CEUET Mathematics — Arithmetic — Multiples, Factors, PEMDAS, Fractions & DecimalsDetailed Explanation
Want to really understand Arithmetic — Multiples, Factors, PEMDAS, Fractions & Decimals before tackling CEUET Mathematics questions? This detailed explanation breaks down every key concept, shows you why it matters for the CEUET 2026, and walks through the reasoning Centro Escolar University expects on high-difficulty questions.
Exam context
The Centro Escolar University Entrance Test is conducted by Centro Escolar University and is scheduled for Q3–Q4 2026. The Mathematics subtest is marked as "Core" in the official pattern, and Arithmetic — Multiples, Factors, PEMDAS, Fractions & Decimals appears in position 1st of 9 in the CEUET Mathematics review rotation. Passing mark: Competitive overall score. Recent CEUET 2026 papers have drawn roughly a meaningful share of questions from this subject.
Arithmetic — Multiples, Factors, PEMDAS, Fractions & Decimals - Detailed explanation
Arithmetic forms the foundation of all mathematics and is crucial for success in college entrance exams like UPCAT. This chapter covers essential arithmetic concepts including multiples and factors, order of operations (PEMDAS), and working with fractions and decimals. These topics appear frequently in entrance exams and are building blocks for more advanced mathematical concepts. Understanding these fundamentals will help you solve problems quickly and accurately during time-pressured exams.
Concepts
Multiples and Factors
A multiple of a number is obtained by multiplying that number by any positive integer. For example, multiples of 5 are: 5, 10, 15, 20, 25, ... A factor of a number is any positive integer that divides the number evenly (with no remainder). For example, factors of 12 are: 1, 2, 3, 4, 6, 12. Understanding the relationship between multiples and factors is essential: if m is a factor of n, then n is a multiple of m.
Examples
We test each number systematically. A more efficient method is to find factor pairs: 1×24, 2×12, 3×8, 4×6. This ensures we don't miss any factors.
Scenario
Find all factors of 24
Solution
Step 1: Start with 1 (always a factor) Step 2: Check each number from 1 to 24 - 24 ÷ 1 = 24 ✓ - 24 ÷ 2 = 12 ✓ - 24 ÷ 3 = 8 ✓ - 24 ÷ 4 = 6 ✓ - 24 ÷ 5 = 4.8 ✗ - 24 ÷ 6 = 4 ✓ - 24 ÷ 8 = 3 ✓ - 24 ÷ 12 = 2 ✓ - 24 ÷ 24 = 1 ✓ Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24
Multiples follow a pattern - they increase by the original number each time. This pattern helps in mental calculations and recognizing multiples quickly.
Scenario
Find the first 6 multiples of 7
Solution
Step 1: Multiply 7 by consecutive positive integers 7 × 1 = 7 7 × 2 = 14 7 × 3 = 21 7 × 4 = 28 7 × 5 = 35 7 × 6 = 42 First 6 multiples of 7: 7, 14, 21, 28, 35, 42
Applications
- Simplifying fractions by finding common factors
- Solving word problems involving repeated patterns
- Finding LCD and GCD for fraction operations
- Determining if numbers are divisible by others
Misconceptions
- Thinking 0 is a factor of any number (0 can never be a factor)
- Confusing multiples and factors (24 is a multiple of 6, not the other way around)
- Forgetting that 1 and the number itself are always factors
Related Concepts
- Prime numbers
- LCM and GCD
- Divisibility rules
- Prime factorization
Common Exam Questions
Example
List all factors of 36. Answer: Find pairs (1,36), (2,18), (3,12), (4,9), (6,6) → Factors: 1, 2, 3, 4, 6, 9, 12, 18, 36
Approach
Use factor pairs method for efficiency
Question Type
Factor listing
Example
Is 84 a multiple of 12? Answer: 84 ÷ 12 = 7 with no remainder, so YES
Approach
Check divisibility without remainder
Question Type
Multiple identification
Key Points To Remember
- The smallest multiple of any number is the number itself
- The largest factor of any number is the number itself
- 1 is a factor of every positive integer
- If m is a factor of n, then n is a multiple of m
- Every number has at least two factors: 1 and itself
Prime Numbers and Composite Numbers
A prime number has exactly two factors: 1 and itself. Examples include 2, 3, 5, 7, 11, 13, 17, 19, 23. A composite number has more than two factors. Examples include 4, 6, 8, 9, 10, 12. Note that 0 and 1 are neither prime nor composite. The number 2 is the only even prime number. Understanding prime numbers is crucial for prime factorization and simplifying fractions.
Examples
We only need to check prime factors up to the square root because if a number has a factor larger than its square root, it must also have a corresponding factor smaller than the square root.
Scenario
Determine if 97 is prime
Solution
Step 1: Find √97 ≈ 9.85, so check primes up to 9 Step 2: Test divisibility by 2, 3, 5, 7 - 97 ÷ 2 = 48.5 (not divisible) - 97 ÷ 3 = 32.33... (not divisible) - 97 ÷ 5 = 19.4 (not divisible) - 97 ÷ 7 = 13.86... (not divisible) Step 3: Since no prime up to √97 divides 97, it is PRIME
Both methods give the same result. The factor tree method is visual, while the division method is systematic. Use whichever method you find easier to follow.
Scenario
Find the prime factorization of 60
Solution
Method 1 - Factor Tree: 60 = 4 × 15 4 = 2 × 2 15 = 3 × 5 Therefore: 60 = 2² × 3 × 5 Method 2 - Division: 60 ÷ 2 = 30 30 ÷ 2 = 15 15 ÷ 3 = 5 5 ÷ 5 = 1 Therefore: 60 = 2² × 3 × 5
Applications
- Simplifying fractions to lowest terms
- Finding LCM and GCD efficiently
- Cryptography and number theory
- Factoring algebraic expressions
Misconceptions
- Thinking 1 is prime (1 has only one factor, not two)
- Forgetting that 2 is prime (it's the only even prime)
- Not checking all primes up to the square root when testing primality
Related Concepts
- Factors and multiples
- LCM and GCD
- Divisibility rules
- Number theory
Common Exam Questions
Example
Is 91 prime? √91 ≈ 9.5, test: 91 ÷ 7 = 13, so 91 = 7 × 13 (composite)
Approach
Test divisibility by small primes up to square root
Question Type
Prime identification
Example
Express 72 as product of primes: 72 = 2³ × 3²
Approach
Use systematic division or factor trees
Question Type
Prime factorization
Key Points To Remember
- 2 is the smallest and only even prime number
- All prime numbers greater than 2 are odd
- 1 is neither prime nor composite
- To check if a number n is prime, test divisibility by primes up to √n
- Every composite number can be expressed as a product of prime factors
Least Common Multiple (LCM) and Greatest Common Factor (GCF)
The LCM of two numbers is the smallest positive number that is divisible by both numbers. The GCF is the largest positive number that divides both numbers evenly. These concepts are essential for adding/subtracting fractions and solving word problems. There are two main methods: listing multiples/factors, and using prime factorization. The relationship LCM × GCF = product of the two numbers is very useful for checking answers.
Examples
For LCM with prime factorization, take the highest power of each prime. For GCF, take the lowest power of each common prime. The verification confirms our answer.
Scenario
Find LCM and GCF of 18 and 24
Solution
Method 1 - Listing: Multiples of 18: 18, 36, 54, 72, 90, 108... Multiples of 24: 24, 48, 72, 96... LCM = 72 Factors of 18: 1, 2, 3, 6, 9, 18 Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24 GCF = 6 Method 2 - Prime factorization: 18 = 2 × 3² 24 = 2³ × 3 LCM = 2³ × 3² = 8 × 9 = 72 GCF = 2¹ × 3¹ = 2 × 3 = 6 Verification: LCM × GCF = 72 × 6 = 432 = 18 × 24 ✓
LCM problems often appear as 'when will events coincide again' questions. The LCM gives us the time interval for simultaneous occurrence.
Scenario
A red light blinks every 6 seconds, a green light every 8 seconds. If they blink together now, when will they next blink together?
Solution
Step 1: This is asking for LCM of 6 and 8 Step 2: Find prime factorizations 6 = 2 × 3 8 = 2³ Step 3: LCM = 2³ × 3 = 8 × 3 = 24 Step 4: They will blink together again after 24 seconds
Applications
- Adding and subtracting fractions (finding common denominators)
- Solving scheduling and timing problems
- Simplifying fractions to lowest terms
- Finding equivalent fractions
Misconceptions
- Thinking LCM is always the product of the numbers (only true for relatively prime numbers)
- Confusing which to use: LCM for 'together again' problems, GCF for 'grouping equally' problems
- Not verifying answers using LCM × GCF = product formula
Related Concepts
- Prime factorization
- Fractions
- Divisibility
- Word problems
Common Exam Questions
Example
Bus A every 15 min, Bus B every 20 min. LCM(15,20) = 60 min = next meeting time
Approach
Identify the pattern, find LCM for 'when together again'
Question Type
Word problems involving cycles
Example
Add 1/6 + 1/8: LCM(6,8) = 24, so 4/24 + 3/24 = 7/24
Approach
Find LCM of denominators for addition/subtraction
Question Type
Fraction operations setup
Key Points To Remember
- LCM × GCF = product of the two original numbers
- If one number divides another, the smaller is the GCF and larger is the LCM
- For relatively prime numbers, GCF = 1 and LCM = product of numbers
- Prime factorization method is more efficient for large numbers
- LCM is used for adding fractions, GCF for simplifying fractions
Divisibility Rules
Divisibility rules help determine if one number divides another without performing actual division. These shortcuts are invaluable for quickly checking factors, simplifying fractions, and solving multiple-choice questions efficiently. The most important rules are for 2, 3, 4, 5, 6, 8, 9, and 10. Mastering these rules will significantly speed up your problem-solving in exams.
Examples
We apply each rule systematically. Notice that divisibility by 6 requires divisibility by both 2 and 3, since 6 = 2 × 3.
Scenario
Check if 4,536 is divisible by 2, 3, 4, 5, 6, 8, 9
Solution
Number: 4,536 Divisible by 2? Last digit is 6 (even) → YES Divisible by 3? 4+5+3+6 = 18, 18÷3 = 6 → YES Divisible by 4? Last two digits: 36, 36÷4 = 9 → YES Divisible by 5? Last digit is 6 (not 0 or 5) → NO Divisible by 6? Divisible by both 2 and 3 → YES Divisible by 8? Last three digits: 536, 536÷8 = 67 → YES Divisible by 9? Sum of digits = 18, 18÷9 = 2 → YES
We start with the smallest 4-digit number and adjust to satisfy the divisibility rule. Adding 8 to make the digit sum equal to 9 gives us the answer.
Scenario
Find the smallest 4-digit number divisible by 9
Solution
Step 1: Smallest 4-digit number is 1000 Step 2: Check if 1000 is divisible by 9 Sum of digits: 1+0+0+0 = 1 1 is not divisible by 9 Step 3: We need sum of digits to be 9 Try 1008: 1+0+0+8 = 9 ✓ Step 4: Answer is 1008
Applications
- Quick factor checking in multiple-choice questions
- Simplifying fractions efficiently
- Mental math calculations
- Prime factorization shortcuts
Misconceptions
- Thinking divisibility by 6 only requires divisibility by 6 (need both 2 AND 3)
- Confusing divisibility by 4 rule (last two digits) with divisibility by 8 rule (last three digits)
- Not continuing the digit sum process until single digit for 3 and 9 rules
Related Concepts
- Factors and multiples
- Prime numbers
- Fraction simplification
- Mental math
Common Exam Questions
Example
Which numbers divide 2,448? Check 2(yes), 3(yes), 4(yes), 6(yes), 8(yes), 9(no)
Approach
Apply rules systematically for each divisor
Question Type
Multiple divisibility checking
Example
Largest 3-digit number divisible by 9: Start with 999, check 9+9+9=27, divisible by 9 ✓
Approach
Use the divisibility rule as a constraint
Question Type
Finding numbers with specific divisibility
Key Points To Remember
- Divisibility by 2: last digit is even (0,2,4,6,8)
- Divisibility by 3: sum of digits is divisible by 3
- Divisibility by 5: last digit is 0 or 5
- Divisibility by 9: sum of digits is divisible by 9
- Divisibility by 10: last digit is 0
Order of Operations (PEMDAS)
PEMDAS stands for Parentheses, Exponents, Multiplication and Division (left to right), Addition and Subtraction (left to right). This rule ensures everyone gets the same answer when evaluating mathematical expressions. In the Philippines, some use BODMAS (Brackets, Orders, Division/Multiplication, Addition/Subtraction) - both mean the same thing. Mastering PEMDAS is crucial for avoiding careless mistakes in calculations.
Examples
Each step follows PEMDAS order strictly. Notice that 3 × 16 and 4 ÷ 2 are both done in step 3 because multiplication and division have equal priority.
Scenario
Evaluate: 2 + 3 × 4² - (10 - 6) ÷ 2
Solution
Step 1: Parentheses first 2 + 3 × 4² - (10 - 6) ÷ 2 = 2 + 3 × 4² - 4 ÷ 2 Step 2: Exponents next = 2 + 3 × 16 - 4 ÷ 2 Step 3: Multiplication and Division (left to right) = 2 + 48 - 2 Step 4: Addition and Subtraction (left to right) = 50 - 2 = 48
Common mistake: doing 6 × 2 = 12 first, then 24 ÷ 12 = 2. Remember: equal priority operations go left to right!
Scenario
Evaluate: 24 ÷ 6 × 2 + 5
Solution
Step 1: No parentheses or exponents Step 2: Division and Multiplication (left to right) 24 ÷ 6 × 2 + 5 = 4 × 2 + 5 (division first because it's leftmost) = 8 + 5 Step 3: Addition = 13
Applications
- Evaluating complex mathematical expressions
- Programming and calculator usage
- Physics and chemistry formulas
- Financial calculations and interest problems
Misconceptions
- Doing operations left to right without considering priority (2+3×4 ≠ 5×4)
- Thinking multiplication always comes before division (they have equal priority)
- Forgetting that operations of equal priority go left to right
Related Concepts
- Algebra
- Equation solving
- Calculator usage
- Mathematical expressions
Common Exam Questions
Example
Evaluate 3(2+4)²÷6-1: = 3(6)²÷6-1 = 3×36÷6-1 = 108÷6-1 = 18-1 = 17
Approach
Apply PEMDAS step by step, showing each stage
Question Type
Expression evaluation
Example
Find error in: 2+3×4=5×4=20. Error: should be 2+(3×4)=2+12=14
Approach
Check if PEMDAS was followed correctly
Question Type
Error identification
Key Points To Remember
- P - Parentheses (or Brackets) first
- E - Exponents (or Orders/Powers) second
- MD - Multiplication and Division from left to right
- AS - Addition and Subtraction from left to right
- Operations of equal priority are done left to right
Fractions
A fraction represents a part of a whole, written as a/b where a is the numerator and b is the denominator (b ≠ 0). Fractions are essential for representing non-whole quantities and are frequently tested in entrance exams. Key operations include simplification, addition, subtraction, multiplication, and division. Understanding equivalent fractions and converting between fractions, decimals, and percentages is also crucial.
Examples
When adding fractions, we need a common denominator. The LCD is the LCM of the denominators. Always simplify the final answer.
Scenario
Add: 2/3 + 5/6
Solution
Step 1: Find LCD of denominators 3 and 6 LCD(3,6) = 6 Step 2: Convert to equivalent fractions 2/3 = (2×2)/(3×2) = 4/6 5/6 = 5/6 (already has denominator 6) Step 3: Add numerators 4/6 + 5/6 = (4+5)/6 = 9/6 Step 4: Simplify 9/6 = 3/2 = 1 1/2
Multiplication of fractions is straightforward - no common denominator needed. Always check if the result can be simplified.
Scenario
Multiply: 3/4 × 2/5
Solution
Step 1: Multiply numerators and denominators directly (3 × 2)/(4 × 5) = 6/20 Step 2: Simplify by finding GCF GCF(6,20) = 2 6/20 = (6÷2)/(20÷2) = 3/10 Answer: 3/10
Formula: improper fraction = (whole × denominator + numerator)/denominator. This conversion is essential for fraction operations.
Scenario
Convert mixed number to improper fraction: 2 3/5
Solution
Step 1: Multiply whole number by denominator 2 × 5 = 10 Step 2: Add the numerator 10 + 3 = 13 Step 3: Write over original denominator 2 3/5 = 13/5
Applications
- Recipe scaling and cooking measurements
- Time calculations (half-hour = 1/2 hour)
- Probability and statistics
- Engineering and construction measurements
Misconceptions
- Adding denominators when adding fractions (1/2 + 1/3 ≠ 2/5)
- Cross-multiplying when adding fractions (only for proportion solving)
- Not simplifying final answers to lowest terms
Related Concepts
- Decimals
- Percentages
- Ratios and proportions
- LCM and GCF
Common Exam Questions
Example
Calculate 1/2 + 1/3 - 1/6: LCD=6, so 3/6 + 2/6 - 1/6 = 4/6 = 2/3
Approach
Find common denominators for +/-, multiply directly for ×/÷
Question Type
Fraction operations
Example
ate 3/8 of pizza, friend ate 1/4. Total eaten? 3/8 + 2/8 = 5/8
Approach
Identify the operation needed, set up fraction equation
Question Type
Word problems with fractions
Key Points To Remember
- Proper fraction: numerator < denominator (3/4)
- Improper fraction: numerator ≥ denominator (5/3)
- Mixed number: whole number + proper fraction (1 2/3)
- To add/subtract: find common denominator
- To multiply: multiply numerators and denominators directly
Decimals
Decimals are another way to represent fractions, using place value with a decimal point. The digits after the decimal point represent tenths, hundredths, thousandths, etc. Understanding decimal operations, conversions to/from fractions, and rounding is essential for practical calculations and exam success. Decimals are often easier to work with than fractions in calculations.
Examples
This is a terminating decimal because the division ends with remainder 0. Fractions with denominators having only factors of 2 and 5 give terminating decimals.
Scenario
Convert 3/8 to decimal
Solution
Step 1: Divide numerator by denominator 3 ÷ 8 Step 2: Perform long division 0.375 8)3.000 24 --- 60 56 --- 40 40 --- 0 Answer: 3/8 = 0.375
Adding zeros to the right doesn't change the value but helps align digits properly. This prevents errors in decimal addition.
Scenario
Add: 12.45 + 3.678 + 0.9
Solution
Step 1: Line up decimal points 12.450 + 3.678 + 0.900 -------- 17.028 Answer: 17.028
When multiplying decimals, the number of decimal places in the answer equals the sum of decimal places in the factors.
Scenario
Multiply: 2.3 × 1.4
Solution
Step 1: Multiply as whole numbers 23 × 14 = 322 Step 2: Count decimal places 2.3 has 1 decimal place 1.4 has 1 decimal place Total: 1 + 1 = 2 decimal places Step 3: Place decimal point 322 → 3.22 Answer: 2.3 × 1.4 = 3.22
Applications
- Money calculations and financial planning
- Scientific measurements and precision
- Sports statistics and averages
- Engineering and construction tolerances
Misconceptions
- Not aligning decimal points when adding/subtracting
- Forgetting to count decimal places when multiplying
- Thinking 0.5 > 0.50 (they're equal)
Related Concepts
- Fractions
- Percentages
- Place value
- Scientific notation
Common Exam Questions
Example
Convert 5/6 to decimal: 5÷6 = 0.8333... = 0.83̄
Approach
Divide numerator by denominator using long division
Question Type
Fraction to decimal conversion
Example
Calculate 4.2 × 0.03 = 0.126 (1 + 2 = 3 decimal places)
Approach
Align decimal points for +/-, count places for ×
Question Type
Decimal operations
Key Points To Remember
- Place values: tenths (0.1), hundredths (0.01), thousandths (0.001)
- Terminating decimals can be converted to fractions easily
- Repeating decimals represent rational numbers
- Line up decimal points when adding/subtracting
- Count decimal places when multiplying
Practice Problems
Take highest powers for LCM, lowest powers for GCF. The verification confirms our answer is correct.
Problem
Find the LCM and GCF of 36 and 48.
Solution
Prime factorization method: 36 = 2² × 3² 48 = 2⁴ × 3 LCM = 2⁴ × 3² = 16 × 9 = 144 GCF = 2² × 3¹ = 4 × 3 = 12 Verification: LCM × GCF = 144 × 12 = 1,728 = 36 × 48 ✓
Follow PEMDAS order carefully. Multiplication has higher priority than subtraction, so 3 × 4 is done before the subtraction.
Problem
Evaluate: 15 - 3 × 2² + (8 - 6) × 4
Solution
Step 1: Parentheses first 15 - 3 × 2² + (8 - 6) × 4 = 15 - 3 × 2² + 2 × 4 Step 2: Exponents = 15 - 3 × 4 + 2 × 4 Step 3: Multiplication (left to right) = 15 - 12 + 8 Step 4: Addition and Subtraction (left to right) = 3 + 8 = 11
Find the LCD first, then convert all fractions. Operations with same denominators are straightforward.
Problem
Calculate: 2/3 + 1/4 - 1/6
Solution
Step 1: Find LCD of 3, 4, and 6 LCM(3, 4, 6) = 12 Step 2: Convert to equivalent fractions 2/3 = 8/12 1/4 = 3/12 1/6 = 2/12 Step 3: Perform operations left to right 8/12 + 3/12 - 2/12 = (8 + 3 - 2)/12 = 9/12 Step 4: Simplify 9/12 = 3/4
LCM problems often involve cycles or schedules. The LCM gives the time when events synchronize again.
Problem
A factory produces items in batches. Machine A completes a batch every 15 minutes, Machine B every 20 minutes. If both start together, when will they next finish batches at the same time?
Solution
Step 1: This asks for LCM of 15 and 20 Step 2: Prime factorization 15 = 3 × 5 20 = 2² × 5 Step 3: LCM = 2² × 3 × 5 = 4 × 3 × 5 = 60 Answer: They will finish together again after 60 minutes (1 hour)
Apply each divisibility rule systematically. For 6, check both 2 and 3 since 6 = 2 × 3.
Problem
Check if 5,472 is divisible by 3, 4, 6, 8, and 9.
Solution
For 5,472: Divisible by 3? Sum of digits: 5+4+7+2 = 18, 18÷3 = 6 ✓ YES Divisible by 4? Last two digits: 72, 72÷4 = 18 ✓ YES Divisible by 6? Divisible by both 2 and 3: Even number ✓ and divisible by 3 ✓ YES Divisible by 8? Last three digits: 472, 472÷8 = 59 ✓ YES Divisible by 9? Sum of digits: 18, 18÷9 = 2 ✓ YES
Exam Preparation Tips
- Memorize prime numbers up to 100 - they appear frequently in factorization problems
- Practice PEMDAS with complex expressions daily to avoid careless mistakes
- Learn divisibility rules thoroughly - they save time in multiple-choice questions
- Master fraction-decimal conversions for common fractions (1/2=0.5, 1/3=0.333...)
- Always simplify fractions to lowest terms unless specified otherwise
- Use the LCM×GCF = product formula to verify your answers quickly
- In word problems, identify whether you need LCM (timing/cycles) or GCF (grouping/sharing)
- Double-check decimal place alignment in addition/subtraction
- For multiple-choice questions, use estimation to eliminate obviously wrong answers
- Practice mental math with common fraction operations to increase speed
In summary
Mastering arithmetic concepts is fundamental for success in college entrance exams and higher-level mathematics. The concepts covered - multiples, factors, prime numbers, LCM/GCF, PEMDAS, fractions, and decimals - form the building blocks for algebra, geometry, and advanced math topics. Regular practice with these concepts will not only improve your exam performance but also develop strong mathematical reasoning skills. Focus on understanding the underlying principles rather than just memorizing procedures. Remember to always show your work step-by-step in exams, verify your answers when possible, and use shortcuts like divisibility rules to save time. With consistent practice and application of the strategies outlined in this chapter, you'll be well-prepared to tackle arithmetic problems confidently in any college entrance examination.
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