Apply each property with its conditions
Arithmetic rules are brief, but their conditions matter. A prime number is an integer greater than 1 with exactly two distinct positive divisors: 1 and itself. An odd integer can be written as 2k + 1, where k is an integer; this form can prove a property that a few examples would only suggest.
For powers, distinguish multiplication from addition: 2³ × 2⁴ = 2⁷, while 2³ + 2⁴ = 24. For real roots, √(x²) = |x|, not always x. In each review, note the omitted condition instead of writing an overly general formula. The next question may change precisely that condition.
Exercise 1: form a perfect square
Prime factorization lets you reason about divisibility and powers without trying many numbers. In the prime factorization of a positive perfect square, every exponent is even.
What is the least positive integer k such that 2⁴ × 3³ × 5 × k is a perfect square?
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The solution, step by step
15.
The method in action
- The exponent of 2 is already even; the exponents of 3 (three) and 5 (one) are odd. So k needs one factor of 3 and one factor of 5: k = 3 × 5 = 15, and 2⁴ × 3⁴ × 5² = (2² × 3² × 5)² = 180².
- k = 3 or k = 5 leaves an odd exponent. Multiplying 15 by a perfect square greater than 1 would also produce a valid multiplier, but a larger one. The prompt’s “least positive integer” excludes zero and requires the smallest value.
Exercise 2: keep the general form
When an integer is divided by a positive integer d, the remainder r satisfies 0 ≤ r < d. If n leaves a remainder of 2 when divided by 5, writing n = 5q + 2, where q is an integer, captures every integer meeting that condition, not just one example.
When positive integer n is divided by 5, the remainder is 2. What is the remainder when 3n + 4 is divided by 5?
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The solution, step by step
0.
The method in action
- Substitute 5q + 2 for n. Then 3n + 4 = 15q + 10 = 5(3q + 2), a multiple of 5.
- 10 is not a remainder when dividing by 5; reduce it further. Testing n = 2 does give 10, but the general form proves the result for every permitted value.
Exercises 3 and 4: check the signs
Parentheses and the domain are part of the problem. Rewrite the expression before calculating when it contains several operations.
What is the value of (3⁴ × 3²) / 3³?
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The solution, step by step
27.
The method in action
- In the numerator, powers with the same base multiply: 3⁶. Dividing by 3³ gives 3³ = 27.
- Adding the exponents of every term would give 3⁹ and ignore the division. Multiplying the exponents would not follow the rule for a product.
x is a negative real number. Which expression equals √(x²)?
Choose an answer.
The solution, step by step
−x.
The method in action
- The principal square root is nonnegative. Since x < 0, |x| = −x.
- When x = −4, √16 = 4 and −x = 4. Answering x would give −4; that confuses the solutions to y² = 16 with the single value represented by √16.
The pitfall
The pitfall to avoid
Treating a few examples as a universal proof is a mistake. The product of two numbers you tested may be even without every possible product being even. For a general rule, write numbers in the forms 2k and 2k + 1 or use a factorization; to disprove a claim, one valid counterexample is enough.
Move from a concept to GRE-style questions
Upsilar tip: redo each exercise with a different question. For perfect squares, compare two multipliers; for remainders, test several possible choices; for roots, remove the condition x < 0 and analyze the cases. This lets you practice the same concept without confusing mastery with remembering an answer.
Start without a timer until you can explain each step. Only then mix concepts in a short set. Track your hesitations: English vocabulary, a forgotten property, calculation, or an unread condition. This classification guides your next session; it does not predict a GRE score.
Key takeaways
- Conditions are part of every property.
- Factoring can replace many trial calculations.
- The principal square root is always nonnegative.