Use Number Line for Number Properties | Data Sufficiency

Here is an often-repeated complaint we hear from test takers – Data Sufficiency questions that deal with number properties are very difficult to handle (even for people who find problem-solving number properties questions manageable)! They feel that such questions are time consuming and often involve too many cases.

Here is our advice – when solving number properties questions, imagine a number line. It reminds us that numbers behave differently “between 0 and 1”, “between -1 and 0”, “less than -1”, and “more than 1”, and that integers occur only at regular intervals and that there are infinite numbers in between them. The integers are, in turn, even and odd. Also, 0, 1 and -1 are special numbers, hence it is always a good idea to consider cases with them.

Let’s see how thinking along these lines can help us on a practice Data Sufficiency question:

If ${a}$ and ${b}$ are non-zero integers, is ${a^b}$ an integer?

Statement 1: ${b^a}$ is negative

Statement 2: ${a^b}$ is negative

The answer to this problem does not lie in actually drawing a number line. The point is that we need to think along these lines: -1, 0, 1, ranges between them, integers, negatives-positives, even-odd, decimals and how each of these comes into play in this case.

What we know from the question stem is that $a$ and $b$ are non-zero integers, which means they occur at regular intervals on the number line. To answer the question, “Is $a^b$ an integer?”, let’s first look at Statement 1:

Statement 1: $b^a$ is negative

For a number to be negative, its base must be negative. But that is not enough – the exponent should not be an even integer. If the exponent is an even integer, the negative signs will cancel out. Since $a$ and $b$ are integers, if $a$ is not an even integer, it must be an odd integer.

We know that the sign of the exponent is immaterial as far as the sign of the result is concerned (since $a^{-n}$ is just $\frac{1}{a^n}$ ). For $b^a$ to be negative, then we know that $b$ must be a negative integer and $a$ must be an odd integer. Does this help us in deducing whether $a^b$ is an integer? Not necessarily!

If $b$ is negative, say -2, $a^{-2} = \frac{1}{a^2}$ . $a$ could be 1, in which case $\frac{1}{a^2} = 1$ (an integer), or $a$ could be 3, in which case $\frac{1}{a^2} = \frac{1}{9}$ (not an integer). Because there are two possible answers, this statement alone is not sufficient.

Let’s look at Statement 2:

Statement 2: $a^b$ is negative

Again, the logic remains the same – for a number to be negative, its base must also be negative and the exponent should not be an even integer. If the exponent is an even integer, the negative signs will cancel out. Since $a$ and $b$ are integers, if $b$ is not an even integer, it must be an odd integer. Again, we know that the sign of the exponent is immaterial as far as the sign of the result is concerned (since $a^{-n}$ is just $\frac{1}{a^n}$ ).

For $a^b$ to be negative, then we know that $a$ must be a negative integer and $b$ must be an odd integer. $a$ could be -1/-2/-3/-4… etc, and $b$ could be 1/3/5… or -1/-3/-5.

If $a = -1$ and $b = 1$ , then $a^b = -1$ (an integer). If $a = -2$ and $b = -3$ , then $a^b = -2^{-3} = \frac{1}{-2^3} = \frac{-1}{8}$ (not an integer). This statement alone is also not sufficient.