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12.2 Number Theory, Prime Factorization, GCF/LCM, and Exponents

Key Takeaways

  • Divisibility rules are mathematically justified by base-ten place value decompositions; numbers are divisible by 3 or 9 if the sum of their digits is divisible by 3 or 9, and by powers of 2 (2, 4, 8) based on their terminal nn digits.

  • Prime numbers are integers p>1p > 1 with exactly two distinct positive divisors (11 and pp); 00 and 11 are neither prime nor composite, and 22 is the unique even prime.

  • The Fundamental Theorem of Arithmetic guarantees that every integer n>1n > 1 has a unique canonical prime factorization N=p1a1p2a2…pkakN = p_1^{a_1} p_2^{a_2} \dots p_k^{a_k}, from which the total number of divisors is given by d(N)=(a1+1)(a2+1)…(ak+1)d(N) = (a_1+1)(a_2+1)\dots(a_k+1).

  • The Greatest Common Factor (GCF) is calculated by taking the lowest power of common prime factors, while the Least Common Multiple (LCM) takes the highest power of all appearing prime factors, satisfying GCF(a,b)×LCM(a,b)=a×b\text{GCF}(a,b) \times \text{LCM}(a,b) = a \times b.

  • Laws of exponents govern product, quotient, power-of-a-power, negative (x−n=1/xnx^{-n} = 1/x^n), and fractional exponent (xm/n=xmnx^{m/n} = \sqrt[n]{x^m}) operations, underpinning scientific notation (a×10na \times 10^n, where 1≤∣a∣<101 \le |a| < 10).

Last updated: August 2026

12.2 Number Theory, Prime Factorization, GCF/LCM, and Exponents

CSET Focus: Number theory questions on CSET Subtest II evaluate your ability to analyze number relationships beyond rote calculation. You are expected to know the algebraic justifications behind divisibility rules, construct canonical prime power factorizations, solve real-world problem scenarios requiring GCF vs. LCM, apply exponent rules across integer and rational powers, and perform operations in scientific notation.


1. Divisibility Rules and Mathematical Justifications

A positive integer nn is divisible by an integer dd (written d∣nd \mid n) if there exists an integer kk such that n=d⋅kn = d \cdot k, meaning division yields an integer quotient with a remainder of zero.

Comprehensive Divisibility Rules & Justifications

DivisorDivisibility ConditionAlgebraic Base-10 Justification / Mathematical ProofWorked Example
2Last digit is even (0,2,4,6,80, 2, 4, 6, 8)Any integer N=10k+d0=2(5k)+d0N = 10k + d_0 = 2(5k) + d_0. Since 2∣102 \mid 10, 2∣N  ⟺  2∣d02 \mid N \iff 2 \mid d_0.4,7384,738: 88 is even   ⟹  \implies Divisible
3Sum of the digits is divisible by 3310n=(99…9)+110^n = (99\dots9) + 1. Expanding N=an10n+⋯+a0=[multiple of 9]+(an+⋯+a0)N = a_n10^n + \dots + a_0 = [\text{multiple of } 9] + (a_n + \dots + a_0). Thus 3∣N  ⟺  3∣∑ai3 \mid N \iff 3 \mid \sum a_i.576576: 5+7+6=185+7+6=18; 3∣18  ⟹  3 \mid 18 \implies Divisible
4The number formed by the last two digits is divisible by 44N=100k+(10d1+d0)N = 100k + (10d_1 + d_0). Since 100=4×25100 = 4 \times 25, 4∣100k4 \mid 100k. Thus 4∣N  ⟺  4∣(10d1+d0)4 \mid N \iff 4 \mid (10d_1 + d_0).15,72415,724: Last two digits 2424; 4∣24  ⟹  4 \mid 24 \implies Divisible
5Last digit is 00 or 55N=10k+d0=5(2k)+d0N = 10k + d_0 = 5(2k) + d_0. Since 5∣105 \mid 10, 5∣N  ⟺  d0∈{0,5}5 \mid N \iff d_0 \in \{0, 5\}.8,9458,945: Ends in 5  ⟹  5 \implies Divisible
6Divisible by both 22 and 33Since gcd⁡(2,3)=1\gcd(2,3) = 1, a number is divisible by 2×3=62 \times 3 = 6 if and only if it is even and the sum of its digits is a multiple of 3.438438: Even and 4+3+8=154+3+8=15 (3∣153 \mid 15)   ⟹  \implies Divisible
8The number formed by the last three digits is divisible by 88N=1,000k+(100d2+10d1+d0)N = 1,000k + (100d_2 + 10d_1 + d_0). Since 1,000=8×1251,000 = 8 \times 125, 8∣1,000k8 \mid 1,000k. Thus 8∣N  ⟺  8∣(last 3 digits)8 \mid N \iff 8 \mid (\text{last 3 digits}).35,12835,128: Last three digits 128128; 128÷8=16  ⟹  128 \div 8 = 16 \implies Divisible
9Sum of the digits is divisible by 99N=[multiple of 9]+∑aiN = [\text{multiple of } 9] + \sum a_i. Since the first term is divisible by 9, 9∣N  ⟺  9∣∑ai9 \mid N \iff 9 \mid \sum a_i.4,8964,896: 4+8+9+6=274+8+9+6=27; 9∣27  ⟹  9 \mid 27 \implies Divisible
10Last digit is 00N=10k+d0N = 10k + d_0. Thus 10∣N  ⟺  d0=010 \mid N \iff d_0 = 0.7,3907,390: Ends in 0  ⟹  0 \implies Divisible

2. Prime and Composite Numbers

Definitions and Core Distinctions

  • Prime Number: An integer p>1p > 1 that has exactly two distinct positive divisors: 11 and pp (e.g., 2,3,5,7,11,13,17,19,23,29,…2, 3, 5, 7, 11, 13, 17, 19, 23, 29, \dots).
  • Composite Number: An integer n>1n > 1 that has more than two distinct positive divisors (e.g., 4,6,8,9,10,12,14,15,…4, 6, 8, 9, 10, 12, 14, 15, \dots).
  • The Numbers 0 and 1:
    • 11 has only one positive divisor (11 itself). Therefore, 11 is neither prime nor composite.
    • 00 has infinitely many divisors (0=0×k0 = 0 \times k for all k∈Z+k \in \mathbb{Z}^+), but division by zero is undefined. 00 is neither prime nor composite.
    • The Number 2: The smallest prime number and the only even prime number in existence.

The Sieve of Eratosthenes

An ancient algorithmic procedure for isolating all prime numbers up to a specified integer NN:

  1. List all consecutive integers from 22 to NN.
  2. Identify the first uncrossed number (22) as prime; cross out all its multiples (4,6,8,…4, 6, 8, \dots).
  3. Advance to the next uncrossed number (33), mark it prime, and cross out all its remaining multiples (9,15,21,…9, 15, 21, \dots).
  4. Repeat this process for each successive prime pp up to N\sqrt{N}.
  5. Stopping Criterion: Any composite number n≤Nn \le N must possess at least one prime factor p≤Np \le \sqrt{N}. If an integer has no prime factors ≤N\le \sqrt{N}, it is guaranteed to be prime.

3. Prime Factorization & The Fundamental Theorem of Arithmetic

The Fundamental Theorem of Arithmetic

Every integer n>1n > 1 either is a prime number itself or can be represented uniquely as a product of prime numbers, up to the order of the prime factors.

Factorization Methods

  1. Factor Tree Method: Decompose the composite number into any two factor branches and continue subdividing each branch until all terminal leaves are prime numbers.
  2. Continuous Division (Ladder / Inverted Division) Method: Successively divide the number by the smallest possible prime divisor until the final quotient is 11.

Canonical Prime Power Factorization Form

Every integer N>1N > 1 can be expressed uniquely as:

N=p1a1⋅p2a2…pkakN = p_1^{a_1} \cdot p_2^{a_2} \dots p_k^{a_k}

where p1<p2<⋯<pkp_1 < p_2 < \dots < p_k are distinct prime numbers and ai∈Z+a_i \in \mathbb{Z}^+ are positive integer exponents.

  • Example: 360=23×32×51360 = 2^3 \times 3^2 \times 5^1.

Formula for the Total Number of Positive Divisors

If N=p1a1p2a2…pkakN = p_1^{a_1} p_2^{a_2} \dots p_k^{a_k}, the total number of positive factors (divisors) of NN, denoted d(N)d(N), is given by the product of each exponent incremented by 11:

d(N)=(a1+1)(a2+1)(a3+1)…(ak+1)d(N) = (a_1 + 1)(a_2 + 1)(a_3 + 1)\dots(a_k + 1)

Derivation: Any divisor of NN must have the form p1e1p2e2…pkekp_1^{e_1} p_2^{e_2} \dots p_k^{e_k}, where each exponent eie_i can independently take any integer value from 00 up to aia_i (yielding ai+1a_i + 1 independent choices).

  • Worked Example for 360=23⋅32⋅51360 = 2^3 \cdot 3^2 \cdot 5^1:
d(360)=(3+1)(2+1)(1+1)=4×3×2=24 positive divisors.d(360) = (3 + 1)(2 + 1)(1 + 1) = 4 \times 3 \times 2 = 24 \text{ positive divisors}.

4. Greatest Common Factor (GCF) and Least Common Multiple (LCM)

Greatest Common Factor (GCF / GCD)

The Greatest Common Factor of two or more integers is the largest positive integer that divides each of the numbers without a remainder.

  • Method 1 (Prime Factorization): Write the canonical prime factorizations of both numbers. The GCF is the product of the lowest power of each common prime factor:
GCF(a,b)=∏pimin⁡(ai,bi)\text{GCF}(a, b) = \prod p_i^{\min(a_i, b_i)}
  • Method 2 (Euclidean Algorithm): Based on the principle that gcd⁡(a,b)=gcd⁡(b,a mod b)\gcd(a, b) = \gcd(b, a \bmod b) where a>ba > b:
    • Divide aa by bb to find quotient qq and remainder rr: a=bq+ra = bq + r.
    • Replace (a,b)(a, b) with (b,r)(b, r) and repeat until remainder r=0r = 0. The last non-zero remainder is gcd⁡(a,b)\gcd(a,b).

Least Common Multiple (LCM)

The Least Common Multiple of two or more integers is the smallest positive integer that is divisible by each of the given numbers.

  • Method (Prime Factorization): The LCM is the product of the highest power of all prime factors occurring in either factorization:
LCM(a,b)=∏pimax⁡(ai,bi)\text{LCM}(a, b) = \prod p_i^{\max(a_i, b_i)}

Fundamental GCF-LCM Identity

For any two positive integers aa and bb:

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

Proof: For each prime pip_i, min⁡(ai,bi)+max⁡(ai,bi)=ai+bi\min(a_i, b_i) + \max(a_i, b_i) = a_i + b_i. Exponent addition yields ∏pimin⁡(ai,bi)×∏pimax⁡(ai,bi)=∏piai+bi=a⋅b\prod p_i^{\min(a_i, b_i)} \times \prod p_i^{\max(a_i, b_i)} = \prod p_i^{a_i + b_i} = a \cdot b.

GCF vs. LCM Problem-Solving Matrix

ConceptMathematical DefinitionPrime Factor Selection RuleKey Problem Contexts & Question PhrasesExemplar Problem
GCF (Greatest Common Factor)Largest integer dividing all numbers evenlyProduct of lowest powers of common primes"Split into equal groups without leftover", "largest identical bouquets/kits", "greatest square tile size"Partitioning 7272 pencils and 108108 markers into identical bags   ⟹  GCF(72,108)=36\implies \text{GCF}(72, 108) = 36 bags
LCM (Least Common Multiple)Smallest integer that all numbers divideProduct of highest powers of all primes"Repeat at the same time", "next synchronous occurrence", "minimum common length", "common denominators"Two bells ring every 1212 min and 1818 min; when will they ring together?   ⟹  LCM(12,18)=36\implies \text{LCM}(12, 18) = 36 min

5. Exponent Laws and Scientific Notation

An exponent denotes repeated multiplication: for n∈Z+n \in \mathbb{Z}^+, xn=x⋅x…x⏟n timesx^n = \underbrace{x \cdot x \dots x}_{n \text{ times}}, where xx is the base and nn is the power/exponent.

The Fundamental Laws of Exponents

Exponent LawMathematical IdentityConceptual JustificationArithmetic Exemplar
Product Rulexa⋅xb=xa+bx^a \cdot x^b = x^{a+b}(x…x)a⋅(x…x)b=(x…x)a+b(x \dots x)_a \cdot (x \dots x)_b = (x \dots x)_{a+b}23⋅24=23+4=27=1282^3 \cdot 2^4 = 2^{3+4} = 2^7 = 128
Quotient Rulexaxb=xa−b\frac{x^a}{x^b} = x^{a-b} (x≠0x \ne 0)Canceling bb factors of xx from aa factors5652=56−2=54=625\frac{5^6}{5^2} = 5^{6-2} = 5^4 = 625
Power of a Power(xa)b=xa⋅b(x^a)^b = x^{a \cdot b}Multiplying bb groups of xax^a(32)3=32×3=36=729(3^2)^3 = 3^{2 \times 3} = 3^6 = 729
Power of a Product(xy)a=xaya(xy)^a = x^a y^aRegrouping factors via Commutative property(2×5)3=23×53=8×125=1,000(2 \times 5)^3 = 2^3 \times 5^3 = 8 \times 125 = 1,000
Power of a Quotient(xy)a=xaya\left(\frac{x}{y}\right)^a = \frac{x^a}{y^a} (y≠0y \ne 0)Multiplying fractions numerator-by-numerator(23)4=2434=1681\left(\frac{2}{3}\right)^4 = \frac{2^4}{3^4} = \frac{16}{81}
Zero Exponentx0=1x^0 = 1 (x≠0x \ne 0)By quotient rule: 1=xnxn=xn−n=x01 = \frac{x^n}{x^n} = x^{n-n} = x^04270=1427^0 = 1; (−8.5)0=1(-8.5)^0 = 1
Negative Exponentx−n=1xnx^{-n} = \frac{1}{x^n} (x≠0x \ne 0)By quotient rule: x0xn=x0−n=x−n=1xn\frac{x^0}{x^n} = x^{0-n} = x^{-n} = \frac{1}{x^n}4−2=142=1164^{-2} = \frac{1}{4^2} = \frac{1}{16}
Rational Exponentxm/n=xmn=(xn)mx^{m/n} = \sqrt[n]{x^m} = (\sqrt[n]{x})^m(x1/n)n=x(1/n)⋅n=x1=x(x^{1/n})^n = x^{(1/n) \cdot n} = x^1 = x272/3=(273)2=32=927^{2/3} = (\sqrt[3]{27})^2 = 3^2 = 9

Scientific Notation

Scientific notation expresses very large or very small real numbers compactly in the standardized form:

a×10na \times 10^n

where 1≤∣a∣<101 \le |a| < 10 (aa is the coefficient/mantissa) and n∈Zn \in \mathbb{Z} is an integer exponent.

  • Converting Standard to Scientific:
    • For numbers ≥10\ge 10, shift the decimal point nn places left (n>0n > 0): 4,500,000=4.5×1064,500,000 = 4.5 \times 10^6.
    • For numbers <1< 1, shift the decimal point nn places right (n<0n < 0): 0.00038=3.8×10−40.00038 = 3.8 \times 10^{-4}.
  • Multiplication and Division in Scientific Notation:
(a×10m)×(b×10n)=(a⋅b)×10m+n(a \times 10^m) \times (b \times 10^n) = (a \cdot b) \times 10^{m+n} a×10mb×10n=(ab)×10m−n\frac{a \times 10^m}{b \times 10^n} = \left(\frac{a}{b}\right) \times 10^{m-n}

Adjustment Rule: If the resulting coefficient a⋅ba \cdot b or a/ba/b falls outside [1,10)[1, 10), re-normalize the coefficient and adjust exponent nn accordingly.

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Prime Factor Selection for GCF, LCM, and Product Identity Verification
Test Your Knowledge

What is the total number of positive integer divisors for the composite number N = 360?

A

12

B

18

C

24

D

36

Test Your Knowledge

A middle school track coach needs to prepare gift bags for student athletes. The coach has 72 energy bars and 108 bottles of electrolyte drinks. If every bag must contain the exact same number of energy bars and the exact same number of electrolyte bottles with none left over, what is the MAXIMUM number of identical gift bags the coach can assemble?

A

36 gift bags

B

18 gift bags

C

72 gift bags

D

216 gift bags

Test Your Knowledge

Evaluate the following expression and express the final result in standard scientific notation:

[(4.8 × 10⁸) × (3.0 × 10⁻³)] / (6.0 × 10²)

A

2.4 × 10⁴

B

2.4 × 10²

C

2.4 × 10⁷

D

2.4 × 10³

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