GMAT number properties strategies

GMAT number properties test whether a statement about even, odd, prime, divisibility, or remainders holds for every number or just for some. These strategies cover testing with concrete numbers, the trap of overgeneralizing from a few examples, and a worked example you can follow step by step.

The core principle

Test with concrete numbers. Number properties questions ask whether a statement is always true, sometimes true, or never true. Reasoning abstractly is slow and error-prone. Picking specific numbers and testing the statement against each one is faster and more reliable. The key is choosing numbers that cover the boundary cases, not just the easy ones.

The testing set: 0, 1, -1, 2, -2

These five numbers catch almost every edge case the GMAT uses. Zero is even but not positive. One is neither prime nor composite. Negative one flips the sign when squared changes the parity. Two is the only even prime. Negative two tests whether a property depends on positivity. If a statement holds for all five, it is likely always true. If it fails for any one, you have your counterexample. Add a fraction like one-half when the question allows non-integers.

The most common trap

Assuming a property holds for all numbers when it only holds for some

You test with 2 and 4, see that the square of each is even, and conclude that the square of an even number is always even. That one happens to be true. But the same logic applied to "the square of a prime is odd" fails for 2, because \(2^2 = 4\) is even. The answer choices include the conclusion you would draw from testing only typical numbers. Test the boundary cases before committing to "must be true."

Step-by-step strategy

When to use: any question that asks "must be true," "could be true," or "is sufficient" about integer properties.

  1. 1Test 0, 1, -1, 2, and -2 against the statement. Add a fraction if the domain allows non-integers.
  2. 2If any number makes the statement false, the answer is "not necessarily" or "could be."
  3. 3If all five confirm the statement, look for a counterexample with a less typical number before committing to "must be true."
  4. 4Remember: 2 is the only even prime, 1 is not prime, and 0 is even.

When not to use: when the question asks for a specific computed value rather than a property judgment.

Worked example

If pp is a prime number greater than 2, then p2−1p^2 - 1 must be divisible by which of the following? Step 1: Identify the property. pp is an odd prime (all primes greater than 2 are odd), and we need to find what p2−1p^2 - 1 is always divisible by. Step 2: Test with concrete primes. Let p=3p = 3: 32−1=83^2 - 1 = 8. Let p=5p = 5: 52−1=245^2 - 1 = 24. Let p=7p = 7: 72−1=487^2 - 1 = 48. Let p=11p = 11: 112−1=12011^2 - 1 = 120. Step 3: Find the common divisor. The values are 8, 24, 48, and 120. The greatest common divisor of 8 and 24 is 8. Check 48: divisible by 8. Check 120: divisible by 8. All are divisible by 8. Step 4: Why it works. An odd number squared is odd, so p2−1p^2 - 1 is even. More precisely, p2−1=(p−1)(p+1)p^2 - 1 = (p - 1)(p + 1). Since pp is odd, both p−1p - 1 and p+1p + 1 are consecutive even numbers, and one of any two consecutive even numbers is divisible by 4. So the product is divisible by 8. The answer is 8.

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