GRE Quantitative Reasoning • Arithmetic and Number Properties

Integer Properties and Divisibility

Master the foundational rules governing factors, multiples, primes, and divisibility that underpin GRE quantitative reasoning.

GRE emblem with a reasoning brain, reading passage, and quantitative graph
Inside this lesson 10 sections 5 questions 8 figures

SECTION 1

Historical Context & Motivation

The study of integers and their divisibility properties is one of the oldest branches of mathematics, stretching back thousands of years to civilizations that needed to divide resources, measure land, and construct calendars. The ancient Egyptians and Babylonians developed practical techniques for factoring and dividing whole numbers, but it was the Greeks who first elevated these observations into a rigorous theoretical framework. Understanding integer properties and divisibility is not merely an academic exercise; these concepts form the bedrock of modern number theory, cryptography, computer science, and—most immediately for your purposes—standardized test reasoning on the GRE.

~300 BCE • Euclid's Elements

Euclid formalized the concept of prime numbers and proved there are infinitely many primes. Books VII–IX of the Elements established the Euclidean algorithm for finding the greatest common divisor, still in use today.

~240 BCE • Sieve of Eratosthenes

Eratosthenes devised an elegant algorithm for listing all prime numbers up to any given limit by systematically eliminating composite numbers—a method that remains a standard pedagogical tool.

1801 • Gauss's Disquisitiones Arithmeticae

Carl Friedrich Gauss published a landmark treatise that systematized number theory, including modular arithmetic and the Fundamental Theorem of Arithmetic, establishing that every integer greater than 1 has a unique prime factorization.

1978 • RSA Cryptography

Rivest, Shamir, and Adleman published the RSA algorithm, which relies on the difficulty of factoring large integers into primes—demonstrating that ancient divisibility concepts underpin modern digital security.

On the GRE, questions about integer properties and divisibility appear in both the Quantitative Comparison and Problem Solving formats. They test your ability to reason quickly about factors, multiples, remainders, and prime decomposition—often without a calculator. The central question this lesson addresses is: How can you systematically determine the relationships between integers—who divides whom, what factors they share, and what structure their prime factorizations reveal?

SECTION 2

Core Principles & Definitions

Before tackling GRE problems, you need a precise vocabulary. An integer is any whole number—positive, negative, or zero—belonging to the set {..., −3, −2, −1, 0, 1, 2, 3, ...}. Divisibility is the relationship that tells us when one integer divides another without leaving a remainder. We say that integer a divides integer b (written a | b) if there exists an integer k such that b = a × k. The following foundational ideas organize everything you need to know.

1

Factors & Multiples

If a | b, then a is a factor (or divisor) of b, and b is a multiple of a. For example, the factors of 12 are 1, 2, 3, 4, 6, and 12.
2

Prime Numbers

A prime number is an integer greater than 1 whose only positive factors are 1 and itself. The first several primes are 2, 3, 5, 7, 11, 13. Note that 2 is the only even prime—a frequent GRE trap.
3

Fundamental Theorem of Arithmetic

Every integer greater than 1 can be expressed as a unique product of prime factors (up to ordering). For instance, 60 = 2² × 3 × 5. This decomposition is the key to solving most divisibility problems.
4

GCD and LCM

The greatest common divisor (GCD) of two integers is the largest integer that divides both. The least common multiple (LCM) is the smallest positive integer that is a multiple of both. For any two positive integers a and b: GCD(a, b) × LCM(a, b) = a × b.
5

The Division Algorithm

For any integer a and positive integer d, there exist unique integers q (quotient) and r (remainder) such that a = d × q + r, where 0 ≤ r < d. Remainder problems on the GRE always rely on this principle.
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KEY TAKEAWAY

Think of prime factorization as the "DNA" of an integer. Just as a biologist can determine an organism's traits from its genetic code, you can determine every divisibility property of a number—its factors, its multiples, its GCD and LCM with another number—from its prime factorization. When you encounter a GRE divisibility problem, your first instinct should be to break the numbers into their prime components.

SECTION 3

Visual Explanation — The Factor Tree & Divisibility Map

Prime Factor Tree of 3603602×1802×9024531535Result360 = 2³ × 3² × 5
The factor tree decomposes 360 step by step. At each branch, a composite number splits into a prime factor (circled) and a remaining composite, until only primes remain. The circled nodes—three 2s, two 3s, and one 5—yield the canonical factorization 360 = 2³ × 3² × 5.

The factor tree above illustrates the most important technique you will use on GRE divisibility problems: reducing a composite number to its prime building blocks. Notice that the order in which you divide does not matter—you could split 360 into 10 × 36 instead—but the final set of prime factors is always the same. This is the Fundamental Theorem of Arithmetic in action: uniqueness of prime factorization guarantees a single canonical decomposition for every positive integer greater than 1. Once you have the prime factorization, you can immediately read off the total number of factors, determine divisibility relationships, and compute the GCD or LCM with any other integer.

SECTION 4

Mathematical Framework

The prime factorization of an integer is the master key to nearly every divisibility question on the GRE. Let us formalize the relationships and formulas that follow from this decomposition.

PRIME FACTORIZATION

n=p1a×p2a×…×pkan = p₁^a₁ × p₂^a₂ × … × pₖ^aₖ

where p₁, p₂, …, pₖ are distinct primes and a₁, a₂, …, aₖ are their respective positive integer exponents. This representation is unique for every n > 1.

NUMBER OF POSITIVE DIVISORS

τ(n)=(a1+1)(a2+1)…(ak+1)τ(n) = (a₁ + 1)(a₂ + 1)…(aₖ + 1)

Each factor of n is formed by choosing an exponent for each prime pᵢ from 0 up to aᵢ. For example, 360 = 2³ × 3² × 5¹ has (3+1)(2+1)(1+1) = 4 × 3 × 2 = 24 divisors.

GCD VIA PRIME FACTORIZATION

GCD(a,b)=p1min(e1,f1)×p2min(e2,f2)×…GCD(a, b) = p₁^min(e₁,f₁) × p₂^min(e₂,f₂) × …

For a = p₁^e₁ × p₂^e₂ × … and b = p₁^f₁ × p₂^f₂ × …, take the minimum exponent for each prime. If a prime doesn't appear in one number, its exponent is 0.

LCM VIA PRIME FACTORIZATION

LCM(a,b)=p1max(e1,f1)×p2max(e2,f2)×…LCM(a, b) = p₁^max(e₁,f₁) × p₂^max(e₂,f₂) × …

Take the maximum exponent for each prime. A useful identity: GCD(a, b) × LCM(a, b) = a × b. This allows you to compute one from the other quickly.

These formulas convert abstract questions about divisibility into concrete arithmetic on exponents. On the GRE, the most common application is counting factors: the test loves to ask "how many positive divisors does n have?" or "which of the following must be a factor of n?" Both reduce to the prime factorization plus the divisor-count formula. Similarly, remainder problems exploit the Division Algorithm: when the GRE asks for the remainder when 247 is divided by 6, you compute 247 = 6 × 41 + 1, so the remainder is 1.

SECTION 5

Divisibility Rules & Classification

Because the GRE does not allow a calculator on the Quantitative Reasoning section, divisibility rules serve as mental shortcuts that let you quickly test whether a given integer is divisible by small primes or composite numbers. Memorizing these rules saves valuable time and reduces computational errors under test pressure.

Quick Divisibility Rules Reference÷ 2Last digit is evene.g. 4,738 → 8 is even ✓÷ 3Sum of digits divisible by 3e.g. 528 → 5+2+8=15 ✓÷ 4Last two digits ÷ 4e.g. 1,324 → 24÷4=6 ✓÷ 5Last digit is 0 or 5e.g. 2,345 → 5 ✓÷ 6Divisible by both 2 and 3e.g. 762 → even & 7+6+2=15 ✓÷ 7Double last digit, subtracte.g. 203 → 20−6=14 ✓÷ 8Last three digits ÷ 8e.g. 7,120 → 120÷8=15 ✓÷ 9Sum of digits divisible by 9e.g. 6,453 → 6+4+5+3=18 ✓÷ 10Last digit is 0e.g. 5,130 → ends in 0 ✓÷ 11Alternating sum of digits ÷ 11e.g. 9,174 → 9−1+7−4=11 ✓÷ 12Divisible by both 3 and 4e.g. 1,296 → digits=18, 96÷4 ✓Composite divisors (6, 12) combine rules of their coprime factor pairs
This reference chart summarizes divisibility rules for integers 2 through 12. Notice the pattern: rules for composite divisors (6, 12) are built by combining the rules for their coprime prime-power components (e.g., 6 = 2 × 3, so check both the rule for 2 and the rule for 3).

A few of these rules deserve special attention for GRE purposes. The rules for 3 and 9 both depend on the digit sum, which exploits the fact that 10 ≡ 1 (mod 3) and 10 ≡ 1 (mod 9). The rule for 4 checks only the last two digits because 100 is divisible by 4, and analogously the rule for 8 checks the last three digits because 1000 is divisible by 8. For the rule for 11, the alternating-sum property arises because 10 ≡ −1 (mod 11), so powers of 10 alternate between +1 and −1 modulo 11. Understanding the "why" behind each rule strengthens your recall and helps you extend the logic to unfamiliar situations on test day.

💡

GRE TIP

When a GRE problem says "n is divisible by 12," you should immediately note that n is divisible by 4 and by 3, and therefore also by 2, by 6, and by 1. Conversely, knowing n is divisible by both 4 and 3 is sufficient to conclude n is divisible by 12—because GCD(4, 3) = 1 (they are coprime). However, knowing n is divisible by 4 and by 6 does not guarantee divisibility by 24, because GCD(4, 6) = 2 ≠ 1. Always check coprimality before combining divisibility conditions.

SECTION 6

Worked Example

Let's work through a GRE-style problem that synthesizes prime factorization, factor counting, and GCD/LCM computation.

ℹ

PROBLEM

If m = 2⁴ × 3² × 7 and n = 2² × 3³ × 5, find: (a) GCD(m, n), (b) LCM(m, n), (c) the number of positive divisors of m, and (d) confirm that GCD(m, n) × LCM(m, n) = m × n.

Complete Solution

  1. 1

    Step 1 — Identify the prime factorizations

    We are given m = 2⁴ × 3² × 7¹ and n = 2² × 3³ × 5¹. The distinct primes across both numbers are {2, 3, 5, 7}. We can write each with all primes included by assigning exponent 0 where a prime is absent: m = 2⁴ × 3² × 5⁰ × 7¹ and n = 2² × 3³ × 5¹ × 7⁰.
    Primes: {2, 3, 5, 7}
  2. 2

    Step 2 — Compute GCD(m, n)

    Take the minimum exponent for each prime: min(4,2) = 2 for prime 2; min(2,3) = 2 for prime 3; min(0,1) = 0 for prime 5; min(1,0) = 0 for prime 7.
    GCD(m, n) = 2² × 3² = 4 × 9 = 36
  3. 3

    Step 3 — Compute LCM(m, n)

    Take the maximum exponent for each prime: max(4,2) = 4 for prime 2; max(2,3) = 3 for prime 3; max(0,1) = 1 for prime 5; max(1,0) = 1 for prime 7.
    LCM(m, n) = 2⁴ × 3³ × 5 × 7 = 16 × 27 × 5 × 7 = 15,120
  4. 4

    Step 4 — Count the positive divisors of m

    Using the divisor-count formula τ(n) = (a₁ + 1)(a₂ + 1)…(aₖ + 1), for m = 2⁴ × 3² × 7¹ we get (4 + 1)(2 + 1)(1 + 1) = 5 × 3 × 2.
    τ(m) = 30 positive divisors
  5. 5

    Step 5 — Verify the GCD × LCM identity

    Compute m × n = (2⁴ × 3² × 7)(2² × 3³ × 5) = 2⁶ × 3⁵ × 5 × 7 = 64 × 243 × 5 × 7 = 544,320. Now GCD × LCM = 36 × 15,120 = 544,320. The identity holds, confirming our calculations.
    36 × 15,120 = 544,320 = m × n ✓

SECTION 7

GRE Strategies & Common Pitfalls

Integer properties questions on the GRE test conceptual understanding as much as computational skill. Knowing the rules is necessary but insufficient—you also need to recognize the common traps the test employs and choose efficient strategies. The table below contrasts effective strategies with the pitfalls they help you avoid.

StrategyCommon PitfallGRE Context
Always factor into primes before attempting divisibility conclusions.Assuming that if a | n and b | n, then ab | n (true only when GCD(a,b) = 1).Quantitative Comparison: "n is divisible by 6 and 10. Is n divisible by 60?" (Not necessarily—LCM(6,10) = 30.)
Remember that 0 is an integer and is divisible by every nonzero integer.Forgetting 0 when asked about "nonnegative integers" or "integers between −5 and 5."Counting problems that include 0 in the valid range inflate the answer by 1.
Recall that 1 is neither prime nor composite.Counting 1 as a prime number, which overstates prime counts and distorts factorizations."How many prime factors does 30 have?" Answer: 3 (not 4). The primes are 2, 3, 5.
Distinguish "distinct prime factors" from "total prime factors" (with multiplicity).Confusing "the number of prime factors of 12" (2, counted as 2 and 3) with "12 has 4 prime factors" (2 × 2 × 3).Read the question carefully: "different" or "distinct" signals unique primes; otherwise, count with multiplicity.
Use the remainder pattern (cycling) for large exponents.Trying to compute large powers directly instead of recognizing repeating remainder cycles."What is the remainder when 7¹⁰⁰ is divided by 5?" Since 7¹ ≡ 2, 7² ≡ 4, 7³ ≡ 3, 7⁴ ≡ 1 (mod 5), the cycle has period 4, and 100 ÷ 4 = 25 exactly, so the remainder is 1.

Key strategies and corresponding pitfalls for GRE integer property questions

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KEY TAKEAWAY

Think of divisibility on the GRE the way a structural engineer thinks about load-bearing: each factor is a support column, and you cannot claim a structure stands unless every required column is verified. When combining divisibility conditions, the prime factorization is your blueprint—it tells you exactly which "columns" are present and at what strength (exponent). Never assume two conditions combine multiplicatively unless their prime supports are completely independent (coprime).

SECTION 8

Connections to Advanced Number Theory

While the GRE tests fundamental integer properties, these concepts serve as the gateway to far deeper mathematics. The skills you develop here—prime factorization, modular reasoning, and divisor analysis—reappear in more sophisticated forms in graduate-level number theory, abstract algebra, and computational complexity. The table below maps each GRE-level concept to its more advanced generalization, providing context for why these foundations matter beyond test preparation.

GRE-Level ConceptAdvanced Generalization
Prime factorization of integersUnique factorization domains (UFDs) in ring theory; prime ideals in algebraic number fields
Divisibility rules using digit sumsModular arithmetic and congruence theory; Chinese Remainder Theorem for systems of congruences
Counting divisors via τ(n)Multiplicative arithmetic functions (σ(n), Euler's totient φ(n)); Dirichlet series and analytic number theory
GCD and LCM via prime exponentsLattice theory of divisors; Bézout's identity and the Extended Euclidean Algorithm
Remainder and quotient (Division Algorithm)Euclidean domains; polynomial division in F[x]; modular exponentiation in cryptography (RSA, Diffie-Hellman)

From GRE fundamentals to advanced number theory

One particularly elegant extension is Euler's totient function φ(n), which counts the number of integers from 1 to n that are coprime to n. For a prime p, φ(p) = p − 1, and for a product of prime powers, φ(n) = n × ∏(1 − 1/p) over all distinct primes p dividing n. This function powers Euler's theorem: a^φ(n) ≡ 1 (mod n) whenever GCD(a, n) = 1, which generalizes Fermat's Little Theorem and is the theoretical backbone of RSA encryption. If you continue to graduate study in mathematics or computer science, you will see these GRE-level ideas elevated into powerful abstract tools.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

If an integer n is divisible by both 8 and 12, then n must be divisible by which of the following? (A) 16 (B) 24 (C) 48 (D) 96 (E) 192
Show answer
The correct answer is 24. To find the smallest number that n must be divisible by, compute LCM(8, 12). The prime factorizations are 8 = 2³ and 12 = 2² × 3, so LCM(8, 12) = 2³ × 3 = 24. Thus any integer divisible by both 8 and 12 must be divisible by 24. The trap arises from incorrectly multiplying 8 × 12 = 96; this product equals LCM(8,12) × GCD(8,12) = 24 × 4, so the product overstates the required divisor when GCD > 1. Always compute the LCM, not the product, when combining divisibility conditions.

Question 2: BASIC CALCULATION

How many positive divisors does 540 have?
Show answer
First, find the prime factorization: 540 = 2² × 3³ × 5¹. Apply the divisor-count formula τ(n) = (a₁ + 1)(a₂ + 1)(a₃ + 1) = (2 + 1)(3 + 1)(1 + 1) = 3 × 4 × 2 = 24. Therefore, 540 has 24 positive divisors. You can verify a few: 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 27, 30, 36, 45, 54, 60, 90, 108, 135, 180, 270, 540—exactly 24.

Question 3: INTERMEDIATE

When the positive integer n is divided by 7, the remainder is 3. When n is divided by 4, the remainder is 2. What is the smallest positive integer n satisfying both conditions?
Show answer
We need n ≡ 3 (mod 7) and n ≡ 2 (mod 4). From the first condition, n can be 3, 10, 17, 24, 31, 38, … Testing each against the second condition: 3 ÷ 4 leaves remainder 3 (no); 10 ÷ 4 leaves remainder 2 (yes!). So the smallest n is 10. We can verify: 10 = 7 × 1 + 3 ✓ and 10 = 4 × 2 + 2 ✓. The general solution is n = 10 + 28k for nonnegative integers k, where 28 = LCM(7, 4), because subsequent solutions repeat every LCM(7, 4) units.

Question 4: APPLIED

A school has 180 students and 120 faculty members. The administration wants to divide everyone into equal-sized committees so that each committee has the same number of students and the same number of faculty members (but not necessarily the same total). What is the greatest number of committees that can be formed?
Show answer
The number of committees must divide both 180 (students) and 120 (faculty) evenly, so we need GCD(180, 120). Factor: 180 = 2² × 3² × 5 and 120 = 2³ × 3 × 5. Taking minimum exponents: GCD = 2² × 3¹ × 5¹ = 4 × 3 × 5 = 60. Thus the greatest number of committees is 60, with each committee containing 180 ÷ 60 = 3 students and 120 ÷ 60 = 2 faculty members (5 per committee total). This is a classic GCD application: the GCD represents the finest equal partition of two quantities.

Question 5: CRITICAL THINKING

Prove that for any integer n, the product n(n + 1)(n + 2) is always divisible by 6. Then determine: for which values of n is n(n + 1)(n + 2) divisible by 24?
Show answer
Divisibility by 6: Among any three consecutive integers n, n+1, n+2, at least one is divisible by 2 and at least one is divisible by 3 (since consecutive integers cycle through all residues mod 2 and mod 3). Since GCD(2, 3) = 1, the product is always divisible by 2 × 3 = 6. ✓ Divisibility by 24 = 8 × 3: We already have a guaranteed factor of 3. We need the product to contribute at least 2³ = 8. Among three consecutive integers, either one or two are even. Case 1 — n is even (two even factors: n and n+2): Since n and n+2 are both even and differ by 2, exactly one of them is divisible by 4. So together they contribute at least 2¹ × 2² = 2³ = 8. The product is divisible by 24. • n = 2: 2 × 3 × 4 = 24 ✓ • n = 4: 4 × 5 × 6 = 120 = 24 × 5 ✓ Case 2 — n is odd (one even factor: n+1): The only even number among n, n+1, n+2 is n+1. For the product to be divisible by 8, n+1 alone must supply at least 2³, meaning n+1 must be divisible by 8, i.e., n ≡ 7 (mod 8). • n = 7: 7 × 8 × 9 = 504 = 24 × 21 ✓ • n = 15: 15 × 16 × 17 = 4080 = 24 × 170 ✓ • n = 1: 1 × 2 × 3 = 6 ✗ (n+1 = 2, only one factor of 2) • n = 3: 3 × 4 × 5 = 60 ✗ (n+1 = 4, only two factors of 2) • n = 5: 5 × 6 × 7 = 210 ✗ (n+1 = 6, only one factor of 2) Conclusion: n(n+1)(n+2) is divisible by 24 if and only if n is even OR n ≡ 7 (mod 8). Equivalently, it is NOT divisible by 24 when n is odd and n ≢ 7 (mod 8), i.e., when n ≡ 1, 3, or 5 (mod 8).

SUMMARY

Lesson Summary

Summary

This lesson established the core framework for reasoning about integer properties and divisibility on the GRE. Every positive integer greater than 1 has a unique prime factorization, which serves as the master key to all divisibility questions. From this factorization, you can count positive divisors using the formula τ(n) = (a₁ + 1)(a₂ + 1)…(aₖ + 1), compute the GCD by taking minimum exponents and the LCM by taking maximum exponents, and verify results using the identity GCD × LCM = a × b.

For efficient test performance, internalize the divisibility rules for 2 through 12, understand that composite-number rules combine the rules of their coprime components, and remember key pitfalls: 1 is not prime, 2 is the only even prime, and 0 is divisible by every nonzero integer. When in doubt, break numbers into primes, compare exponents, and let the Division Algorithm handle remainder questions. These tools, practiced to fluency, will make GRE integer-property questions among the fastest and most reliable points you can earn.

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