Study guide
12.2 Number Theory, Prime Factorization, GCF/LCM, and Exponents
California Subject Examinations for Teachers CSET
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12 Chapter 12: Number Sense, Operations, and Fractions
12.1 Number Classifications, Place Value, Properties of Operations, and Integers 12.2 Number Theory, Prime Factorization, GCF/LCM, and Exponents 12.3 Fractions, Decimals, Percentages, and Proportional Problem Solving
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12.2 Number Theory, Prime Factorization, GCF/LCM, and Exponents
Key Takeaways Explain This Topic AI
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 n n n digits.
Prime numbers are integers p > 1 p > 1 p > 1 with exactly two distinct positive divisors ( 1 1 1 and p p p ); 0 0 0 and 1 1 1 are neither prime nor composite, and 2 2 2 is the unique even prime.
The Fundamental Theorem of Arithmetic guarantees that every integer n > 1 n > 1 n > 1 has a unique canonical prime factorization N = p 1 a 1 p 2 a 2 … p k a k N = p_1^{a_1} p_2^{a_2} \dots p_k^{a_k} N = p 1 a 1 p 2 a 2 … p k a k , from which the total number of divisors is given by d ( N ) = ( a 1 + 1 ) ( a 2 + 1 ) … ( a k + 1 ) d(N) = (a_1+1)(a_2+1)\dots(a_k+1) d ( N ) = ( a 1 + 1 ) ( a 2 + 1 ) … ( 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 GCF ( a , b ) × LCM ( a , b ) = a × b .
Laws of exponents govern product, quotient, power-of-a-power, negative ( x − n = 1 / x n x^{-n} = 1/x^n x − n = 1/ x n ), and fractional exponent ( x m / n = x m n x^{m/n} = \sqrt[n]{x^m} x m / n = n x m ) operations, underpinning scientific notation ( a × 10 n a \times 10^n a × 1 0 n , where 1 ≤ ∣ a ∣ < 10 1 \le |a| < 10 1 ≤ ∣ 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 n n n is divisible by an integer d d d (written d ∣ n d \mid n d ∣ n ) if there exists an integer k k k such that n = d ⋅ k n = d \cdot k n = d ⋅ k , meaning division yields an integer quotient with a remainder of zero.
Comprehensive Divisibility Rules & Justifications
Divisor Divisibility Condition Algebraic Base-10 Justification / Mathematical Proof Worked Example
2 Last digit is even ( 0 , 2 , 4 , 6 , 8 0, 2, 4, 6, 8 0 , 2 , 4 , 6 , 8 ) Any integer N = 10 k + d 0 = 2 ( 5 k ) + d 0 N = 10k + d_0 = 2(5k) + d_0 N = 10 k + d 0 = 2 ( 5 k ) + d 0 . Since 2 ∣ 10 2 \mid 10 2 ∣ 10 , 2 ∣ N ⟺ 2 ∣ d 0 2 \mid N \iff 2 \mid d_0 2 ∣ N ⟺ 2 ∣ d 0 . 4 , 738 4,738 4 , 738 : 8 8 8 is even ⟹ \implies ⟹ Divisible
3 Sum of the digits is divisible by 3 3 3 10 n = ( 99 … 9 ) + 1 10^n = (99\dots9) + 1 1 0 n = ( 99 … 9 ) + 1 . Expanding N = a n 10 n + ⋯ + a 0 = [ multiple of 9 ] + ( a n + ⋯ + a 0 ) N = a_n10^n + \dots + a_0 = [\text{multiple of } 9] + (a_n + \dots + a_0) N = a n 1 0 n + ⋯ + a 0 = [ multiple of 9 ] + ( a n + ⋯ + a 0 ) . Thus 3 ∣ N ⟺ 3 ∣ ∑ a i 3 \mid N \iff 3 \mid \sum a_i 3 ∣ N ⟺ 3 ∣ ∑ a i . 576 576 576 : 5 + 7 + 6 = 18 5+7+6=18 5 + 7 + 6 = 18 ; 3 ∣ 18 ⟹ 3 \mid 18 \implies 3 ∣ 18 ⟹ Divisible
4 The number formed by the last two digits is divisible by 4 4 4 N = 100 k + ( 10 d 1 + d 0 ) N = 100k + (10d_1 + d_0) N = 100 k + ( 10 d 1 + d 0 ) . Since 100 = 4 × 25 100 = 4 \times 25 100 = 4 × 25 , 4 ∣ 100 k 4 \mid 100k 4 ∣ 100 k . Thus 4 ∣ N ⟺ 4 ∣ ( 10 d 1 + d 0 ) 4 \mid N \iff 4 \mid (10d_1 + d_0) 4 ∣ N ⟺ 4 ∣ ( 10 d 1 + d 0 ) . 15 , 724 15,724 15 , 724 : Last two digits 24 24 24 ; 4 ∣ 24 ⟹ 4 \mid 24 \implies 4 ∣ 24 ⟹ Divisible
5 Last digit is 0 0 0 or 5 5 5 N = 10 k + d 0 = 5 ( 2 k ) + d 0 N = 10k + d_0 = 5(2k) + d_0 N = 10 k + d 0 = 5 ( 2 k ) + d 0 . Since 5 ∣ 10 5 \mid 10 5 ∣ 10 , 5 ∣ N ⟺ d 0 ∈ { 0 , 5 } 5 \mid N \iff d_0 \in \{0, 5\} 5 ∣ N ⟺ d 0 ∈ { 0 , 5 } . 8 , 945 8,945 8 , 945 : Ends in 5 ⟹ 5 \implies 5 ⟹ Divisible
6 Divisible by both 2 2 2 and 3 3 3 Since gcd ( 2 , 3 ) = 1 \gcd(2,3) = 1 g cd ( 2 , 3 ) = 1 , a number is divisible by 2 × 3 = 6 2 \times 3 = 6 2 × 3 = 6 if and only if it is even and the sum of its digits is a multiple of 3. 438 438 438 : Even and 4 + 3 + 8 = 15 4+3+8=15 4 + 3 + 8 = 15 ( 3 ∣ 15 3 \mid 15 3 ∣ 15 ) ⟹ \implies ⟹ Divisible
8 The number formed by the last three digits is divisible by 8 8 8 N = 1 , 000 k + ( 100 d 2 + 10 d 1 + d 0 ) N = 1,000k + (100d_2 + 10d_1 + d_0) N = 1 , 000 k + ( 100 d 2 + 10 d 1 + d 0 ) . Since 1 , 000 = 8 × 125 1,000 = 8 \times 125 1 , 000 = 8 × 125 , 8 ∣ 1 , 000 k 8 \mid 1,000k 8 ∣ 1 , 000 k . Thus 8 ∣ N ⟺ 8 ∣ ( last 3 digits ) 8 \mid N \iff 8 \mid (\text{last 3 digits}) 8 ∣ N ⟺ 8 ∣ ( last 3 digits ) . 35 , 128 35,128 35 , 128 : Last three digits 128 128 128 ; 128 ÷ 8 = 16 ⟹ 128 \div 8 = 16 \implies 128 ÷ 8 = 16 ⟹ Divisible
9 Sum of the digits is divisible by 9 9 9 N = [ multiple of 9 ] + ∑ a i N = [\text{multiple of } 9] + \sum a_i N = [ multiple of 9 ] + ∑ a i . Since the first term is divisible by 9, 9 ∣ N ⟺ 9 ∣ ∑ a i 9 \mid N \iff 9 \mid \sum a_i 9 ∣ N ⟺ 9 ∣ ∑ a i . 4 , 896 4,896 4 , 896 : 4 + 8 + 9 + 6 = 27 4+8+9+6=27 4 + 8 + 9 + 6 = 27 ; 9 ∣ 27 ⟹ 9 \mid 27 \implies 9 ∣ 27 ⟹ Divisible
10 Last digit is 0 0 0 N = 10 k + d 0 N = 10k + d_0 N = 10 k + d 0 . Thus 10 ∣ N ⟺ d 0 = 0 10 \mid N \iff d_0 = 0 10 ∣ N ⟺ d 0 = 0 . 7 , 390 7,390 7 , 390 : Ends in 0 ⟹ 0 \implies 0 ⟹ Divisible
- 2. Prime and Composite Numbers
- Definitions and Core Distinctions
Prime Number: An integer p > 1 p > 1 p > 1 that has exactly two distinct positive divisors : 1 1 1 and p p p (e.g., 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , … 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, \dots 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , … ).
Composite Number: An integer n > 1 n > 1 n > 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 4 , 6 , 8 , 9 , 10 , 12 , 14 , 15 , … ).
The Numbers 0 and 1:
1 1 1 has only one positive divisor ( 1 1 1 itself). Therefore, 1 1 1 is neither prime nor composite .
0 0 0 has infinitely many divisors ( 0 = 0 × k 0 = 0 \times k 0 = 0 × k for all k ∈ Z + k \in \mathbb{Z}^+ k ∈ Z + ), but division by zero is undefined. 0 0 0 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 N N N :
List all consecutive integers from 2 2 2 to N N N .
Identify the first uncrossed number ( 2 2 2 ) as prime; cross out all its multiples ( 4 , 6 , 8 , … 4, 6, 8, \dots 4 , 6 , 8 , … ).
Advance to the next uncrossed number ( 3 3 3 ), mark it prime, and cross out all its remaining multiples ( 9 , 15 , 21 , … 9, 15, 21, \dots 9 , 15 , 21 , … ).
Repeat this process for each successive prime p p p up to N \sqrt{N} N .
Stopping Criterion: Any composite number n ≤ N n \le N n ≤ N must possess at least one prime factor p ≤ N p \le \sqrt{N} p ≤ N . If an integer has no prime factors ≤ N \le \sqrt{N} ≤ N , it is guaranteed to be prime.
- 3. Prime Factorization & The Fundamental Theorem of Arithmetic
- The Fundamental Theorem of Arithmetic
Every integer n > 1 n > 1 n > 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
Factor Tree Method: Decompose the composite number into any two factor branches and continue subdividing each branch until all terminal leaves are prime numbers.
Continuous Division (Ladder / Inverted Division) Method: Successively divide the number by the smallest possible prime divisor until the final quotient is 1 1 1 .
Canonical Prime Power Factorization Form
Every integer N > 1 N > 1 N > 1 can be expressed uniquely as: N = p 1 a 1 ⋅ p 2 a 2 … p k a k N = p_1^{a_1} \cdot p_2^{a_2} \dots p_k^{a_k} N = p 1 a 1 ⋅ p 2 a 2 … p k a k
where p 1 < p 2 < ⋯ < p k p_1 < p_2 < \dots < p_k p 1 < p 2 < ⋯ < p k are distinct prime numbers and a i ∈ Z + a_i \in \mathbb{Z}^+ a i ∈ Z + are positive integer exponents.
Example: 360 = 2 3 × 3 2 × 5 1 360 = 2^3 \times 3^2 \times 5^1 360 = 2 3 × 3 2 × 5 1 .
Formula for the Total Number of Positive Divisors
If N = p 1 a 1 p 2 a 2 … p k a k N = p_1^{a_1} p_2^{a_2} \dots p_k^{a_k} N = p 1 a 1 p 2 a 2 … p k a k , the total number of positive factors (divisors) of N N N , denoted d ( N ) d(N) d ( N ) , is given by the product of each exponent incremented by 1 1 1 : d ( N ) = ( a 1 + 1 ) ( a 2 + 1 ) ( a 3 + 1 ) … ( a k + 1 ) d(N) = (a_1 + 1)(a_2 + 1)(a_3 + 1)\dots(a_k + 1) d ( N ) = ( a 1 + 1 ) ( a 2 + 1 ) ( a 3 + 1 ) … ( a k + 1 )
Derivation: Any divisor of N N N must have the form p 1 e 1 p 2 e 2 … p k e k p_1^{e_1} p_2^{e_2} \dots p_k^{e_k} p 1 e 1 p 2 e 2 … p k e k , where each exponent e i e_i e i can independently take any integer value from 0 0 0 up to a i a_i a i (yielding a i + 1 a_i + 1 a i + 1 independent choices).
Worked Example for 360 = 2 3 ⋅ 3 2 ⋅ 5 1 360 = 2^3 \cdot 3^2 \cdot 5^1 360 = 2 3 ⋅ 3 2 ⋅ 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}. d ( 360 ) = ( 3 + 1 ) ( 2 + 1 ) ( 1 + 1 ) = 4 × 3 × 2 = 24 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 ) = ∏ p i min ( a i , b i ) \text{GCF}(a, b) = \prod p_i^{\min(a_i, b_i)} GCF ( a , b ) = ∏ p i m i n ( a i , b i )
Method 2 (Euclidean Algorithm): Based on the principle that gcd ( a , b ) = gcd ( b , a m o d b ) \gcd(a, b) = \gcd(b, a \bmod b) g cd ( a , b ) = g cd ( b , a mod b ) where a > b a > b a > b :
Divide a a a by b b b to find quotient q q q and remainder r r r : a = b q + r a = bq + r a = b q + r .
Replace ( a , b ) (a, b) ( a , b ) with ( b , r ) (b, r) ( b , r ) and repeat until remainder r = 0 r = 0 r = 0 . The last non-zero remainder is gcd ( a , b ) \gcd(a,b) g cd ( 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 ) = ∏ p i max ( a i , b i ) \text{LCM}(a, b) = \prod p_i^{\max(a_i, b_i)} LCM ( a , b ) = ∏ p i m a x ( a i , b i )
Fundamental GCF-LCM Identity
For any two positive integers a a a and b b b : GCF ( a , b ) × LCM ( a , b ) = a × b \text{GCF}(a, b) \times \text{LCM}(a, b) = a \times b GCF ( a , b ) × LCM ( a , b ) = a × b
Proof: For each prime p i p_i p i , min ( a i , b i ) + max ( a i , b i ) = a i + b i \min(a_i, b_i) + \max(a_i, b_i) = a_i + b_i min ( a i , b i ) + max ( a i , b i ) = a i + b i . Exponent addition yields ∏ p i min ( a i , b i ) × ∏ p i max ( a i , b i ) = ∏ p i a i + b i = 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 ∏ p i m i n ( a i , b i ) × ∏ p i m a x ( a i , b i ) = ∏ p i a i + b i = a ⋅ b .
GCF vs. LCM Problem-Solving Matrix
Concept Mathematical Definition Prime Factor Selection Rule Key Problem Contexts & Question Phrases Exemplar Problem
GCF (Greatest Common Factor) Largest integer dividing all numbers evenly Product of lowest powers of common primes "Split into equal groups without leftover", "largest identical bouquets/kits", "greatest square tile size" Partitioning 72 72 72 pencils and 108 108 108 markers into identical bags ⟹ GCF ( 72 , 108 ) = 36 \implies \text{GCF}(72, 108) = 36 ⟹ GCF ( 72 , 108 ) = 36 bags
LCM (Least Common Multiple) Smallest integer that all numbers divide Product of highest powers of all primes "Repeat at the same time", "next synchronous occurrence", "minimum common length", "common denominators" Two bells ring every 12 12 12 min and 18 18 18 min; when will they ring together? ⟹ LCM ( 12 , 18 ) = 36 \implies \text{LCM}(12, 18) = 36 ⟹ LCM ( 12 , 18 ) = 36 min
5. Exponent Laws and Scientific Notation
An exponent denotes repeated multiplication: for n ∈ Z + n \in \mathbb{Z}^+ n ∈ Z + , x n = x ⋅ x … x ⏟ n times x^n = \underbrace{x \cdot x \dots x}_{n \text{ times}} x n = n times x ⋅ x … x , where x x x is the base and n n n is the power/exponent.
The Fundamental Laws of Exponents
Exponent Law Mathematical Identity Conceptual Justification Arithmetic Exemplar
Product Rule x a ⋅ x b = x a + b x^a \cdot x^b = x^{a+b} x a ⋅ 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} ( x … x ) a ⋅ ( x … x ) b = ( x … x ) a + b 2 3 ⋅ 2 4 = 2 3 + 4 = 2 7 = 128 2^3 \cdot 2^4 = 2^{3+4} = 2^7 = 128 2 3 ⋅ 2 4 = 2 3 + 4 = 2 7 = 128
Quotient Rule x a x b = x a − b \frac{x^a}{x^b} = x^{a-b} x b x a = x a − b ( x ≠ 0 x \ne 0 x = 0 ) Canceling b b b factors of x x x from a a a factors 5 6 5 2 = 5 6 − 2 = 5 4 = 625 \frac{5^6}{5^2} = 5^{6-2} = 5^4 = 625 5 2 5 6 = 5 6 − 2 = 5 4 = 625
Power of a Power ( x a ) b = x a ⋅ b (x^a)^b = x^{a \cdot b} ( x a ) b = x a ⋅ b Multiplying b b b groups of x a x^a x a ( 3 2 ) 3 = 3 2 × 3 = 3 6 = 729 (3^2)^3 = 3^{2 \times 3} = 3^6 = 729 ( 3 2 ) 3 = 3 2 × 3 = 3 6 = 729
Power of a Product ( x y ) a = x a y a (xy)^a = x^a y^a ( x y ) a = x a y a Regrouping factors via Commutative property ( 2 × 5 ) 3 = 2 3 × 5 3 = 8 × 125 = 1 , 000 (2 \times 5)^3 = 2^3 \times 5^3 = 8 \times 125 = 1,000 ( 2 × 5 ) 3 = 2 3 × 5 3 = 8 × 125 = 1 , 000
Power of a Quotient ( x y ) a = x a y a \left(\frac{x}{y}\right)^a = \frac{x^a}{y^a} ( y x ) a = y a x a ( y ≠ 0 y \ne 0 y = 0 ) Multiplying fractions numerator-by-numerator ( 2 3 ) 4 = 2 4 3 4 = 16 81 \left(\frac{2}{3}\right)^4 = \frac{2^4}{3^4} = \frac{16}{81} ( 3 2 ) 4 = 3 4 2 4 = 81 16
Zero Exponent x 0 = 1 x^0 = 1 x 0 = 1 ( x ≠ 0 x \ne 0 x = 0 ) By quotient rule: 1 = x n x n = x n − n = x 0 1 = \frac{x^n}{x^n} = x^{n-n} = x^0 1 = x n x n = x n − n = x 0 427 0 = 1 427^0 = 1 42 7 0 = 1 ; ( − 8.5 ) 0 = 1 (-8.5)^0 = 1 ( − 8.5 ) 0 = 1
Negative Exponent x − n = 1 x n x^{-n} = \frac{1}{x^n} x − n = x n 1 ( x ≠ 0 x \ne 0 x = 0 ) By quotient rule: x 0 x n = x 0 − n = x − n = 1 x n \frac{x^0}{x^n} = x^{0-n} = x^{-n} = \frac{1}{x^n} x n x 0 = x 0 − n = x − n = x n 1 4 − 2 = 1 4 2 = 1 16 4^{-2} = \frac{1}{4^2} = \frac{1}{16} 4 − 2 = 4 2 1 = 16 1
Rational Exponent x m / n = x m n = ( x n ) m x^{m/n} = \sqrt[n]{x^m} = (\sqrt[n]{x})^m x m / n = n x m = ( n x ) m ( x 1 / n ) n = x ( 1 / n ) ⋅ n = x 1 = x (x^{1/n})^n = x^{(1/n) \cdot n} = x^1 = x ( x 1/ n ) n = x ( 1/ n ) ⋅ n = x 1 = x 27 2 / 3 = ( 27 3 ) 2 = 3 2 = 9 27^{2/3} = (\sqrt[3]{27})^2 = 3^2 = 9 2 7 2/3 = ( 3 27 ) 2 = 3 2 = 9
Scientific Notation
Scientific notation expresses very large or very small real numbers compactly in the standardized form: a × 10 n a \times 10^n a × 1 0 n
where 1 ≤ ∣ a ∣ < 10 1 \le |a| < 10 1 ≤ ∣ a ∣ < 10 ( a a a is the coefficient/mantissa) and n ∈ Z n \in \mathbb{Z} n ∈ Z is an integer exponent.
Converting Standard to Scientific:
For numbers ≥ 10 \ge 10 ≥ 10 , shift the decimal point n n n places left ( n > 0 n > 0 n > 0 ): 4 , 500 , 000 = 4.5 × 10 6 4,500,000 = 4.5 \times 10^6 4 , 500 , 000 = 4.5 × 1 0 6 .
For numbers < 1 < 1 < 1 , shift the decimal point n n n places right ( n < 0 n < 0 n < 0 ): 0.00038 = 3.8 × 10 − 4 0.00038 = 3.8 \times 10^{-4} 0.00038 = 3.8 × 1 0 − 4 .
Multiplication and Division in Scientific Notation: ( a × 10 m ) × ( b × 10 n ) = ( a ⋅ b ) × 10 m + n (a \times 10^m) \times (b \times 10^n) = (a \cdot b) \times 10^{m+n} ( a × 1 0 m ) × ( b × 1 0 n ) = ( a ⋅ b ) × 1 0 m + n a × 10 m b × 10 n = ( a b ) × 10 m − n \frac{a \times 10^m}{b \times 10^n} = \left(\frac{a}{b}\right) \times 10^{m-n} b × 1 0 n a × 1 0 m = ( b a ) × 1 0 m − n
Adjustment Rule: If the resulting coefficient a ⋅ b a \cdot b a ⋅ b or a / b a/b a / b falls outside [ 1 , 10 ) [1, 10) [ 1 , 10 ) , re-normalize the coefficient and adjust exponent n n n accordingly.