Here is an oft repeated complaint we hear from test takers - Data Sufficiency questions on Number Properties are very difficult to handle (The problem solving questions are still manageable.) They feel that such questions are time consuming and often involve too many cases.

Here is our advice to them - when solving number properties questions, think of the 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 only occur at regular intervals and there are infinite numbers in between them. The integers are, in turn, even and odd. Also, that 0, 1 and -1 are special numbers. It is always a good idea to consider cases with them because results with them are often special. Let’s see how thinking on these lines can help us.

*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*

Solution: The answer to this problem does not lie in drawing the number line. The point is that we need to think on 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: a and b are non zero integers.

So they occur at regular intervals on the number line.

Question: Is a^b an integer?

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 be not be an even integer. If the exponent is an even integer, the negative sign will cancel off. Since a and b are integers, if a is not an even integer, it is 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 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) = 1/a^2

a could be 1 in which case 1/a^2 = 1 (an integer)

a could be 3 in which case 1/3^2 = 1/9 (not an integer)

So this statement alone is not sufficient.

Statement 2: a^b is negative

Again, the logic remains the same - for a number to be negative, its base must be negative and the exponent should not be an even integer. If the exponent is an even integer, the negative sign will cancel off. Since a and b are integers, if b is not an even integer, it is 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 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

b could be 1/3/5… or -1/-3/-5

If a = -1, b = 1, then a^b = -1 (an integer)

If a = -2, b = -3, then a^b = (-2)^(-3) = 1/(-2)^3 = -1/8 (not an integer)

So this statement alone is not sufficient.

We hope you see how we are using values of 1 and -1 to enumerate our cases.

Now, let’s consider using both statements.

a is a negative odd integer so it can take values such as -1, -3, -5, -7, …

b is a negative odd integer too so it can take values such as -1, -3, -5, -7, …

If a = -1, b = -1, a^b = -1 (an integer)

If a = -3, b = -3, a^b = (-3)^(-3) = -1/27 (not an integer)

So even using both statements together, we do not know whether a^b is an integer.

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