $\left | x\right | > 1$

Tap to reveal answer

Answer

Statement 1 alone is inconclusive, as can be demonstrated by examining two negative values of other than .

Case 1: .

Then

$x^{4} = (-2)^{4} = 16$

$x^{5} = (-2)^{5} = -32$

$x^{6} = (-2)^{6} = 64$

$x^{6}$ is the greatest of these values.

Case 2: $x = -\frac{1}{2}$

Then

$x ^{4}=\left ( -\frac{1}{2} \right )^{4} = \frac{1}{16}$

$x ^{5}=\left ( -\frac{1}{2} \right )^{5} = -\frac{1}{32}$

$x ^{6}=\left ( -\frac{1}{2} \right )^{6} = \frac{1}{64}$

$x^{4}$ is the greatest of these values.

Now assume Statement 2 alone. Either or .

Case 1: .

Then $x \cdot x^{4} > 1\cdot x^{4}$ , so $x^{5} > x^{4}$ ; similarly, $x^{6} > x^{5}$ .

$x^{6}$ is the greatest of the three.

Case 2: .

Odd power $x^{5}$ is negative, and even powers $x^{4}$ and $x^{6}$ are positive, so one of the latter two is the greatest. Since , it follows that $x^{2} > 1$ . It then follows that $x^{2} \cdot x^{4} > 1 \cdot x^{4}$ , or $x^{6} > x^{4}$ .

Again, $x^{6}$ is the greatest of the three.

Statement 2 alone is sufficient, but not Statement 1.

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Question

is a number not in the set $\left\{-1, 0, 1 \right\}$ .

Of the elements $\left \{ x^{2}, x^{3}, x^{4}\right \}$ , which is the greatest?

Statement 1: is a negative number.

Statement 2: $\left | x\right |>\frac{1}{2}$

Tap to reveal answer

Answer

Assume both statements are known. The greatest of the three numbers must be $x^{2}$ or $x^{4}$ , since even powers of negative numbers are positive and odd powers of negative numbers are negative.

Case 1: $x = -\frac{2}{3}$

$x^{2} = \left ( -\frac{2}{3} \right )^{2} = \frac{4}{9} = \frac{36}{81}$

$x^{4} = \left ( -\frac{2}{3} \right )^{4} = \frac{16}{81}$

$x^{2} > x^{4}> x^{3}$

Case 2: ,

then

$x^{2} = \left ( -2 \right )^{2} =4$

$x^{4} = \left (-2 \right )^{4} = 16$

$x^{4} > x^{2} > x^{3}$

In both cases, is negative and $\left | x\right |>\frac{1}{2}$ , but in one case, $x^{2}$ is the greatest number, and in the other, $x^{4}$ is. The two statements together are inconclusive.

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Question

is a number not in the set $\left\{-1, 0, 1 \right\}$ .

Of the elements $\left \{ x^{4}, x^{5}, x^{6}\right \}$ , which is the greatest?

Statement 1: is a negative number.

Statement 2: $\left | x\right | > 1$

Tap to reveal answer

Answer

Statement 1 alone is inconclusive, as can be demonstrated by examining two negative values of other than .

Case 1: .

Then

$x^{4} = (-2)^{4} = 16$

$x^{5} = (-2)^{5} = -32$

$x^{6} = (-2)^{6} = 64$

$x^{6}$ is the greatest of these values.

Case 2: $x = -\frac{1}{2}$

Then

$x ^{4}=\left ( -\frac{1}{2} \right )^{4} = \frac{1}{16}$

$x ^{5}=\left ( -\frac{1}{2} \right )^{5} = -\frac{1}{32}$

$x ^{6}=\left ( -\frac{1}{2} \right )^{6} = \frac{1}{64}$

$x^{4}$ is the greatest of these values.

Now assume Statement 2 alone. Either or .

Case 1: .

Then $x \cdot x^{4} > 1\cdot x^{4}$ , so $x^{5} > x^{4}$ ; similarly, $x^{6} > x^{5}$ .

$x^{6}$ is the greatest of the three.

Case 2: .

Odd power $x^{5}$ is negative, and even powers $x^{4}$ and $x^{6}$ are positive, so one of the latter two is the greatest. Since , it follows that $x^{2} > 1$ . It then follows that $x^{2} \cdot x^{4} > 1 \cdot x^{4}$ , or $x^{6} > x^{4}$ .

Again, $x^{6}$ is the greatest of the three.

Statement 2 alone is sufficient, but not Statement 1.

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Question

is a number not in the set $\left\{-1, 0, 1 \right\}$ .

Of the elements $\left \{ x^{4}, x^{5}, x^{6}\right \}$ , which is the greatest?

Statement 1: is a negative number.

Statement 2: $\left | x\right | > 1$

Tap to reveal answer

Answer

Statement 1 alone is inconclusive, as can be demonstrated by examining two negative values of other than .

Case 1: .

Then

$x^{4} = (-2)^{4} = 16$

$x^{5} = (-2)^{5} = -32$

$x^{6} = (-2)^{6} = 64$

$x^{6}$ is the greatest of these values.

Case 2: $x = -\frac{1}{2}$

Then

$x ^{4}=\left ( -\frac{1}{2} \right )^{4} = \frac{1}{16}$

$x ^{5}=\left ( -\frac{1}{2} \right )^{5} = -\frac{1}{32}$

$x ^{6}=\left ( -\frac{1}{2} \right )^{6} = \frac{1}{64}$

$x^{4}$ is the greatest of these values.

Now assume Statement 2 alone. Either or .

Case 1: .

Then $x \cdot x^{4} > 1\cdot x^{4}$ , so $x^{5} > x^{4}$ ; similarly, $x^{6} > x^{5}$ .

$x^{6}$ is the greatest of the three.

Case 2: .

Odd power $x^{5}$ is negative, and even powers $x^{4}$ and $x^{6}$ are positive, so one of the latter two is the greatest. Since , it follows that $x^{2} > 1$ . It then follows that $x^{2} \cdot x^{4} > 1 \cdot x^{4}$ , or $x^{6} > x^{4}$ .

Again, $x^{6}$ is the greatest of the three.

Statement 2 alone is sufficient, but not Statement 1.

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Question

is a number not in the set $\left\{-1, 0, 1 \right\}$ .

Of the elements $\left \{ x^{2}, x^{3}, x^{4}\right \}$ , which is the greatest?

Statement 1: is a negative number.

Statement 2: $\left | x\right |>\frac{1}{2}$

Tap to reveal answer

Answer

Assume both statements are known. The greatest of the three numbers must be $x^{2}$ or $x^{4}$ , since even powers of negative numbers are positive and odd powers of negative numbers are negative.

Case 1: $x = -\frac{2}{3}$

$x^{2} = \left ( -\frac{2}{3} \right )^{2} = \frac{4}{9} = \frac{36}{81}$

$x^{4} = \left ( -\frac{2}{3} \right )^{4} = \frac{16}{81}$

$x^{2} > x^{4}> x^{3}$

Case 2: ,

then

$x^{2} = \left ( -2 \right )^{2} =4$

$x^{4} = \left (-2 \right )^{4} = 16$

$x^{4} > x^{2} > x^{3}$

In both cases, is negative and $\left | x\right |>\frac{1}{2}$ , but in one case, $x^{2}$ is the greatest number, and in the other, $x^{4}$ is. The two statements together are inconclusive.

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Question

is a number not in the set $\left\{-1, 0, 1 \right\}$ .

Of the elements $\left \{ x^{4}, x^{5}, x^{6}\right \}$ , which is the greatest?

Statement 1: is a negative number.

Statement 2: $\left | x\right | > 1$

Tap to reveal answer

Answer

Statement 1 alone is inconclusive, as can be demonstrated by examining two negative values of other than .

Case 1: .

Then

$x^{4} = (-2)^{4} = 16$

$x^{5} = (-2)^{5} = -32$

$x^{6} = (-2)^{6} = 64$

$x^{6}$ is the greatest of these values.

Case 2: $x = -\frac{1}{2}$

Then

$x ^{4}=\left ( -\frac{1}{2} \right )^{4} = \frac{1}{16}$

$x ^{5}=\left ( -\frac{1}{2} \right )^{5} = -\frac{1}{32}$

$x ^{6}=\left ( -\frac{1}{2} \right )^{6} = \frac{1}{64}$

$x^{4}$ is the greatest of these values.

Now assume Statement 2 alone. Either or .

Case 1: .

Then $x \cdot x^{4} > 1\cdot x^{4}$ , so $x^{5} > x^{4}$ ; similarly, $x^{6} > x^{5}$ .

$x^{6}$ is the greatest of the three.

Case 2: .

Odd power $x^{5}$ is negative, and even powers $x^{4}$ and $x^{6}$ are positive, so one of the latter two is the greatest. Since , it follows that $x^{2} > 1$ . It then follows that $x^{2} \cdot x^{4} > 1 \cdot x^{4}$ , or $x^{6} > x^{4}$ .

Again, $x^{6}$ is the greatest of the three.

Statement 2 alone is sufficient, but not Statement 1.

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Question

is a number not in the set $\left\{-1, 0, 1 \right\}$ .

Of the elements $\left \{ x^{2}, x^{3}, x^{4}\right \}$ , which is the greatest?

Statement 1: is a negative number.

Statement 2: $\left | x\right |>\frac{1}{2}$

Tap to reveal answer

Answer

Assume both statements are known. The greatest of the three numbers must be $x^{2}$ or $x^{4}$ , since even powers of negative numbers are positive and odd powers of negative numbers are negative.

Case 1: $x = -\frac{2}{3}$

$x^{2} = \left ( -\frac{2}{3} \right )^{2} = \frac{4}{9} = \frac{36}{81}$

$x^{4} = \left ( -\frac{2}{3} \right )^{4} = \frac{16}{81}$

$x^{2} > x^{4}> x^{3}$

Case 2: ,

then

$x^{2} = \left ( -2 \right )^{2} =4$

$x^{4} = \left (-2 \right )^{4} = 16$

$x^{4} > x^{2} > x^{3}$

In both cases, is negative and $\left | x\right |>\frac{1}{2}$ , but in one case, $x^{2}$ is the greatest number, and in the other, $x^{4}$ is. The two statements together are inconclusive.

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Question

is a number not in the set $\left\{-1, 0, 1 \right\}$ .

Of the elements $\left \{ x^{2}, x^{3}, x^{4}\right \}$ , which is the greatest?

Statement 1: is a negative number.

Statement 2: $\left | x\right |>\frac{1}{2}$

Tap to reveal answer

Answer

Assume both statements are known. The greatest of the three numbers must be $x^{2}$ or $x^{4}$ , since even powers of negative numbers are positive and odd powers of negative numbers are negative.

Case 1: $x = -\frac{2}{3}$

$x^{2} = \left ( -\frac{2}{3} \right )^{2} = \frac{4}{9} = \frac{36}{81}$

$x^{4} = \left ( -\frac{2}{3} \right )^{4} = \frac{16}{81}$

$x^{2} > x^{4}> x^{3}$

Case 2: ,

then

$x^{2} = \left ( -2 \right )^{2} =4$

$x^{4} = \left (-2 \right )^{4} = 16$

$x^{4} > x^{2} > x^{3}$

In both cases, is negative and $\left | x\right |>\frac{1}{2}$ , but in one case, $x^{2}$ is the greatest number, and in the other, $x^{4}$ is. The two statements together are inconclusive.

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Question

is a number not in the set $\left\{-1, 0, 1 \right\}$ .

Of the elements $\left \{ x^{4}, x^{5}, x^{6}\right \}$ , which is the greatest?

Statement 1: is a negative number.

Statement 2: $\left | x\right | > 1$

Tap to reveal answer

Answer

Statement 1 alone is inconclusive, as can be demonstrated by examining two negative values of other than .

Case 1: .

Then

$x^{4} = (-2)^{4} = 16$

$x^{5} = (-2)^{5} = -32$

$x^{6} = (-2)^{6} = 64$

$x^{6}$ is the greatest of these values.

Case 2: $x = -\frac{1}{2}$

Then

$x ^{4}=\left ( -\frac{1}{2} \right )^{4} = \frac{1}{16}$

$x ^{5}=\left ( -\frac{1}{2} \right )^{5} = -\frac{1}{32}$

$x ^{6}=\left ( -\frac{1}{2} \right )^{6} = \frac{1}{64}$

$x^{4}$ is the greatest of these values.

Now assume Statement 2 alone. Either or .

Case 1: .

Then $x \cdot x^{4} > 1\cdot x^{4}$ , so $x^{5} > x^{4}$ ; similarly, $x^{6} > x^{5}$ .

$x^{6}$ is the greatest of the three.

Case 2: .

Odd power $x^{5}$ is negative, and even powers $x^{4}$ and $x^{6}$ are positive, so one of the latter two is the greatest. Since , it follows that $x^{2} > 1$ . It then follows that $x^{2} \cdot x^{4} > 1 \cdot x^{4}$ , or $x^{6} > x^{4}$ .

Again, $x^{6}$ is the greatest of the three.

Statement 2 alone is sufficient, but not Statement 1.

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Question

is a number not in the set $\left\{-1, 0, 1 \right\}$ .

Of the elements $\left \{ x^{2}, x^{3}, x^{4}\right \}$ , which is the greatest?

Statement 1: is a negative number.

Statement 2: $\left | x\right |>\frac{1}{2}$

Tap to reveal answer

Answer

Assume both statements are known. The greatest of the three numbers must be $x^{2}$ or $x^{4}$ , since even powers of negative numbers are positive and odd powers of negative numbers are negative.

Case 1: $x = -\frac{2}{3}$

$x^{2} = \left ( -\frac{2}{3} \right )^{2} = \frac{4}{9} = \frac{36}{81}$

$x^{4} = \left ( -\frac{2}{3} \right )^{4} = \frac{16}{81}$

$x^{2} > x^{4}> x^{3}$

Case 2: ,

then

$x^{2} = \left ( -2 \right )^{2} =4$

$x^{4} = \left (-2 \right )^{4} = 16$

$x^{4} > x^{2} > x^{3}$

In both cases, is negative and $\left | x\right |>\frac{1}{2}$ , but in one case, $x^{2}$ is the greatest number, and in the other, $x^{4}$ is. The two statements together are inconclusive.

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Question

is a number not in the set $\left\{-1, 0, 1 \right\}$ .

Of the elements $\left \{ x^{2}, x^{3}, x^{4}\right \}$ , which is the greatest?

Statement 1: is a negative number.

Statement 2: $\left | x\right |>\frac{1}{2}$

Tap to reveal answer

Answer

Assume both statements are known. The greatest of the three numbers must be $x^{2}$ or $x^{4}$ , since even powers of negative numbers are positive and odd powers of negative numbers are negative.

Case 1: $x = -\frac{2}{3}$

$x^{2} = \left ( -\frac{2}{3} \right )^{2} = \frac{4}{9} = \frac{36}{81}$

$x^{4} = \left ( -\frac{2}{3} \right )^{4} = \frac{16}{81}$

$x^{2} > x^{4}> x^{3}$

Case 2: ,

then

$x^{2} = \left ( -2 \right )^{2} =4$

$x^{4} = \left (-2 \right )^{4} = 16$

$x^{4} > x^{2} > x^{3}$

In both cases, is negative and $\left | x\right |>\frac{1}{2}$ , but in one case, $x^{2}$ is the greatest number, and in the other, $x^{4}$ is. The two statements together are inconclusive.

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Question

is a number not in the set $\left\{-1, 0, 1 \right\}$ .

Of the elements $\left \{ x^{2}, x^{3}, x^{4}\right \}$ , which is the greatest?

Statement 1: is a negative number.

Statement 2: $\left | x\right |>\frac{1}{2}$

Tap to reveal answer

Answer

Assume both statements are known. The greatest of the three numbers must be $x^{2}$ or $x^{4}$ , since even powers of negative numbers are positive and odd powers of negative numbers are negative.

Case 1: $x = -\frac{2}{3}$

$x^{2} = \left ( -\frac{2}{3} \right )^{2} = \frac{4}{9} = \frac{36}{81}$

$x^{4} = \left ( -\frac{2}{3} \right )^{4} = \frac{16}{81}$

$x^{2} > x^{4}> x^{3}$

Case 2: ,

then

$x^{2} = \left ( -2 \right )^{2} =4$

$x^{4} = \left (-2 \right )^{4} = 16$

$x^{4} > x^{2} > x^{3}$

In both cases, is negative and $\left | x\right |>\frac{1}{2}$ , but in one case, $x^{2}$ is the greatest number, and in the other, $x^{4}$ is. The two statements together are inconclusive.

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Question

is a number not in the set $\left\{-1, 0, 1 \right\}$ .

Of the elements $\left \{ x^{4}, x^{5}, x^{6}\right \}$ , which is the greatest?

Statement 1: is a negative number.

Statement 2: $\left | x\right | > 1$

Tap to reveal answer

Answer

Statement 1 alone is inconclusive, as can be demonstrated by examining two negative values of other than .

Case 1: .

Then

$x^{4} = (-2)^{4} = 16$

$x^{5} = (-2)^{5} = -32$

$x^{6} = (-2)^{6} = 64$

$x^{6}$ is the greatest of these values.

Case 2: $x = -\frac{1}{2}$

Then

$x ^{4}=\left ( -\frac{1}{2} \right )^{4} = \frac{1}{16}$

$x ^{5}=\left ( -\frac{1}{2} \right )^{5} = -\frac{1}{32}$

$x ^{6}=\left ( -\frac{1}{2} \right )^{6} = \frac{1}{64}$

$x^{4}$ is the greatest of these values.

Now assume Statement 2 alone. Either or .

Case 1: .

Then $x \cdot x^{4} > 1\cdot x^{4}$ , so $x^{5} > x^{4}$ ; similarly, $x^{6} > x^{5}$ .

$x^{6}$ is the greatest of the three.

Case 2: .

Odd power $x^{5}$ is negative, and even powers $x^{4}$ and $x^{6}$ are positive, so one of the latter two is the greatest. Since , it follows that $x^{2} > 1$ . It then follows that $x^{2} \cdot x^{4} > 1 \cdot x^{4}$ , or $x^{6} > x^{4}$ .

Again, $x^{6}$ is the greatest of the three.

Statement 2 alone is sufficient, but not Statement 1.

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Question

is a number not in the set $\left\{-1, 0, 1 \right\}$ .

Of the elements $\left \{ x^{2}, x^{3}, x^{4}\right \}$ , which is the greatest?

Statement 1: is a negative number.

Statement 2: $\left | x\right |>\frac{1}{2}$

Tap to reveal answer

Answer

Assume both statements are known. The greatest of the three numbers must be $x^{2}$ or $x^{4}$ , since even powers of negative numbers are positive and odd powers of negative numbers are negative.

Case 1: $x = -\frac{2}{3}$

$x^{2} = \left ( -\frac{2}{3} \right )^{2} = \frac{4}{9} = \frac{36}{81}$

$x^{4} = \left ( -\frac{2}{3} \right )^{4} = \frac{16}{81}$

$x^{2} > x^{4}> x^{3}$

Case 2: ,

then

$x^{2} = \left ( -2 \right )^{2} =4$

$x^{4} = \left (-2 \right )^{4} = 16$

$x^{4} > x^{2} > x^{3}$

In both cases, is negative and $\left | x\right |>\frac{1}{2}$ , but in one case, $x^{2}$ is the greatest number, and in the other, $x^{4}$ is. The two statements together are inconclusive.

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Question

is an integer. Is there a real number such that $x^{A}=B$ ?

Statement 1: is negative

Statement 2: is even

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Answer

The equivalent question is "does have a real $A \textrm{th }$ root?"

If you know only that is negative, you need to know whether is even or odd; negative numbers have real odd-numbered roots, but not real even-numbered roots.

If you know only that is even, you need to know whether is negative or nonnegative; negative numbers do not have real even-numbered roots, but nonnegative numbers do.

If you know both, however, then you know that the answer is no, since as stated before, negative numbers do not have real even-numbered roots.

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

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Question

Which is the greater quantity, $3^{x}$ or $9^{y}$ - or are they equal?

Statement 1:

Statement 2:

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Answer

From Statement 1 alone,

$3^{x} = 3^{2y+2} = 3^{2y} \cdot 3^{2} =\left ( 3^{2} \right ) ^{y}\cdot 9 = 9^{y} \cdot 9>9^{y}$

Now assume Statement 2 alone. We show that this is insufficient with two cases:

Case 1:

$3^{3} = 27$ ; $9^{1} = 9$ ; therefore, $3^{x} >9^{y}$

Case 1:

$3^{-3} = \frac{1}{27}$ ; $9^{-1} = \frac{1}{9}$ ; therefore, $3^{x}<9^{y}$

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Question

Does $\log_{2} \left ( MN \right )$ exist?

Statement 1: and are both negative.

Statement 2: divided by 2 yields an integer.

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Answer

A logarithm can be taken of a number if and only if the number is positive. If Statement 1 alone is true, then , being the product of two negative numbers, must be positive, and $\log_{2} \left ( MN \right )$ exists.

Statement 2 is irrelevant; 4 and both yield integers when divided by 2, but $\log_{2} 4 = 2$ and $\log_{2}\left ( -4 \right )$ does not exist.

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Question

Johnny was assigned to write a number in scientific notation by filling the circle and the square in the pattern below with two numbers.

$\bigcirc \times 10 ^{\square }$

Johnny filled in both shapes with numbers. Did he succeed?

Statement 1: He filled in the circle with the number "10".

Statement 2: He filled in the square with a negative integer.

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Answer

The number $a \times 10 ^{n}$ is a number written in scientific notation if and only of two conditions are true:

$1 \leq \left | a \right | < 10$

is an integer

By Statement 1 Johnny filled in the circle incorrectly, since it makes .

By Statement 2, Johnny filled in the square correctly, but the statement says nothing about how he filled in the circle; Statement 2 leaves the question open.

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