Gauss 7 · Past Contest Questions

Number Theory

Module 02 of 06 · Full Solution Edition

Module 02 of 06
Problems 28 total
Source Gauss 7 Past Questions
Format Solution Edition
§ 2.1 · Building Blocks

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
Definition Check
Warm-up

(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

2
2016 Gauss 7 · Q3
Explorer

Which of the following is a multiple of 7?

  • A. 75
  • B. 76
  • C. 77
  • D. 78
  • E. 79

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

3
2015 Gauss 7 · Q9
Explorer

The number 6 has exactly four positive divisors: 1, 2, 3, and 6. How many positive divisors does 20 have?

  • A. 2
  • B. 6
  • C. 3
  • D. 5
  • E. 8

Strategy

List divisors as pairs that multiply to 20. Each pair contributes 2 divisors (unless it's a perfect-square pair).

Steps

  1. Pair (1, 20): factors 1 and 20.
  2. Pair (2, 10): factors 2 and 10.
  3. Pair (4, 5): factors 4 and 5.
  4. 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

4
2020 Gauss 7 · Q10
Explorer

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?

  • A. 8
  • B. 10
  • C. 14
  • D. 18
  • E. 22

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

5
2021 Gauss 7 · Q16
Practice

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?

  • A. 1
  • B. 2
  • C. 3
  • D. 4
  • E. 5

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

§ 2.2 · The Indivisibles

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!)
6
UK 11+ Entry · Q21
Explorer

How many prime numbers are there between 20 and 30?

  • A. 0
  • B. 1
  • C. 2
  • D. 3
  • E. 4

Strategy

Skip all even numbers and all multiples of 5 — those are obviously composite. Check the rest.

Steps

  1. Candidates: 21, 23, 27, 29 (even numbers + 25 ruled out).
  2. 21 = 3 × 7 ✗
  3. 23: not divisible by 2, 3, 5, 7 — and 7² = 49 > 23, so 23 is prime.
  4. 27 = 3³ ✗
  5. 29: not divisible by 2, 3, 5 — prime.
  6. 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

7
2025 Gauss 7 · Q19
Explorer

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?

  • A. 10
  • B. 9
  • C. 8
  • D. 7
  • E. 6

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

  1. For each number 10–19, baseline = 2 boxes (one parity + one prime/composite).
  2. 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

8
2023 Gauss 7 · Q14
Practice

The sum of two different prime numbers is 10. The product of these two numbers is:

  • A. 24
  • B. 16
  • C. 4
  • D. 21
  • E. 9

Strategy

List primes < 10 and find a pair that sums to 10.

Steps

  1. Primes < 10: 2, 3, 5, 7.
  2. Pairs summing to 10: (3, 7). (Not 5+5, since they must be different.)
  3. Product = 3 × 7 = 21.

Pitfall

"Different primes" rules out 5 + 5. Always check that condition carefully.

Answer

D — 21

§ 2.3 · Breaking Numbers Apart

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).

36 = 2 × 2 × 3 × 3 = 2² × 3² 60 = 2 × 2 × 3 × 5 = 2² × 3 × 5 385 = 5 × 7 × 11

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).

9
2024 Gauss 7 · Q14
Explorer

The number 385 has three prime factors. The sum of these prime factors is:

  • A. 21
  • B. 26
  • C. 25
  • D. 23
  • E. 22

Strategy

385 ends in 5 → divisible by 5. Start factoring from there.

Steps

  1. 385 ÷ 5 = 77.
  2. 77 = 7 × 11.
  3. So 385 = 5 × 7 × 11.
  4. 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

10
2019 Gauss 7 · Q20
Explorer

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?

  • A. 35
  • B. 47
  • C. 53
  • D. 96
  • E. 103

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

  1. 42 = 2 × 3 × 7. Number of divisors = (1+1)(1+1)(1+1) = 8 ✓
  2. So n = 42 works! Its divisors are 1, 2, 3, 6, 7, 14, 21, 42.
  3. 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

11
2022 Gauss 7 · Q11
Practice

The sum of the prime factors of 42 is:

  • A. 23
  • B. 43
  • C. 12
  • D. 17
  • E. 13

Steps

  1. 42 = 2 × 3 × 7.
  2. 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

§ 2.4 · Common Ground

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.

12
2013 Gauss 7 · Q12
Explorer

Which of the following pairs of numbers has a greatest common factor of 20?

  • A. 200 and 2000
  • B. 40 and 50
  • C. 20 and 40
  • D. 20 and 25
  • E. 40 and 80

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

13
HCF Word Problem
Explorer

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?

  • A. 4
  • B. 5
  • C. 6
  • D. 7
  • E. 8

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

  1. 56 = 2³ × 7.   72 = 2³ × 3².
  2. HCF = 2³ = 8 (the shared part).
  3. Use 8 bags and 8 boxes.
  4. Apples per bag = 56 ÷ 8 = 7.

Pitfall

"Least apples per bag" corresponds to the greatest number of bags. Translate carefully!

Answer

D — 7 apples

14
2025 Gauss 7 · Q12
Explorer

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:

  • A. 6:28
  • B. 6:30
  • C. 6:33
  • D. 6:40
  • E. 6:55

Concept

"Next time both happen together" = LCM problem.

Steps

  1. LCM(3, 5) = 15 minutes.
  2. 6:25 + 15 min = 6:40.

Answer

D — 6:40 a.m.

15
LCM Practice
Practice

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?

  • A. 21
  • B. 23
  • C. 24
  • D. 48
  • E. 14

Steps

  1. 6 = 2 × 3.   8 = 2³.
  2. 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

16
2021 Gauss 7 · Q24
Challenge

How many different pairs of positive whole numbers have a greatest common factor of 4 and a lowest common multiple of 4620?

  • A. 4
  • B. 5
  • C. 7
  • D. 8
  • E. 11

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

  1. a × b = 4 × 4620 = 18480. Also a × b = 16xy, so xy = 1155.
  2. Factor 1155: 1155 = 3 × 5 × 7 × 11 — four distinct primes.
  3. 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 = 16 ordered ways.
  4. 1155 is not a perfect square (no possibility of x = y), so we divide by 2 to get unordered pairs: 16 ÷ 2 = 8 pairs.

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

§ 2.5 · Quick Checks

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.
17
1990 Math League · Grade 5 Q9
Explorer

Which of the following numbers is divisible by 3?

  • A. 49
  • B. 50
  • C. 51
  • D. 52

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

18
2015 Gauss 7 · Q13
Explorer

How many even whole numbers between 1 and 99 are multiples of 5?

  • A. 5
  • B. 7
  • C. 9
  • D. 11
  • E. 13

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

§ 2.6 · Mixed Review · 10 problems

Homework

19
Factors of 42
Homework

Which of the following is not a factor of 42?

  • A. 1
  • B. 6
  • C. 7
  • D. 8

Steps

Factors of 42 = {1, 2, 3, 6, 7, 14, 21, 42}. 8 is missing.

Answer

D — 8

20
Factors of 56
Homework

Which of the following is not a factor of 56?

  • A. 6
  • B. 2
  • C. 8
  • D. 4

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

21
Common Multiple
Homework

Which of the following is a multiple of both 5 and 6?

  • A. 11
  • B. 24
  • C. 25
  • D. 30
  • E. 35

Concept

Multiple of both = multiple of LCM(5, 6) = 30.

Steps

Only 30 is a multiple of 30 in the list.

Answer

D — 30

22
Factor and Multiple
Homework

Which of the following is both a factor of 16 and a multiple of 4?

  • A. 1
  • B. 2
  • C. 6
  • D. 4
  • E. 10

Steps

  1. Factors of 16: {1, 2, 4, 8, 16}.
  2. Multiples of 4: {4, 8, 12, 16, …}.
  3. Intersection: {4, 8, 16}. From the options, only 4 appears.

Answer

D — 4

23
2015 Junior Math Challenge · Q11
Homework

What is the smallest prime number that is the sum of three different prime numbers?

  • A. 11
  • B. 15
  • C. 17
  • D. 19
  • E. 23

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

  1. 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.
  2. Three distinct odd primes — try 3+5+7 = 15 (not prime).
  3. Next: 3+5+11 = 19 (prime ✓).

Pitfall

15 = 3 × 5 isn't prime. Skip it and continue.

Answer

D — 19

24
2013 Math League · Grade 6 Q7
Homework

Which of the following is the sum of two prime numbers?

  • A. 11
  • B. 17
  • C. 23
  • D. 31

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

25
2023 P4 Nanhua SA2 · Q19
Homework

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?

  • A. 10:12
  • B. 10:13
  • C. 10:24
  • D. 10:26

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

  1. 1st joint lighting: 10:00.
  2. 2nd: 10:12.
  3. 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

26
LCM Practice
Homework

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?

  • A. 14 min
  • B. 24 min
  • C. 28 min
  • D. 58 min

Steps

  1. 2 = 2   ·   4 = 2²   ·   7 = 7.
  2. LCM = 2² × 7 = 28 minutes.

Pitfall

Don't multiply 2 × 4 × 7 = 56. The factor of 2 is already inside 4.

Answer

C — 28 minutes

27
1988 Math League · Grade 6 Q16
Homework

Which of the following numbers is divisible by 15?

  • A. 225
  • B. 425
  • C. 515
  • D. 1550

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

28
2001 Math League · Grade 6 Q7
Homework

Which number is divisible by 3 × 3 (i.e., 9)?

  • A. 663
  • B. 603
  • C. 336
  • D. 303

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

Gauss 7 — Past Contest Questions · Number Theory · 28 Problems