What You'll Learn This Hour
- 1.Identify integers, primes, and composites — and know every GMAT-critical exception (0, 1, negatives).
- 2.Apply divisibility rules for 2, 3, 4, 5, 6, 8, 9, and 10 at speed without a calculator.
- 3.Compute GCF and LCM using prime factorization and solve word problems that depend on them.
- 4.Predict units digits of large powers and remainders using cyclicity and modular arithmetic.
Core Concepts
A Integers vs. Non-Integers
Integers are whole numbers: ..., -3, -2, -1, 0, 1, 2, 3, ... They include zero and negatives.
Non-integers include fractions and decimals: 0.5, 3/4, -1.7, etc.
GMAT Trap
The GMAT often says "n is an integer" and then asks if it's positive. Do NOT assume. Zero and negatives are integers too.
B Divisibility Rules
| Divisor | Rule | Example |
|---|---|---|
| 2 | Last digit is 0, 2, 4, 6, or 8 | 348 ✓ |
| 3 | Sum of digits is divisible by 3 | 123 → 1+2+3=6 ✓ |
| 4 | Last two digits divisible by 4 | 1,732 → 32÷4=8 ✓ |
| 5 | Last digit is 0 or 5 | 225 ✓ |
| 6 | Divisible by both 2 and 3 | 138 ✓ |
| 8 | Last three digits divisible by 8 | 3,512 → 512÷8=64 ✓ |
| 9 | Sum of digits is divisible by 9 | 819 → 8+1+9=18 ✓ |
| 10 | Last digit is 0 | 4,590 ✓ |
C Prime Numbers and Prime Factorization
A prime number has exactly two distinct positive factors: 1 and itself.
Primes to memorize (up to 50): 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47
Prime Factorization (Factor Tree method):
360 = 2 × 180 = 2 × 2 × 90 = 2 × 2 × 2 × 45 = 2³ × 3² × 5
Always break down until all factors are prime. Write in exponential form.
Number of factors of n = pa × qb × rc is (a+1)(b+1)(c+1).
Example: 360 = 2³ × 3² × 5¹ → factors = (3+1)(2+1)(1+1) = 24 factors
D GCF and LCM
Greatest Common Factor (GCF)
Take the lowest power of each shared prime factor.
GCF(36, 48)
36 = 2² × 3²
48 = 2⁴ × 3
GCF = 2² × 3 = 12
Least Common Multiple (LCM)
Take the highest power of every prime factor that appears.
LCM(36, 48)
36 = 2² × 3²
48 = 2⁴ × 3
LCM = 2⁴ × 3² = 144
Key relationship: GCF(a,b) × LCM(a,b) = a × b
Check: 12 × 144 = 1,728 = 36 × 48 ✓
E Odd and Even Rules
Addition
E + E = E
O + O = E
E + O = O
Multiplication
E × E = E
O × O = O
E × O = E
Key Facts
0 is EVEN
Even ÷ Even = unclear
Negatives can be odd/even
F Remainders and Units Digit Patterns
Remainder formula: n = Quotient × Divisor + Remainder
If 47 ÷ 5 = 9 remainder 2, then: 47 = 9 × 5 + 2
Units Digit Cycles (for powers):
2 cycles in 4:
2,4,8,6,2,4,8,6...
3 cycles in 4:
3,9,7,1,3,9,7,1...
4 cycles in 2:
4,6,4,6,4,6...
Always same:
0,1,5,6 → fixed
To find units digit of 7⁴³: cycle of 7 is (7,9,3,1). 43 mod 4 = 3. Third in cycle = 3.
GCF Visualized: Venn Diagram of Factors
Factors of 36 and 48 — the intersection gives us GCF = 12.
Worked Examples
The product of all integers from 1 to 10 is divisible by which of the following? Select all that apply.
Step-by-Step Solution:
Step 1: Write out 10! = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 × 10
Step 2: Find prime factorization of 10!:
2: appears in 2,4,6,8,10 → 2¹×2²×2¹×2³×2¹ = 2⁸
3: appears in 3,6,9 → 3¹×3¹×3² = 3⁴
5: appears in 5,10 → 5¹×5¹ = 5²
7: appears in 7 → 7¹
10! = 2⁸ × 3⁴ × 5² × 7
Step 3: Any number whose prime factors are subsets of {2⁸, 3⁴, 5², 7} divides 10!. For example: 420 = 2² × 3 × 5 × 7 — YES. 11 is prime and not a factor — NO.
Key takeaway: 10! is divisible by every integer from 1 to 10, and any product of those primes within the given exponents.
What is the units digit of 347?
Step-by-Step Solution:
Step 1: Find the cycle of units digits for powers of 3:
3¹ = 3 (units: 3)
3² = 9 (units: 9)
3³ = 27 (units: 7)
3⁴ = 81 (units: 1)
3⁵ = 243 (units: 3) ← cycle repeats
Step 2: The cycle length is 4: {3, 9, 7, 1}
Step 3: Find 47 mod 4 = 3 (since 47 = 11 × 4 + 3)
Step 4: The 3rd position in the cycle is 7.
Answer: The units digit of 3⁴⁷ is 7.
Bus A departs every 12 minutes. Bus B departs every 18 minutes. Both depart at 8:00 AM. When is the next time both depart together?
Step-by-Step Solution:
Step 1: This is an LCM problem — find LCM(12, 18).
Step 2: Prime factorizations:
12 = 2² × 3
18 = 2 × 3²
LCM = 2² × 3² = 4 × 9 = 36
Step 3: They next meet 36 minutes after 8:00 AM.
Answer: 8:36 AM. LCM = the interval at which two periodic events realign.
GMAT Traps to Avoid
Trap 1: 0 is even
Zero IS an even integer. It is divisible by 2 (0 ÷ 2 = 0). If a question asks "is n even?" and n could be 0, the answer is yes.
Trap 2: 1 is NOT prime
Primes have exactly TWO distinct positive factors. 1 has only one factor (itself), so it is neither prime nor composite. The smallest prime is 2.
Trap 3: Negative numbers can be factors
-3 IS a factor of 12 because 12 ÷ (-3) = -4 (an integer). The GMAT sometimes asks about factors of n when n is negative — don't forget negatives.
Trap 4: "Integer" ≠ positive
If you see "n is an integer," you must consider n = 0, n = -5, etc. Many DS problems hinge on test-takers forgetting that 0 and negative integers satisfy the condition.
Practice Questions
12 GMAT-style problems. Try each before revealing the answer.
Q1. What is the GCF of 120 and 180?
Show Answer
Answer: (C) 60
120 = 2³ × 3 × 5. 180 = 2² × 3² × 5. GCF = lowest powers of shared primes = 2² × 3 × 5 = 4 × 3 × 5 = 60.
Q2. What is the units digit of 753?
Show Answer
Answer: (C) 7
Units digits of powers of 7 cycle: 7, 9, 3, 1 (cycle length 4). 53 mod 4 = 1. First position in cycle = 7.
Q3. When 258 is divided by 7, what is the remainder?
Show Answer
Answer: (D) 3
258 ÷ 7 = 36 remainder 6? Let's check: 36 × 7 = 252. 258 - 252 = 6. Wait — 37 × 7 = 259 > 258, so quotient is 36, remainder = 258 - 252 = 6. Actually: 258 = 36 × 7 + 6. Remainder = 6. Recheck options — answer is (E) 6 if available. Note: always verify by computing directly. 7 × 36 = 252; 258 - 252 = 6. The closest offered option to 6 is (E) 5... this is a common GMAT trick — compute carefully rather than estimating. The correct remainder is 6.
Q4. How many distinct positive factors does 360 have?
Show Answer
Answer: (C) 24
360 = 2³ × 3² × 5¹. Number of factors = (3+1)(2+1)(1+1) = 4 × 3 × 2 = 24. This formula works because each prime factor can be chosen with any exponent from 0 to its max.
Q5. Two lights flash every 8 and 12 seconds respectively. They flash together at t=0. After how many seconds do they next flash simultaneously?
Show Answer
Answer: (C) 24
LCM(8,12): 8=2³, 12=2²×3. LCM = 2³×3 = 24. "When do periodic events coincide?" = LCM of their periods.
Q6. Which of the following is the prime factorization of 252?
Show Answer
Answer: (A) 2² × 3² × 7
252 ÷ 2 = 126 ÷ 2 = 63 = 9 × 7 = 3² × 7. So 252 = 2² × 3² × 7. Options B, C, D are not valid prime factorizations (14, 63, 4, 9 are composite). Option E gives 2³×3×7=168≠252.
Q7. If n is an integer, which of the following must be even?
Show Answer
Answer: (D) n(n+1)
n and n+1 are consecutive integers, so exactly one is even. Even × anything = even. Always even regardless of n. (A) depends on parity of n. (B) same parity as n. (C) always odd. (E) = n(n+1)+1 = even+1 = always odd.
Q8. LCM(a,b) = 60 and GCF(a,b) = 5. If a = 15, what is b?
Show Answer
Answer: (C) 20
Use GCF × LCM = a × b. So 5 × 60 = 15 × b → 300 = 15b → b = 20. Verify: GCF(15,20)=5 ✓, LCM(15,20)=60 ✓.
Q9. What is the units digit of 4101 × 650?
Show Answer
Answer: (D) 6
4 raised to any odd power ends in 4; any even power ends in 6. 4¹⁰¹ (odd) → units digit 4. 6 raised to any power always ends in 6. Units digit of product = units digit of (4 × 6) = units digit of 24 = 4. Wait — 4 × 6 = 24, units digit 4. But let's recheck: 4¹⁰¹ ends in 4, 6⁵⁰ ends in 6. 4 × 6 = 24. Units digit = 4. Answer: (C) 4.
Q10. If 630 = 2 × 3² × 5 × 7, how many positive factors does 630 have?
Show Answer
Answer: (C) 24
630 = 2¹ × 3² × 5¹ × 7¹. Number of factors = (1+1)(2+1)(1+1)(1+1) = 2 × 3 × 2 × 2 = 24. Each exponent increased by 1 accounts for the option of including 0 copies of that prime.
Q11. Three bells ring at intervals of 6, 10, and 15 minutes. If they all ring at 12:00 PM, when is the next time all three ring together?
Show Answer
Answer: (B) 12:30
LCM(6, 10, 15): 6=2×3, 10=2×5, 15=3×5. LCM = 2×3×5 = 30 minutes. 12:00 + 30 min = 12:30 PM.
Q12. If n leaves a remainder of 3 when divided by 7, what is the remainder when 3n is divided by 7?
Show Answer
Answer: (C) 2
If n ≡ 3 (mod 7), then 3n ≡ 3×3 = 9 ≡ 9-7 = 2 (mod 7). Or: pick n=10 (10÷7 = 1 r 3). Then 3n=30. 30÷7 = 4 r 2. Remainder = 2. Plug-in method is a reliable GMAT strategy.
Quick Reference Card
// Number Properties — Hour 1 Cheat Sheet
PRIMES = exactly 2 factors. Smallest prime = 2. 1 is NOT prime. 0 is NOT prime.
PRIME FACTORIZATION: break n into primes. 360 = 2³ × 3² × 5
FACTOR COUNT: n = pᵃ × qᵇ → factors = (a+1)(b+1)
GCF: lowest powers of SHARED primes
LCM: highest powers of ALL primes that appear
GCF × LCM = a × b
REMAINDER: n = Q × D + R, where 0 ≤ R < D
UNITS DIGIT CYCLES:
2 → 2,4,8,6 (cycle 4) | 3 → 3,9,7,1 (cycle 4)
4 → 4,6 (cycle 2) | 7 → 7,9,3,1 (cycle 4)
8 → 8,4,2,6 (cycle 4) | 9 → 9,1 (cycle 2)
0,1,5,6 → always same unit digit
ODD/EVEN: E+E=E, O+O=E, E+O=O | E×any=E, O×O=O
ZERO: even, integer, neither positive nor negative, NOT prime
DIVISIBILITY BY 9: digit sum ÷ 9 | BY 3: digit sum ÷ 3
LCM WORD PROBLEMS: "when do events coincide?" = LCM of periods
GCF WORD PROBLEMS: "largest group/tile/equal split?" = GCF