3.3 Number Theory: Prime Factorization, GCF, LCM & Divisibility Rules

Prime and Composite Numbers: Structural Foundations

Number theory explores the properties of the integers ($\mathbb{Z}$), with prime numbers serving as the foundational building blocks of multiplicative arithmetic.

Definitions and Special Status of 0 and 1

• A prime number is an integer $p > 1$ that has exactly two distinct positive divisors: 1 and $p$ itself.
• A composite number is an integer $n > 1$ that has more than two distinct positive divisors (i.e., it can be factored into $n = a \times b$ where $1 < a, b < n$).
• The Number 1: By mathematical convention, the number 1 is neither prime nor composite. It is designated as a unit (it has a multiplicative inverse in the integers, namely itself). Defining 1 as prime would break the uniqueness clause of the Fundamental Theorem of Arithmetic, because any integer could then have infinitely many prime factorizations (e.g., $12 = 2^2 \times 3 = 1 \times 2^2 \times 3 = 1^2 \times 2^2 \times 3$).
• The Number 0: The integer 0 is neither prime nor composite. Every non-zero integer divides 0 (since $0 = k \times 0$), giving 0 infinitely many divisors.

Prime Identification and the Sieve of Eratosthenes

The Sieve of Eratosthenes is an ancient, systematic algorithm for finding all primes up to a given bound $N$:

1. List all natural numbers from 2 to $N$.
2. Identify the first unmarked number as prime (2) and cross out all its proper multiples ($4, 6, 8, \dots$).
3. Advance to the next unmarked number (3), declare it prime, and cross out all its multiples ($6, 9, 12, \dots$).
4. Repeat this procedure for each successive prime $p$.
5. The process terminates when $p > \sqrt{N}$, as all remaining unmarked numbers are guaranteed to be prime.

There are exactly 25 prime numbers less than 100:

$$\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97\}$$

Note that 2 is the only even prime number; all other even integers are multiples of 2 and therefore composite.

The Primality Testing Boundary

To determine whether a specific integer $N$ is prime, one only needs to test for divisibility by prime numbers $p \le \sqrt{N}$. Proof: If $N$ is composite, it can be written as $N = a \times b$ where $1 < a \le b < N$. If both $a > \sqrt{N}$ and $b > \sqrt{N}$, then $a \times b > \sqrt{N} \times \sqrt{N} = N$, which is a contradiction. Thus, at least one factor must satisfy $a \le \sqrt{N}$. If no prime less than or equal to $\sqrt{N}$ divides $N$, then $N$ must be prime.

For example, to test whether 173 is prime, note that $13^2 = 169 < 173 < 14^2 = 196$, so $\sqrt{173} \approx 13.15$. We only test primes $\{2, 3, 5, 7, 11, 13\}$:

• $173$ is odd (not divisible by 2).
• Sum of digits $1 + 7 + 3 = 11$ (not divisible by 3).
• Ends in 3 (not divisible by 5).
• $173 \div 7 = 24.71$ (not divisible by 7).
• $173 \div 11 = 15.72$ (not divisible by 11).
• $173 \div 13 = 13.30$ (not divisible by 13). Because none of these primes divide 173, 173 is prime.

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The Fundamental Theorem of Arithmetic

The Fundamental Theorem of Arithmetic (Unique Factorization Theorem) states that:

Every integer $n > 1$ can be expressed uniquely as a product of prime numbers, up to the order of the factors: $n = p_1^{a_1} p_2^{a_2} p_3^{a_3} \cdots p_k^{a_k}$ where each $p_i$ is a prime number such that $p_1 < p_2 < \dots < p_k$, and each $a_i$ is a positive integer.

This expression is the canonical prime factorization of $n$.

Determining the Total Number of Positive Divisors

The canonical prime factorization allows one to directly compute the total number of positive divisors $d(n)$ without enumerating each factor. Any divisor $d$ of $n$ must take the form:

$$d = p_1^{b_1} p_2^{b_2} \cdots p_k^{b_k} \quad \text{where } 0 \le b_i \le a_i$$

Because there are $(a_i + 1)$ independent choices for each exponent $b_i$ (ranging from 0 to $a_i$), the Multiplication Principle of counting gives:

$$d(n) = (a_1 + 1)(a_2 + 1)(a_3 + 1) \cdots (a_k + 1)$$

For example, consider $n = 360$:

$$360 = 2^3 \times 3^2 \times 5^1$$

The total number of positive divisors is:

$$d(360) = (3 + 1)(2 + 1)(1 + 1) = 4 \times 3 \times 2 = 24\text{ divisors}$$

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Divisibility Rules and Modular Algebraic Proofs

Divisibility tests are mental shortcuts based on our base-10 positional notation. Any integer $N$ with digits $d_k d_{k-1} \dots d_1 d_0$ can be represented as a polynomial in powers of 10:

$$N = d_k 10^k + d_{k-1} 10^{k-1} + \dots + d_2 10^2 + d_1 10 + d_0$$

1. Rules for 2, 5, and 10 (Last Digit Tests)

Because the base $10 = 2 \times 5$, 10 is divisible by 2, 5, and 10. Consequently, every term $d_j 10^j$ for $j \ge 1$ is a multiple of 10 and contributes zero to the remainder modulo 2, 5, or 10:

$$N = 10(d_k 10^{k-1} + \dots + d_1) + d_0 \equiv d_0 \pmod{2, 5, 10}$$

• Divisibility by 2: The units digit $d_0$ must be even ($0, 2, 4, 6, 8$).
• Divisibility by 5: The units digit $d_0$ must be $0$ or $5$.
• Divisibility by 10: The units digit $d_0$ must be $0$.

2. Rules for 4 and 8 (Sub-Block Tests)

• Divisibility by 4: Since $10^2 = 100 = 4 \times 25$, all terms for $j \ge 2$ are divisible by 4. Thus: $$N = 100(d_k 10^{k-2} + \dots + d_2) + (10 d_1 + d_0) \equiv (10 d_1 + d_0) \pmod 4$$ An integer is divisible by 4 if and only if its last two digits form a number divisible by 4 (e.g., in 73,528, $28 = 4 \times 7$, so 73,528 is divisible by 4).
• Divisibility by 8: Since $10^3 = 1000 = 8 \times 125$, all terms for $j \ge 3$ are divisible by 8. An integer is divisible by 8 if and only if its last three digits form a number divisible by 8 (e.g., in 54,832, $832 \div 8 = 104$, so 54,832 is divisible by 8).

3. Rules for 3 and 9 (Sum of Digits Tests)

Notice that $10 = 9 + 1$, $100 = 99 + 1$, and generally $10^j = (10^j - 1) + 1$, where $(10^j - 1) = 999\dots9$ is divisible by 9 (and thus by 3):

$$10^j \equiv 1 \pmod 9 \quad \text{and} \quad 10^j \equiv 1 \pmod 3$$

Substituting this into the base-10 polynomial expansion:

$$N = \sum_{j=0}^k d_j 10^j = \sum_{j=0}^k d_j (99\dots9 + 1) = \sum_{j=0}^k d_j (99\dots9) + \sum_{j=0}^k d_j$$

Because the first sum is a multiple of 9, $N$ leaves the exact same remainder as the sum of its digits:

$$N \equiv \sum_{j=0}^k d_j \pmod 9 \quad \text{and} \quad N \equiv \sum_{j=0}^k d_j \pmod 3$$

• Rule for 3: An integer is divisible by 3 if the sum of its digits is divisible by 3.
• Rule for 9: An integer is divisible by 9 if the sum of its digits is divisible by 9.

4. Rule for 6 (Composite Divisibility)

Because $6 = 2 \times 3$ and $\gcd(2, 3) = 1$, an integer is divisible by 6 if and only if it is divisible by both 2 and 3 simultaneously. Thus, the number must be even and the sum of its digits must be a multiple of 3.

5. Rule for 11 (Alternating Sum of Digits)

Notice that $10 = 11 - 1 \equiv -1 \pmod{11}$. Therefore, powers of 10 alternate signs modulo 11:

$$10^0 \equiv 1, \quad 10^1 \equiv -1, \quad 10^2 \equiv 1, \quad 10^3 \equiv -1, \quad \dots, \quad 10^j \equiv (-1)^j \pmod{11}$$

Substituting into the expansion yields:

$$N \equiv d_0 - d_1 + d_2 - d_3 + d_4 - \dots \pmod{11}$$

An integer is divisible by 11 if and only if the alternating sum of its digits (starting with the units digit as positive) is a multiple of 11 (including 0). For example, test $85,976$:

$$6 - 7 + 9 - 5 + 8 = 11$$

Since 11 is divisible by 11, $85,976$ is divisible by 11 ($85,976 = 11 \times 7,816$).

6. Rule for 7 (Truncation and Doubling)

To test an integer $N = 10a + b$ (where $b$ is the last digit and $a$ is the truncated prefix), subtract twice the last digit from the prefix: $a - 2b$. Proof: Consider the linear combination:

$$10(a - 2b) = 10a - 20b = (10a + b) - 21b = N - 21b$$

Because $21b$ is a multiple of 7, $10(a - 2b) \equiv N \pmod 7$. Because $\gcd(10, 7) = 1$, $N$ is divisible by 7 if and only if $a - 2b$ is divisible by 7. For example, test 2,464:

1. Prefix 246, last digit 4: $246 - 2(4) = 246 - 8 = 238$.
2. Repeat for 238: Prefix 23, last digit 8: $23 - 2(8) = 23 - 16 = 7$. Since 7 is divisible by 7, 2,464 is divisible by 7 ($2,464 = 7 \times 352$).

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Reference Table: Divisibility Rules

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Algorithmic Methods for GCF and LCM

• The Greatest Common Factor (GCF), or Greatest Common Divisor (GCD), of two positive integers $a$ and $b$ is the largest positive integer that divides both $a$ and $b$ without remainder. Note that $\text{GCF}(a, b) \le \min(a, b)$.
• The Least Common Multiple (LCM) of two positive integers $a$ and $b$ is the smallest positive integer that is a multiple of both $a$ and $b$. Note that $\text{LCM}(a, b) \ge \max(a, b)$.

Method 1: Prime Factorization and Min/Max Exponent Rules

Express $a$ and $b$ in their canonical prime factorizations using all prime factors present in either number (assigning exponent 0 if a prime does not divide a number):

$$a = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k} \quad \text{and} \quad b = p_1^{b_1} p_2^{b_2} \cdots p_k^{b_k}$$

• GCF Formula: Take the minimum exponent for each shared prime factor: $$\text{GCF}(a, b) = p_1^{\min(a_1, b_1)} p_2^{\min(a_2, b_2)} \cdots p_k^{\min(a_k, b_k)}$$
• LCM Formula: Take the maximum exponent across all prime factors: $$\text{LCM}(a, b) = p_1^{\max(a_1, b_1)} p_2^{\max(a_2, b_2)} \cdots p_k^{\max(a_k, b_k)}$$

For example, to find the GCF and LCM of $72$ and $120$:

$$72 = 2^3 \times 3^2 \times 5^0$$ $$120 = 2^3 \times 3^1 \times 5^1$$

• $\text{GCF}(72, 120) = 2^{\min(3,3)} \times 3^{\min(2,1)} \times 5^{\min(0,1)} = 2^3 \times 3^1 \times 5^0 = 8 \times 3 = 24$.
• $\text{LCM}(72, 120) = 2^{\max(3,3)} \times 3^{\max(2,1)} \times 5^{\max(0,1)} = 2^3 \times 3^2 \times 5^1 = 8 \times 9 \times 5 = 360$.

Method 2: The Venn Diagram Representation

Draw two overlapping circles representing numbers $a$ and $b$:

1. Place each instance of a common prime factor in the intersection ($A \cap B$).
2. Place remaining prime factors unique to $a$ in the left region, and those unique to $b$ in the right region.
3. GCF is the product of all prime factors in the intersection.
4. LCM is the product of all prime factors in the entire union ($A \cup B$).

Method 3: The Ladder (Cake / Upside-Down Division) Method

Write $a$ and $b$ side by side and divide both concurrently by a common prime factor. Repeat until the resulting quotients are coprime (share no common factor other than 1):

$$\begin{array}{r|rr} 2 & 72 & 120 \\ \hline 2 & 36 & 60 \\ \hline 2 & 18 & 30 \\ \hline 3 & 9 & 15 \\ \hline & 3 & 5 \end{array}$$

• The GCF is the product of the divisors along the left vertical column: $2 \times 2 \times 2 \times 3 = 24$.
• The LCM is the product of the left vertical column and the bottom quotient row (forming an "L" shape): $24 \times (3 \times 5) = 24 \times 15 = 360$.

Method 4: The Euclidean Algorithm

For large integers where prime factorization is computationally difficult, the Euclidean Algorithm computes the GCF in a few steps. It relies on the principle that the greatest common divisor of two integers divides their difference and remainder:

$$\gcd(a, b) = \gcd(b, a \bmod b)$$

Applying successive division with remainder: $a = bq + r$, then $\gcd(a, b) = \gcd(b, r)$. The process terminates when the remainder reaches 0; the last non-zero remainder is the GCF.

The Fundamental Product Identity

For any two positive integers $a$ and $b$:

$$\text{GCF}(a, b) \times \text{LCM}(a, b) = a \times b$$

Proof: In the prime factorization formulas, each prime $p_i$ contributes $p_i^{\min(a_i, b_i)}$ to the GCF and $p_i^{\max(a_i, b_i)}$ to the LCM. Because for any two real numbers $\min(x, y) + \max(x, y) = x + y$, multiplying the two yields:

$$p_i^{\min(a_i, b_i)} \cdot p_i^{\max(a_i, b_i)} = p_i^{\min(a_i, b_i) + \max(a_i, b_i)} = p_i^{a_i + b_i} = p_i^{a_i} \cdot p_i^{b_i}$$

Taking the product across all primes reproduces $a \times b$. Note: this product identity holds strictly for two numbers and does not generalize directly to three or more numbers without inclusion-exclusion adjustments.

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Real-World Applications and Problem Solving

A critical competency on educator certification exams is distinguishing between word problems that require finding the GCF versus those that require finding the LCM.

Contexts Requiring GCF (Partitioning and Tiling)

GCF problems typically involve breaking down quantities into smaller, equal subsets without leftovers:

• Equal Partitioning: Creating identical gift bags, medical kits, or care packages from different quantities of items (e.g., distributing 48 pens and 64 notebooks into identical kits).
• Cutting Materials: Cutting boards, fabrics, or wires of lengths $L_1$ and $L_2$ into pieces of the greatest possible equal length without any waste.
• Geometric Tiling: Paving a rectangular floor of dimensions $L \times W$ with the largest possible congruent square tiles without cutting any tiles.

Contexts Requiring LCM (Synchronization and Cycles)

LCM problems typically involve periodic events that repeat at fixed time or distance intervals and ask when the events will synchronize again:

• Periodic Synchronization: Two or more blinking lighthouses, bell chimes, or traffic signals that flash or chime at intervals of $t_1, t_2,$ and $t_3$ seconds, asking when they will flash simultaneously next.
• Track Laps: Runners running around a circular track who complete one lap in $m_1$ and $m_2$ minutes, asking when they will meet again at the starting line.
• Packaging Discrepancies: Hot dogs packaged in packs of 10 and hot dog buns in packs of 8, asking for the minimum number of each package to buy so that no hot dogs or buns remain unmatched.

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Worked Number Theory Examples

Worked Example 1: Divisor Count and Factor Analysis

Problem: Find the canonical prime factorization and total number of positive divisors for 720.

Solution: Step 1: Factor 720 using a factor tree or successive division:

$$720 = 72 \times 10 = (8 \times 9) \times (2 \times 5) = (2^3 \times 3^2) \times (2^1 \times 5^1) = 2^4 \times 3^2 \times 5^1$$

Step 2: Apply the divisor formula $d(n) = (a_1 + 1)(a_2 + 1)(a_3 + 1)$:

$$d(720) = (4 + 1)(2 + 1)(1 + 1) = 5 \times 3 \times 2 = 30$$

Therefore, 720 has exactly 30 positive divisors.

Worked Example 2: The Euclidean Algorithm for Large Integers

Problem: Find the GCF of 546 and 126 using the Euclidean algorithm, and then compute their LCM using the product identity.

Solution: Step 1: Execute successive divisions with remainder:

$$\begin{aligned} 546 &= 126 \times 4 + 42 \\ 126 &= 42 \times 3 + 0 \end{aligned}$$

Because the remainder is 0, the last non-zero remainder is 42. Thus:

$$\text{GCF}(546, 126) = 42$$

Step 2: Use the identity $\text{GCF}(a, b) \times \text{LCM}(a, b) = a \times b$:

$$42 \times \text{LCM}(546, 126) = 546 \times 126$$ $$\text{LCM}(546, 126) = \frac{546 \times 126}{42} = 546 \times 3 = 1638$$

Worked Example 3: Floor Tiling Application (GCF)

Problem: A rectangular craft table measures 140 cm by 84 cm. A teacher wishes to cover the entire tabletop completely with congruent square tiles of the largest possible whole-number side length, without cutting any tiles. What is the side length of the tile, and how many tiles are required?

Solution: Step 1: Identify that tiling without cutting requires a common divisor of both length and width. The largest tile requires the GCF of 140 and 84. Factor both numbers:

$$140 = 14 \times 10 = 2^2 \times 5 \times 7$$ $$84 = 12 \times 7 = 2^2 \times 3 \times 7$$ $$\text{GCF}(140, 84) = 2^2 \times 7 = 4 \times 7 = 28\text{ cm}$$

The side length of the largest square tile is 28 cm.

Step 2: Determine total tile count:

• Number of tiles along length: $140 \div 28 = 5$.
• Number of tiles along width: $84 \div 28 = 3$.
• Total tiles needed: $5 \times 3 = 15\text{ tiles}$.

Worked Example 4: Cyclic Synchronization Application (LCM)

Problem: Three city transit buses depart a central terminal at 6:00 AM. Route A bus departs every 12 minutes, Route B bus departs every 18 minutes, and Route C bus departs every 30 minutes. At what time will all three buses depart the central terminal together again?

Solution: Step 1: Identify that finding the next simultaneous departure requires finding the Least Common Multiple of the three cycle periods: $\text{LCM}(12, 18, 30)$.

Step 2: Prime factorize each time interval:

$$12 = 2^2 \times 3^1$$ $$18 = 2^1 \times 3^2$$ $$30 = 2^1 \times 3^1 \times 5^1$$

Step 3: Apply the maximum exponent formula across all three prime factorizations:

$$\text{LCM}(12, 18, 30) = 2^{\max(2,1,1)} \times 3^{\max(1,2,1)} \times 5^{\max(0,0,1)} = 2^2 \times 3^2 \times 5^1 = 4 \times 9 \times 5 = 180\text{ minutes}$$

Step 4: Convert minutes to hours:

$$180\text{ minutes} = 3\text{ hours}$$

Adding 3 hours to 6:00 AM gives 9:00 AM as the next simultaneous departure.