Number Theory
Module 02 of 06 · Full Solution Edition
Factors & Multiples
Core Concept: Two sides of the same divisibility coin
If A ÷ B leaves no remainder, then:
- B is a factor of A — it divides evenly into A.
- A is a multiple of B — it’s B times some whole number.
Example: 30 ÷ 5 = 6, so 5 is a factor of 30 and 30 is a multiple of 5. Every number is a factor of itself; 1 is a factor of every number.
Counting factors: always pair them up. For 20, the pairs are (1, 20), (2, 10), (4, 5) → 6 factors total.
(1) Circle the factors of 28: A. 2 B. 7 C. 12 D. 14 E. 28 F. 56
(2) Circle the multiples of 12: A. 4 B. 5 C. 8 D. 12 E. 20 F. 48
Strategy
For factors: ask “does this number divide 28 evenly?” For multiples: ask “is this number 12 × something?”
Steps
Part (1) — Factors of 28 are 1, 2, 4, 7, 14, 28.
- A. 2 ✓ B. 7 ✓ C. 12 ✗ (28 ÷ 12 not integer) D. 14 ✓ E. 28 ✓ F. 56 ✗ (too big — 56 is a multiple, not a factor)
Part (2) — Multiples of 12 are 12, 24, 36, 48, 60, …
- A. 4 ✗ B. 5 ✗ C. 8 ✗ D. 12 ✓ E. 20 ✗ F. 48 ✓
Pitfall
Don’t confuse factors with multiples! Factors are small (they divide the number); multiples are big (the number divides them). 56 is a multiple of 28, not a factor.
Answer
(1) A, B, D, E · (2) D, F
Which of the following is a multiple of 7?
Strategy
Recall the times table for 7: 7, 14, 21, 28, …, 70, 77, 84.
Steps
7 × 11 = 77. The other options aren’t divisible by 7 (e.g., 75 ÷ 7 = 10.71…).
Answer
C — 77
The number 6 has exactly four positive divisors: 1, 2, 3, and 6. How many positive divisors does 20 have?
Strategy
List divisors as pairs that multiply to 20. Each pair contributes 2 divisors (unless it’s a perfect-square pair).
Steps
- Pair (1, 20): factors 1 and 20.
- Pair (2, 10): factors 2 and 10.
- Pair (4, 5): factors 4 and 5.
- Total: 1, 2, 4, 5, 10, 20 → 6 divisors.
Variation
For a perfect square like 36, the pair (6, 6) only contributes one factor — so 36 has an odd number of divisors (9 total).
Answer
B — 6 divisors
The proper divisors of 12 are 1, 2, 3, 4, 6 (all divisors other than 12 itself). Their sum is 1+2+3+4+6 = 16 > 12, so 12 is called an abundant number. Which of the following is also abundant?
Concept
Abundant ⇔ sum of proper divisors > the number.
Steps
Check each:
- 8: divisors 1, 2, 4 → sum 7 < 8 ✗
- 10: 1, 2, 5 → sum 8 < 10 ✗
- 14: 1, 2, 7 → sum 10 < 14 ✗
- 18: 1, 2, 3, 6, 9 → sum 21 > 18 ✓
- 22: 1, 2, 11 → sum 14 < 22 ✗
Pitfall
“Proper divisors” exclude the number itself. Don’t accidentally add 18 to its own divisor sum.
Answer
D — 18
The number 6 has exactly 4 positive factors and the number 9 has exactly 3 positive factors. How many numbers in the list 14, 21, 28, 35, 42 have exactly 4 positive factors?
Concept
A number has exactly 4 factors when it is either (a) p³ for a prime p, or (b) p × q for two distinct primes p, q.
Steps
List factors for each:
- 14 = 2 × 7 → factors {1, 2, 7, 14} → 4 ✓
- 21 = 3 × 7 → factors {1, 3, 7, 21} → 4 ✓
- 28 = 2² × 7 → factors {1, 2, 4, 7, 14, 28} → 6 ✗
- 35 = 5 × 7 → factors {1, 5, 7, 35} → 4 ✓
- 42 = 2 × 3 × 7 → factors {1, 2, 3, 6, 7, 14, 21, 42} → 8 ✗
Three numbers (14, 21, 35) have exactly 4 factors.
Variation
If you need exactly 3 factors, the number must be the square of a prime: 4 = 2², 9 = 3², 25 = 5², 49 = 7², …
Answer
C — 3 numbers
Prime & Composite Numbers
Core Concept: Memorize these primes once and for all
- Prime: exactly 2 factors (1 and itself). Examples: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.
- Composite: more than 2 factors. Examples: 4, 6, 8, 9, 10, 12, …
- Neither: 1 is neither prime nor composite (it has only 1 factor).
Three common traps:
- 1 is not prime.
- 2 is the only even prime.
- 9, 15, 21, 25, 27, 33, 35, 39 are all composite, not prime. (9 = 3×3 trips up the most students!)
How many prime numbers are there between 20 and 30?
Strategy
Skip all even numbers and all multiples of 5 — those are obviously composite. Check the rest.
Steps
- Candidates: 21, 23, 27, 29 (even numbers + 25 ruled out).
- 21 = 3 × 7 ✗
- 23: not divisible by 2, 3, 5, 7 — and 7² = 49 > 23, so 23 is prime.
- 27 = 3³ ✗
- 29: not divisible by 2, 3, 5 — prime.
- Total: 2 primes (23 and 29).
Pitfall
Don’t include 27 (= 3³) — it’s a common slip. Also be careful with 21 (= 3×7).
Answer
C — 2 primes
Ten students each receive a card numbered with a different integer from 10 to 19. They each check off boxes that describe their number: Odd Number, Even Number, Prime Number, Composite Number, Perfect Square. How many students check off exactly two boxes?
Strategy
Every number 10–19 checks exactly one of (Odd/Even) and exactly one of (Prime/Composite). That’s 2 boxes by default. A perfect square adds a 3rd box.
Steps
- For each number 10–19, baseline = 2 boxes (one parity + one prime/composite).
- Perfect squares in 10–19: just 16. So 16 checks 3 boxes; all 9 others check exactly 2.
Pitfall
None of 10–19 is “1”, so none falls outside both prime and composite categories. (If the range included 1, that would be a special case.)
Answer
B — 9 students
The sum of two different prime numbers is 10. The product of these two numbers is:
Strategy
List primes < 10 and find a pair that sums to 10.
Steps
- Primes < 10: 2, 3, 5, 7.
- Pairs summing to 10: (3, 7). (Not 5+5, since they must be different.)
- Product =
3 × 7 = 21.
Pitfall
“Different primes” rules out 5 + 5. Always check that condition carefully.
Answer
D — 21
Prime Factorization
Core Concept: Every number has a unique prime DNA
The Fundamental Theorem of Arithmetic says: every integer > 1 can be written as a product of primes in exactly one way (up to order).
Two factoring methods:
- Factor tree: split into any two factors, then keep splitting until all branches are prime.
- Repeated division: divide by the smallest prime repeatedly. 36 ÷ 2 = 18, ÷ 2 = 9, ÷ 3 = 3, ÷ 3 = 1. Done.
Divisor-count formula: if n = p^a × q^b × r^c, then number of divisors = (a+1)(b+1)(c+1).
The number 385 has three prime factors. The sum of these prime factors is:
Strategy
385 ends in 5 → divisible by 5. Start factoring from there.
Steps
-
385 ÷ 5 = 77. -
77 = 7 × 11. - So
385 = 5 × 7 × 11. - Sum =
5 + 7 + 11 = 23.
Variation
Memorize a few common “double-prime” products: 21 = 3×7, 33 = 3×11, 35 = 5×7, 39 = 3×13, 77 = 7×11, 91 = 7×13.
Answer
D — 23
The positive integer n has exactly 8 positive divisors including 1 and n. Two of these divisors are 14 and 21. What is the sum of all 8 positive divisors of n?
Concept
If 14 and 21 both divide n, then so does LCM(14, 21) = 42. So n is a multiple of 42.
Strategy
Find the smallest n with exactly 8 divisors that includes 14 and 21.
Steps
- 42 = 2 × 3 × 7. Number of divisors =
(1+1)(1+1)(1+1) = 8✓ - So n = 42 works! Its divisors are 1, 2, 3, 6, 7, 14, 21, 42.
- Sum =
1 + 2 + 3 + 6 + 7 + 14 + 21 + 42 = 96.
Variation
If a larger multiple like 84 were forced, it would have (2+1)(1+1)(1+1) = 12 divisors — too many.
Answer
D — 96
The sum of the prime factors of 42 is:
Steps
-
42 = 2 × 3 × 7. - Sum =
2 + 3 + 7 = 12.
Pitfall
“Prime factors” means just the primes, not all factors. Don’t sum 1+2+3+6+7+14+21+42!
Answer
C — 12
HCF & LCM
Core Concept: The greatest shared factor and the smallest shared multiple
- HCF (or GCD): the biggest number that divides both. Use it when splitting into equal groups: “What’s the largest bag size?”
- LCM: the smallest number that both divide. Use it for timing/cycling: “When will both events coincide?”
Word-problem tells:
- HCF ⇒ “equal groups”, “largest possible”, “splits evenly”.
- LCM ⇒ “next time both”, “least common”, “synchronize”.
Useful identity: HCF(a, b) × LCM(a, b) = a × b.
Which of the following pairs of numbers has a greatest common factor of 20?
Steps
Compute GCD for each:
- A. GCD(200, 2000) = 200 ✗
- B. GCD(40, 50) = 10 ✗
- C. GCD(20, 40) = 20 ✓
- D. GCD(20, 25) = 5 ✗
- E. GCD(40, 80) = 40 ✗
Pitfall
When one number is a multiple of the other, the GCD is the smaller one. GCD(20, 40) = 20, not 10 or 40.
Answer
C — 20 and 40
A farmer packed 56 apples into bags and 72 pears into boxes. Each bag contained the same number of apples and each box the same number of pears, with no fruit left over. He used an equal number of bags and boxes for both fruits. What is the least possible number of apples in each bag?
Concept
“Equal number of bags and boxes” + “no fruit left over” means this count must divide BOTH 56 and 72. The largest such count = HCF(56, 72), which gives the smallest apples-per-bag.
Steps
- 56 = 2³ × 7. 72 = 2³ × 3².
- HCF = 2³ = 8 (the shared part).
- Use 8 bags and 8 boxes.
- Apples per bag =
56 ÷ 8 = 7.
Pitfall
“Least apples per bag” corresponds to the greatest number of bags. Translate carefully!
Answer
D — 7 apples
A train stops at Waterloo Station every 3 minutes. A bus stops at Waterloo Station every 5 minutes. A train and a bus both stop at Waterloo Station at 6:25 a.m. The next time they both stop together is:
Concept
“Next time both happen together” = LCM problem.
Steps
- LCM(3, 5) = 15 minutes.
- 6:25 + 15 min = 6:40.
Answer
D — 6:40 a.m.
Jolie decorated the room with green and blue lights. The green light blinks every 6 minutes and the blue light blinks every 8 minutes. Both lights were turned on at the same time. After how many minutes did all the lights blink together for the next time?
Steps
- 6 = 2 × 3. 8 = 2³.
- LCM =
2³ × 3 = 24.
Pitfall
Don’t just multiply 6 × 8 = 48. That’s a multiple of both, but not the least common multiple.
Answer
C — 24 minutes
How many different pairs of positive whole numbers have a greatest common factor of 4 and a lowest common multiple of 4620?
Concept
If HCF(a, b) = 4, write a = 4x, b = 4y where HCF(x, y) = 1 (the leftover parts share no factors).
Strategy
Use HCF × LCM = a × b to find xy, then count coprime factor pairs.
Steps
-
a × b = 4 × 4620 = 18480. Alsoa × b = 16xy, soxy = 1155. - Factor 1155:
1155 = 3 × 5 × 7 × 11— four distinct primes. - For each prime, decide: does it go into x or into y? (It must go fully into one, since gcd(x,y)=1.) That’s
2⁴ = 16ordered ways. - 1155 is not a perfect square (no possibility of x = y), so we divide by 2 to get unordered pairs:
16 ÷ 2 = 8pairs.
Variation
General rule: if HCF and LCM are given, count = 2^(k−1) where k = number of distinct prime factors in LCM ÷ HCF.
Answer
D — 8 pairs
Divisibility Rules
Core Concept: Shortcut tests for common divisors
- By 2: last digit is 0, 2, 4, 6, or 8.
- By 3: digit sum is divisible by 3. (E.g., 123 → 1+2+3 = 6 ✓)
- By 4: last two digits form a number divisible by 4. (E.g., 312 → 12 ÷ 4 ✓)
- By 5: last digit is 0 or 5.
- By 6: divisible by both 2 AND 3.
- By 8: last three digits divisible by 8.
- By 9: digit sum divisible by 9.
- By 10: last digit is 0.
- By 25: last two digits are 00, 25, 50, or 75.
Which of the following numbers is divisible by 3?
Strategy
Check each digit sum.
Steps
- 49 → 4+9 = 13 ✗
- 50 → 5+0 = 5 ✗
- 51 → 5+1 = 6 ✓
- 52 → 5+2 = 7 ✗
Answer
C — 51
How many even whole numbers between 1 and 99 are multiples of 5?
Concept
“Even AND multiple of 5” = “multiple of LCM(2, 5)” = “multiple of 10”.
Steps
Multiples of 10 between 1 and 99: 10, 20, 30, 40, 50, 60, 70, 80, 90 → 9 numbers.
Answer
C — 9 numbers
Homework
Which of the following is not a factor of 42?
Steps
Factors of 42 = {1, 2, 3, 6, 7, 14, 21, 42}. 8 is missing.
Answer
D — 8
Which of the following is not a factor of 56?
Steps
56 = 2³ × 7. Factors = {1, 2, 4, 7, 8, 14, 28, 56}. 6 is missing.
Pitfall
6 = 2 × 3, but 56 has no factor of 3 (digit sum 5+6 = 11, not divisible by 3). So 6 ∤ 56.
Answer
A — 6
Which of the following is a multiple of both 5 and 6?
Concept
Multiple of both = multiple of LCM(5, 6) = 30.
Steps
Only 30 is a multiple of 30 in the list.
Answer
D — 30
Which of the following is both a factor of 16 and a multiple of 4?
Steps
- Factors of 16: {1, 2, 4, 8, 16}.
- Multiples of 4: {4, 8, 12, 16, …}.
- Intersection: {4, 8, 16}. From the options, only 4 appears.
Answer
D — 4
What is the smallest prime number that is the sum of three different prime numbers?
Concept
The sum of three primes is odd iff all three are odd, OR exactly one is even (2). For the sum to be a prime > 2, it must be odd.
Strategy
Try all-odd combinations starting from the smallest primes.
Steps
- If 2 is included: 2 + odd + odd = even. The only even prime is 2 itself, but our sum is ≥ 2+3+5 = 10. So 2 can’t be one of the three.
- Three distinct odd primes — try 3+5+7 = 15 (not prime).
- Next: 3+5+11 = 19 (prime ✓).
Pitfall
15 = 3 × 5 isn’t prime. Skip it and continue.
Answer
D — 19
Which of the following is the sum of two prime numbers?
Strategy
Sum of two primes that is odd ⇒ one of them must be 2 (the only even prime). Check whether n − 2 is prime.
Steps
- 11 − 2 = 9 = 3² (not prime) ✗
- 17 − 2 = 15 = 3 × 5 ✗
- 23 − 2 = 21 = 3 × 7 ✗
- 31 − 2 = 29 (prime ✓) → 31 = 2 + 29
Variation
This is a peek at Goldbach’s conjecture: every even integer > 2 is the sum of two primes. (Unproven, but true for all numbers ever checked.)
Answer
D — 31
There are three electric stars on a Christmas tree. They light up every 3, 4, and 6 minutes respectively. They are all switched on at 10:00. At what time will the stars light up together for the 3rd time?
Concept
The interval between consecutive joint lightings = LCM(3, 4, 6) = 12 minutes.
Strategy
Count carefully: 10:00 is the 1st joint lighting. The 2nd is 12 min later, the 3rd is 24 min later.
Steps
- 1st joint lighting: 10:00.
- 2nd: 10:12.
- 3rd: 10:24.
Pitfall
Off-by-one error: if you count 10:00 as the “1st joint lighting”, then 3rd is 2 × 12 = 24 min later, not 3 × 12 = 36 min.
Answer
C — 10:24
Lamp A flashes red light once every 2 minutes. Lamp B flashes yellow every 4 minutes. Lamp C flashes blue every 7 minutes. What is the shortest possible time taken for all the lamps to flash light at the same time?
Steps
- 2 = 2 · 4 = 2² · 7 = 7.
- LCM =
2² × 7 = 28minutes.
Pitfall
Don’t multiply 2 × 4 × 7 = 56. The factor of 2 is already inside 4.
Answer
C — 28 minutes
Which of the following numbers is divisible by 15?
Concept
Divisible by 15 ⇔ divisible by both 3 AND 5. Use digit-sum for 3, last-digit for 5.
Steps
All end in 5 or 0, so all are divisible by 5. Check digit sums for divisibility by 3:
- 225 → 2+2+5 = 9 ✓
- 425 → 4+2+5 = 11 ✗
- 515 → 5+1+5 = 11 ✗
- 1550 → 1+5+5+0 = 11 ✗
Answer
A — 225
Which number is divisible by 3 × 3 (i.e., 9)?
Concept
Divisibility by 9: digit sum is divisible by 9.
Steps
- 663 → 6+6+3 = 15 ✗ (15/9 isn’t integer)
- 603 → 6+0+3 = 9 ✓
- 336 → 3+3+6 = 12 ✗
- 303 → 3+0+3 = 6 ✗
Pitfall
Don’t confuse “divisible by 3” with “divisible by 9”. 663 has digit sum 15 (divisible by 3, not by 9).
Answer
B — 603
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