Prime Time

Given: In the idli-vada game, 'idli' is said for multiples of 3, 'vada' for multiples of 5, and 'idli-vada' for common multiples (multiples of both 3 and 5, i.e., multiples of 15).

Concept: 'Idli-vada' is said at every common multiple of 3 and 5. The LCM of 3 and 5 is 15, so 'idli-vada' is said at every multiple of 15.

The multiples of 15 are: 15, 30, 45, 60, 75, 90, 105, 120, 135, 150, …

Counting: 1st → 15, 2nd → 30, 3rd → 45, …, 10th → 150.

$$10 \times 15 = 150$$

'Idli-vada' is said for the 10th time at the number 150.

Given: Game played from 1 to 90. 'Idli' for multiples of 3, 'vada' for multiples of 5, 'idli-vada' for multiples of both (i.e., multiples of 15).

Part a — Number of times 'idli' is said:
Multiples of 3 up to 90: $\lfloor 90 \div 3 \rfloor = 30$
$$\textbf{Answer: 30 times}$$

Part b — Number of times 'vada' is said:
Multiples of 5 up to 90: $\lfloor 90 \div 5 \rfloor = 18$
$$\textbf{Answer: 18 times}$$

Part c — Number of times 'idli-vada' is said:
Multiples of 15 up to 90: $\lfloor 90 \div 15 \rfloor = 6$
$$\textbf{Answer: 6 times}$$
(These occur at 15, 30, 45, 60, 75, 90.)

Given: Game played from 1 to 900.

Number of times 'idli' is said (multiples of 3 up to 900):
$$\lfloor 900 \div 3 \rfloor = 300$$

Number of times 'vada' is said (multiples of 5 up to 900):
$$\lfloor 900 \div 5 \rfloor = 180$$

Number of times 'idli-vada' is said (multiples of 15 up to 900):
$$\lfloor 900 \div 15 \rfloor = 60$$

Observation: Each answer is exactly 10 times the answer for the game up to 90, because 900 = 10 × 90.

Answers: 'idli' → 300 times; 'vada' → 180 times; 'idli-vada' → 60 times.

Yes, the figure is related to the idli-vada game.

In Fig. 5.1, the numbers are arranged in a grid. The shaded numbers are multiples of 3 (idli), the circled numbers are multiples of 5 (vada), and the numbers that are both shaded and circled are multiples of 15 (idli-vada).

When the game is played till 30, the common multiples (idli-vada) are 15 and 30.
When the game is played till 60, the common multiples are 15, 30, 45, and 60.

To draw the figure for the game up to 60: arrange numbers 1 to 60 in rows of 10. Shade all multiples of 3, circle all multiples of 5, and mark both shading and circle on multiples of 15 (i.e., 15, 30, 45, 60).

The figure visually represents the pattern of multiples of 3 and 5, and their common multiples, exactly as in the idli-vada game.

For each pair, 'idli' = multiples of the smaller number, 'vada' = multiples of the larger number, 'idli-vada' = common multiples (multiples of LCM).

a. 2 and 5 (LCM = 10):
- 'Idli' (multiples of 2 up to 60): 2,4,6,8,10,12,…,60 → 30 times
- 'Vada' (multiples of 5 up to 60): 5,10,15,20,…,60 → 12 times
- 'Idli-vada' (multiples of 10 up to 60): 10,20,30,40,50,60 → 6 times

For the figure: arrange 1–60 in rows of 10; shade even numbers, circle multiples of 5, both shade and circle multiples of 10.

b. 3 and 7 (LCM = 21):
- 'Idli' (multiples of 3 up to 60): 3,6,9,…,60 → 20 times
- 'Vada' (multiples of 7 up to 60): 7,14,21,28,35,42,49,56 → 8 times
- 'Idli-vada' (multiples of 21 up to 60): 21, 42 → 2 times

For the figure: shade multiples of 3, circle multiples of 7, both shade and circle 21 and 42.

c. 4 and 6 (LCM = 12):
- 'Idli' (multiples of 4 up to 60): 4,8,12,…,60 → 15 times
- 'Vada' (multiples of 6 up to 60): 6,12,18,…,60 → 10 times
- 'Idli-vada' (multiples of 12 up to 60): 12,24,36,48,60 → 5 times

For the figure: shade multiples of 4, circle multiples of 6, both shade and circle multiples of 12.

Note: For drawing, arrange numbers 1–60 in a 6×10 grid and mark accordingly as described above.

(Based on the standard table used in this chapter where multiples of 3 are shaded and multiples of 5 are circled.)

1. Shaded numbers are all multiples of 3. They share the common property of being divisible by 3.

2. Circled numbers are all multiples of 5. They share the common property of being divisible by 5 (their units digit is 0 or 5).

3. Numbers that are both shaded and circled are multiples of both 3 and 5, i.e., multiples of 15. In the range shown (31–70), these are 45 and 60. These numbers are called common multiples of 3 and 5.

Given: Find multiples of 40 between 310 and 410 (not including 310 and 410).

Multiples of 40: …, 280, 320, 360, 400, 440, …

Check:
- $40 \times 8 = 320$ ✓ (310 < 320 < 410)
- $40 \times 9 = 360$ ✓ (310 < 360 < 410)
- $40 \times 10 = 400$ ✓ (310 < 400 < 410)
- $40 \times 11 = 440$ ✗ (greater than 410)

The multiples of 40 between 310 and 410 are: 320, 360, and 400.

Part a:
Given: Number < 40, one factor is 7, sum of digits = 8.

Multiples of 7 less than 40: 7, 14, 21, 28, 35.

Check sum of digits:
- 7 → 7 (not 8)
- 14 → 1+4 = 5 (not 8)
- 21 → 2+1 = 3 (not 8)
- 28 → 2+8 = 10 (not 8)
- 35 → 3+5 = 8 ✓

The number is 35.

Part b:
Given: Number < 100, factors include 3 and 5, one digit is 1 more than the other.

Since 3 and 5 are both factors, the number must be a multiple of LCM(3,5) = 15.

Multiples of 15 less than 100: 15, 30, 45, 60, 75, 90.

Check condition (one digit is 1 more than the other):
- 15 → digits 1 and 5; 5 – 1 = 4 (not 1)
- 30 → digits 3 and 0; 3 – 0 = 3 (not 1)
- 45 → digits 4 and 5; 5 – 4 = 1 ✓
- 60 → digits 6 and 0; 6 – 0 = 6 (not 1)
- 75 → digits 7 and 5; 7 – 5 = 2 (not 1)
- 90 → digits 9 and 0; 9 – 0 = 9 (not 1)

The number is 45.

Given: A perfect number has the sum of all its factors equal to twice the number.

Check numbers between 1 and 10:

- 6: Factors are 1, 2, 3, 6. Sum = 1+2+3+6 = 12 = 2×6 ✓

Verification: $2 \times 6 = 12$ and $1+2+3+6 = 12$. ✓

The perfect number between 1 and 10 is 6.

Part a: 20 and 28
Factors of 20: 1, 2, 4, 5, 10, 20
Factors of 28: 1, 2, 4, 7, 14, 28
Common factors: 1, 2, 4

Part b: 35 and 50
Factors of 35: 1, 5, 7, 35
Factors of 50: 1, 2, 5, 10, 25, 50
Common factors: 1, 5

Part c: 4, 8 and 12
Factors of 4: 1, 2, 4
Factors of 8: 1, 2, 4, 8
Factors of 12: 1, 2, 3, 4, 6, 12
Common factors: 1, 2, 4

Part d: 5, 15 and 25
Factors of 5: 1, 5
Factors of 15: 1, 3, 5, 15
Factors of 25: 1, 5, 25
Common factors: 1, 5

A multiple of 25 that is NOT a multiple of 50 must be an odd multiple of 25 (i.e., $25 \times$ an odd number).

- $25 \times 1 = 25$ (not a multiple of 50 ✓)
- $25 \times 3 = 75$ (not a multiple of 50 ✓)
- $25 \times 5 = 125$ (not a multiple of 50 ✓)

Three such numbers are: 25, 75, and 125.

Given: Both numbers are less than 10. The first common multiple (LCM) is greater than 50.

We need two numbers, both less than 10, whose LCM is greater than 50.

Check pairs with LCM > 50:
- 7 and 8: LCM = 56 > 50 ✓ (both < 10)
- 7 and 9: LCM = 63 > 50 ✓ (both < 10)
- 8 and 9: LCM = 72 > 50 ✓ (both < 10)
- 6 and 7: LCM = 42 < 50 ✗
- 7 and 8: LCM = 56 ✓

The problem says the first 'idli-vada' is said after number 50, meaning the LCM > 50.

Possible pairs: (7 and 8) with LCM 56, (7 and 9) with LCM 63, or (8 and 9) with LCM 72.

Given: Treasures at 28 and 70. Jump sizes that land on both must be common factors of 28 and 70.

Factors of 28: 1, 2, 4, 7, 14, 28
Factors of 70: 1, 2, 5, 7, 10, 14, 35, 70

Common factors of 28 and 70: 1, 2, 7, 14

Jump sizes that will land on both 28 and 70 are: 1, 2, 7, and 14.

Note: The diagram (Fig. img_3) shows a Venn-diagram style figure with two overlapping circles. The common multiples (in the intersection) are visible but the individual multiples in each circle are erased. Since the figure is not fully visible, we solve based on the standard version of this problem.

In the standard version of this problem in the textbook, the common multiples shown are 12 and 24 (multiples of both 4 and 6, for example).

If the two numbers are 4 and 6 (LCM = 12):
- Multiples of 4 only (not 6): 4, 8, 16, 20, 28, 32, …
- Multiples of 6 only (not 4): 6, 18, 30, …
- Common multiples (multiples of 12): 12, 24, 36, …

Students should identify the two numbers from the given common multiples and then list the remaining multiples in each region accordingly.

Method: Identify the LCM from the common multiples shown, then find the two original numbers, and fill in their individual multiples in the respective regions.

We need the LCM of 1, 2, 3, 4, 5, 6, 8, 9, 10 (excluding 7).

Prime factorisations:
- $1 = 1$
- $2 = 2$
- $3 = 3$
- $4 = 2^2$
- $5 = 5$
- $6 = 2 \times 3$
- $8 = 2^3$
- $9 = 3^2$
- $10 = 2 \times 5$

LCM = highest power of each prime:
$$\text{LCM} = 2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = 360$$

The smallest number that is a multiple of all numbers from 1 to 10 except 7 is 360.

We need the LCM of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.

Prime factorisations:
- $2 = 2$
- $3 = 3$
- $4 = 2^2$
- $5 = 5$
- $6 = 2 \times 3$
- $7 = 7$
- $8 = 2^3$
- $9 = 3^2$
- $10 = 2 \times 5$

LCM = highest power of each prime present:
$$\text{LCM} = 2^3 \times 3^2 \times 5 \times 7 = 8 \times 9 \times 5 \times 7 = 2520$$

The smallest number that is a multiple of all numbers from 1 to 10 is 2520.

Given: 2 is a prime and even number.

An even number is divisible by 2. If an even number is greater than 2, it has at least three factors: 1, 2, and itself. Therefore, it cannot be prime.

No, there is no other even prime number. 2 is the only even prime number.

Primes up to 100: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

Differences between successive primes:
- 3−2=1, 5−3=2, 7−5=2, 11−7=4, 13−11=2, 17−13=4, 19−17=2, 23−19=4, 29−23=6, 31−29=2, 37−31=6, 41−37=4, 43−41=2, 47−43=4, 53−47=6, 59−53=6, 61−59=2, 67−61=6, 71−67=4, 73−71=2, 79−73=6, 83−79=4, 89−83=6, 97−89=8.

Smallest difference = 1 (between 2 and 3).
Largest difference = 8 (between 89 and 97).

The table arranges numbers 1–100 in rows of 10 (decades).

Primes in each decade:
- 1–10: 2, 3, 5, 7 → 4 primes
- 11–20: 11, 13, 17, 19 → 4 primes
- 21–30: 23, 29 → 2 primes
- 31–40: 31, 37 → 2 primes
- 41–50: 41, 43, 47 → 3 primes
- 51–60: 53, 59 → 2 primes
- 61–70: 61, 67 → 2 primes
- 71–80: 71, 73, 79 → 3 primes
- 81–90: 83, 89 → 2 primes
- 91–100: 97 → 1 prime

No, there are not an equal number of primes in every row.

Least primes: 91–100 (only 1 prime).
Most primes: 1–10 and 11–20 (4 primes each).

A prime number has exactly two factors: 1 and itself.

- 23: Check divisibility by primes up to $\sqrt{23} \approx 4.8$, i.e., 2 and 3. 23 is odd; 2+3=5, not divisible by 3. 23 is prime. ✓
- 51: 5+1=6, divisible by 3. $51 = 3 \times 17$. 51 is not prime.
- 37: Check primes up to $\sqrt{37} \approx 6.1$, i.e., 2, 3, 5. 37 is odd; 3+7=10, not divisible by 3; does not end in 0 or 5. 37 is prime. ✓
- 26: Even number, $26 = 2 \times 13$. 26 is not prime.

The prime numbers are 23 and 37.

Primes less than 20: 2, 3, 5, 7, 11, 13, 17, 19.

We need pairs whose sum is divisible by 5:

- $2 + 3 = 5$ ✓ (multiple of 5) → Pair: (2, 3)
- $2 + 13 = 15$ ✓ (multiple of 5) → Pair: (2, 13)
- $7 + 13 = 20$ ✓ (multiple of 5) → Pair: (7, 13)
- (Other valid pairs: (3,7)=10 ✓, (2,3)=5 ✓, (11,19)=30 ✓)

Three pairs: (2, 3), (2, 13), and (7, 13). [Other valid answers are also acceptable.]

We need pairs of prime numbers up to 100 that use the same digits (just in different order — i.e., reversals of each other).

Checking all two-digit primes and their reversal:
- 13 and 31: both prime ✓
- 17 and 71: both prime ✓
- 37 and 73: both prime ✓
- 79 and 97: both prime ✓
- 11 and 11: same number, not a pair
- 12 reversed is 21 = 3×7, not prime
- 14 reversed is 41 (prime) but 14 is not prime

Pairs of prime numbers up to 100 with the same digits:
$$(13, 31),\ (17, 71),\ (37, 73),\ (79, 97)$$

We need 7 consecutive numbers, all composite (none prime).

Consider numbers 90 to 96:
- 90 = 2×45 (composite)
- 91 = 7×13 (composite)
- 92 = 4×23 (composite)
- 93 = 3×31 (composite)
- 94 = 2×47 (composite)
- 95 = 5×19 (composite)
- 96 = 2×48 (composite)

All seven numbers 90, 91, 92, 93, 94, 95, 96 are composite.

Seven consecutive composite numbers between 1 and 100: 90, 91, 92, 93, 94, 95, 96.

Twin primes are pairs of primes differing by 2.

Primes up to 100: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

Checking pairs with difference 2:
- (3, 5) ✓ — given
- (5, 7) ✓
- (11, 13) ✓
- (17, 19) ✓ — given
- (29, 31) ✓
- (41, 43) ✓
- (59, 61) ✓
- (71, 73) ✓

Other twin prime pairs between 1 and 100 (besides (3,5) and (17,19)):
$$(5,7),\ (11,13),\ (29,31),\ (41,43),\ (59,61),\ (71,73)$$

a. There is no prime number whose units digit is 4.
TRUE. Any number ending in 4 is even (divisible by 2). If it is greater than 2, it has at least three factors (1, 2, and itself), so it cannot be prime.

b. A product of primes can also be prime.
FALSE. A product of two or more primes has at least four factors (1, each prime, and the product itself), so it is composite. For example, $2 \times 3 = 6$, which is not prime.

c. Prime numbers do not have any factors.
FALSE. Every prime number has exactly two factors: 1 and itself. For example, 7 has factors 1 and 7.

d. All even numbers are composite numbers.
FALSE. The number 2 is even but it is prime (its only factors are 1 and 2).

e. 2 is a prime and so is the next number, 3. For every other prime, the next number is composite.
TRUE. Every prime greater than 3 is odd. The number after any odd number is even (and greater than 2), hence divisible by 2, hence composite. So for every prime p &gt; 3, the number $p+1$ is even and greater than 2, making it composite.

We find the prime factorisation of each:

- $45 = 3 \times 3 \times 5 = 3^2 \times 5$ → only 2 distinct primes (3 and 5)
- $60 = 2 \times 2 \times 3 \times 5 = 2^2 \times 3 \times 5$ → 3 distinct primes, but not a product of exactly three (has repeated factor)
- $91 = 7 \times 13$ → only 2 distinct primes
- $105 = 3 \times 5 \times 7$ → exactly 3 distinct primes, each appearing once ✓
- $330 = 2 \times 3 \times 5 \times 11$ → 4 distinct primes

The number that is the product of exactly three distinct prime numbers is 105 $(= 3 \times 5 \times 7)$.

Using digits 2, 4, 5 each exactly once, the possible three-digit numbers are:
245, 254, 425, 452, 524, 542.

Check each for primality:
- 245 = 5 × 49 = 5 × 7² → composite
- 254 = 2 × 127 → composite (even)
- 425 = 5 × 85 = 5 × 5 × 17 → composite
- 452 = 4 × 113 → composite (even)
- 524 = 4 × 131 → composite (even)
- 542 = 2 × 271 → composite (even)

All even numbers (ending in 2 or 4) are composite. Numbers ending in 5 are divisible by 5, hence composite.

No three-digit prime number can be made using each of 2, 4, and 5 exactly once. The answer is 0.

We need primes $p$ such that $2p + 1$ is also prime.

- $p = 3$: $2(3)+1 = 7$ ✓ (prime) — given
- $p = 5$: $2(5)+1 = 11$ ✓ (prime)
- $p = 7$: $2(7)+1 = 15 = 3 \times 5$ ✗ (not prime)
- $p = 11$: $2(11)+1 = 23$ ✓ (prime)
- $p = 13$: $2(13)+1 = 27 = 3^3$ ✗ (not prime)
- $p = 23$: $2(23)+1 = 47$ ✓ (prime)
- $p = 29$: $2(29)+1 = 59$ ✓ (prime)
- $p = 41$: $2(41)+1 = 83$ ✓ (prime)

Five examples: (3,7), (5,11), (11,23), (23,47), (29,59).

A pair is 'safe' (co-prime) if the only common factor is 1 (i.e., no jump size other than 1 can reach both).

a. 15 and 39:
Factors of 15: 1, 3, 5, 15
Factors of 39: 1, 3, 13, 39
Common factors: 1 and 3. Since they share the factor 3, not safe (Jumpy can use jump size 3).

b. 4 and 15:
Factors of 4: 1, 2, 4
Factors of 15: 1, 3, 5, 15
Common factors: only 1. Safe ✓ (co-prime)

c. 18 and 29:
Factors of 18: 1, 2, 3, 6, 9, 18
Factors of 29: 1, 29 (29 is prime)
Common factors: only 1. Safe ✓ (co-prime)

d. 20 and 55:
Factors of 20: 1, 2, 4, 5, 10, 20
Factors of 55: 1, 5, 11, 55
Common factors: 1 and 5. Since they share the factor 5, not safe (Jumpy can use jump size 5).

Jump sizes that reach both 15 and 30 must be common factors of 15 and 30.

Factors of 15: 1, 3, 5, 15
Factors of 30: 1, 2, 3, 5, 6, 10, 15, 30

Common factors of 15 and 30: 1, 3, 5, 15

All jump sizes that can reach both 15 and 30 are: 1, 3, 5, and 15.

Using repeated division by prime numbers:

64:
$64 = 2 \times 32 = 2 \times 2 \times 16 = 2^6$
$$64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^6$$

104:
$104 = 2 \times 52 = 2 \times 2 \times 26 = 2 \times 2 \times 2 \times 13$
$$104 = 2^3 \times 13$$

105:
$105 = 3 \times 35 = 3 \times 5 \times 7$
$$105 = 3 \times 5 \times 7$$

243:
$243 = 3 \times 81 = 3 \times 3 \times 27 = 3 \times 3 \times 3 \times 9 = 3^5$
$$243 = 3^5$$

320:
$320 = 2 \times 160 = 2^2 \times 80 = 2^3 \times 40 = 2^4 \times 20 = 2^5 \times 10 = 2^6 \times 5$
$$320 = 2^6 \times 5$$

141:
$141 = 3 \times 47$ (47 is prime)
$$141 = 3 \times 47$$

1728:
$1728 = 2 \times 864 = 2^2 \times 432 = 2^3 \times 216 = 2^3 \times 8 \times 27 = 2^6 \times 27 = 2^6 \times 3^3$
$$1728 = 2^6 \times 3^3$$

729:
$729 = 3^6$ (since $3^6 = 729$)
$$729 = 3^6$$

1024:
$1024 = 2^{10}$ (since $2^{10} = 1024$)
$$1024 = 2^{10}$$

1331:
$1331 = 11^3$ (since $11^3 = 1331$)
$$1331 = 11^3$$

1000:
$1000 = 8 \times 125 = 2^3 \times 5^3$
$$1000 = 2^3 \times 5^3$$

Given: Prime factorisation has one 2, two 3s, and one 11.

$$\text{Number} = 2^1 \times 3^2 \times 11^1 = 2 \times 9 \times 11 = 198$$

The number is 198.

Given: Product of three primes (all < 30) = 1955.

Step 1: Check divisibility. $1955$ ends in 5, so divisible by 5.
$$1955 \div 5 = 391$$

Step 2: Factorise 391.
$391 \div 7 = 55.8...$ (no)
$391 \div 11 = 35.5...$ (no)
$391 \div 17 = 23$ ✓
$$391 = 17 \times 23$$

So $1955 = 5 \times 17 \times 23$.

Verify: all three (5, 17, 23) are prime and less than 30. ✓

The three prime numbers are 5, 17, and 23.

To get the smallest such number, use the smallest available primes.

Part a: Three different prime numbers
Smallest three primes: 2, 3, 5
$$\text{Smallest number} = 2 \times 3 \times 5 = 30$$

Part b: Four different prime numbers
Smallest four primes: 2, 3, 5, 7
$$\text{Smallest number} = 2 \times 3 \times 5 \times 7 = 210$$

Answers: a. 30, b. 210.

Two numbers are co-prime if they have no common prime factor.

a. 30 and 45:
Guess: Not co-prime (both divisible by 5).
$30 = 2 \times 3 \times 5$
$45 = 3^2 \times 5$
Common prime factors: 3 and 5.
Not co-prime.

b. 57 and 85:
Guess: Not co-prime (both seem divisible by some common factor).
$57 = 3 \times 19$
$85 = 5 \times 17$
No common prime factors.
Co-prime. ✓

c. 121 and 1331:
Guess: Not co-prime ($121 = 11^2$, $1331 = 11^3$).
$121 = 11^2$
$1331 = 11^3$
Common prime factor: 11.
Not co-prime.

d. 343 and 216:
Guess: Co-prime ($343 = 7^3$, $216 = 6^3$).
$343 = 7^3$
$216 = 2^3 \times 3^3$
No common prime factors.
Co-prime. ✓

A number is divisible by another if the prime factorisation of the second is included in (contained within) the prime factorisation of the first.

a. 225 and 27:
$225 = 3^2 \times 5^2$
$27 = 3^3$
$27$ requires $3^3$ but $225$ has only $3^2$. The factorisation of 27 is NOT included in 225.
225 is NOT divisible by 27.

b. 96 and 24:
$96 = 2^5 \times 3$
$24 = 2^3 \times 3$
$24$ requires $2^3$ and $3^1$; $96$ has $2^5$ and $3^1$. The factorisation of 24 IS included in 96.
96 IS divisible by 24. ($96 \div 24 = 4$)

c. 343 and 17:
$343 = 7^3$
$17 = 17$ (prime)
17 is not a factor of 343.
343 is NOT divisible by 17.

d. 999 and 99:
$999 = 3^3 \times 37$
$99 = 3^2 \times 11$
$99$ requires the prime factor 11, but 11 is not in the factorisation of 999.
999 is NOT divisible by 99.

First number: $A = 2 \times 3 \times 7 = 42$
Second number: $B = 3 \times 7 \times 11 = 231$

Are they co-prime?
Common prime factors: 3 and 7.
Since they share prime factors, they are NOT co-prime.

Does one divide the other?
For $A$ to divide $B$: the factorisation of $A$ ($2 \times 3 \times 7$) must be included in $B$ ($3 \times 7 \times 11$). But 2 is in $A$ and not in $B$, so $A$ does not divide $B$.

For $B$ to divide $A$: the factorisation of $B$ ($3 \times 7 \times 11$) must be included in $A$ ($2 \times 3 \times 7$). But 11 is in $B$ and not in $A$, so $B$ does not divide $A$.

Neither number divides the other.

Given: Claim — any two prime numbers are co-prime.

A prime number has only two factors: 1 and itself. If $p$ and $q$ are two different prime numbers, then the only factors of $p$ are 1 and $p$, and the only factors of $q$ are 1 and $q$. Since $p \neq q$, the only common factor is 1.

Therefore, any two distinct prime numbers are co-prime.

(Note: If both primes are the same, e.g., $p = q = 5$, then they are not co-prime since 5 is a common factor. But typically 'two prime numbers' implies two different primes.)

Yes, Guna is right. Any two distinct prime numbers are always co-prime.

Numbers between 120 and 140 divisible by 8:
$8 \times 15 = 120$, $8 \times 16 = 128$, $8 \times 17 = 136$, 8 \times 18 = 144 &gt; 140.
So: 128 and 136.

Numbers between 1120 and 1140 divisible by 8:
$8 \times 140 = 1120$, $8 \times 141 = 1128$, $8 \times 142 = 1136$, 8 \times 143 = 1144 &gt; 1140.
So: 1128 and 1136.

Numbers between 3120 and 3140 divisible by 8:
$8 \times 390 = 3120$, $8 \times 391 = 3128$, $8 \times 392 = 3136$, 8 \times 393 = 3144 &gt; 3140.
So: 3128 and 3136.

Observation: In each range, the numbers divisible by 8 end in the same last three digits (128 and 136). This shows that divisibility by 8 depends only on the last three digits.

Changing last two digits of 8560:
$8560$: last three digits are 560. $560 \div 8 = 70$. So 8560 is already divisible by 8.
To make a different multiple: change last two digits so last three digits form a multiple of 8. For example, change to 856→ 8568 ($568 \div 8 = 71$ ✓) or 8576 ($576 \div 8 = 72$ ✓).

Statement 1: Only the last three digits matter when deciding if a given number is divisible by 8.
Agree. Any number can be written as $1000k + r$ where $r$ is the number formed by the last three digits. Since $1000 = 8 \times 125$, the number $1000k$ is always divisible by 8. Therefore, the original number is divisible by 8 if and only if $r$ (the last three digits) is divisible by 8.

Statement 2: If the number formed by the last three digits is divisible by 8, then the original number is divisible by 8.
Agree. As shown above, original number $= 1000k + r$. If $8 \mid r$ and $8 \mid 1000k$, then $8 \mid (1000k + r)$.

Statement 3: If the original number is divisible by 8, then the number formed by the last three digits is divisible by 8.
Agree. If $8 \mid (1000k + r)$ and $8 \mid 1000k$, then $8 \mid r$.

Part a:
This depends on the student's birth year. For example, if born in 2012:
Leap years from 2012 to 2024: 2012, 2016, 2020, 2024.
(Students should list multiples of 4 from their birth year to the current year, excluding centennial years not divisible by 400.)

Part b: Leap years from 2024 to 2099:
Multiples of 4 from 2024 to 2099:
First: 2024, Last: 2096.
Count: $\frac{2096 - 2024}{4} + 1 = \frac{72}{4} + 1 = 18 + 1 = 19$.

Check for exceptions: 2100 is divisible by 100 but not 400, so it would not be a leap year — but 2100 is outside our range. No year between 2024 and 2099 is divisible by 100 (the only candidate would be 2100).

There are 19 leap years from 2024 to 2099.
(They are: 2024, 2028, 2032, 2036, 2040, 2044, 2048, 2052, 2056, 2060, 2064, 2068, 2072, 2076, 2080, 2084, 2088, 2092, 2096.)

A palindrome reads the same forwards and backwards. A 4-digit palindrome has the form $\overline{abba}$ where $a \neq 0$.

For divisibility by 4, only the last two digits matter. The last two digits of $\overline{abba}$ are $\overline{ba}$.

We need $\overline{ba}$ (i.e., the two-digit number $10b + a$) to be divisible by 4.

Smallest 4-digit palindrome divisible by 4:
Smallest 4-digit palindromes: 1001, 1111, 1221, 1331, 1441, 1551, 1661, 1771, 1881, 1991, 2002, …

Check last two digits for divisibility by 4:
- 1001: last two digits 01 → 1 ÷ 4 = not divisible
- 1111: last two digits 11 → not divisible
- 1221: last two digits 21 → not divisible
- 1331: last two digits 31 → not divisible
- 1441: last two digits 41 → not divisible
- 1551: last two digits 51 → not divisible
- 1661: last two digits 61 → not divisible
- 1771: last two digits 71 → not divisible
- 1881: last two digits 81 → not divisible
- 1991: last two digits 91 → not divisible
- 2002: last two digits 02 → 2 ÷ 4 = not divisible
- 2112: last two digits 12 → 12 ÷ 4 = 3 ✓

Smallest = 2112

Largest 4-digit palindrome divisible by 4:
Largest 4-digit palindromes (descending): 9999, 9889, 9779, 9669, 9559, 9449, 9339, 9229, 9119, 9009, 8998, 8888, 8778, 8668, …

Check last two digits:
- 9999: 99 ÷ 4 → not divisible
- 9889: 89 ÷ 4 → not divisible
- 9779: 79 ÷ 4 → not divisible
- 9669: 69 ÷ 4 → not divisible
- 9559: 59 ÷ 4 → not divisible
- 9449: 49 ÷ 4 → not divisible
- 9339: 39 ÷ 4 → not divisible
- 9229: 29 ÷ 4 → not divisible
- 9119: 19 ÷ 4 → not divisible
- 9009: 09 ÷ 4 → not divisible
- 8998: 98 ÷ 4 → not divisible (98 = 4×24+2)
- 8888: 88 ÷ 4 = 22 ✓

Largest = 8888

Smallest 4-digit palindrome divisible by 4: 2112; Largest: 8888.

Part a: Sum of two even numbers gives a multiple of 4.
Sometimes true.

Examples:
- $2 + 2 = 4$ ✓ (multiple of 4)
- $2 + 4 = 6$ ✗ (not a multiple of 4)
- $4 + 8 = 12$ ✓ (multiple of 4)
- $2 + 6 = 8$ ✓ (multiple of 4)
- $2 + 8 = 10$ ✗ (not a multiple of 4)

The sum of two even numbers is always even, but not always a multiple of 4. Sometimes true.

Part b: Sum of two odd numbers gives a multiple of 4.
Sometimes true.

Examples:
- $1 + 3 = 4$ ✓ (multiple of 4)
- $1 + 5 = 6$ ✗ (not a multiple of 4)
- $3 + 5 = 8$ ✓ (multiple of 4)
- $1 + 7 = 8$ ✓ (multiple of 4)
- $1 + 9 = 10$ ✗ (not a multiple of 4)

The sum of two odd numbers is always even, but not always a multiple of 4. Sometimes true.

Key rules:
- Remainder when divided by 10 = units digit
- Remainder when divided by 5: if units digit is 0 or 5, remainder = 0; if units digit is 1 or 6, remainder = 1; if 2 or 7, remainder = 2; if 3 or 8, remainder = 3; if 4 or 9, remainder = 4.
- Remainder when divided by 2: 0 if even, 1 if odd.

| Number | ÷ 10 | ÷ 5 | ÷ 2 |
|--------|------|-----|-----|
| 78 | 8 | 3 | 0 |
| 99 | 9 | 4 | 1 |
| 173 | 3 | 3 | 1 |
| 572 | 2 | 2 | 0 |
| 980 | 0 | 0 | 0 |
| 1111 | 1 | 1 | 1 |
| 2345 | 5 | 0 | 1 |

Given: 14560. We need to find two numbers from {2, 4, 5, 8, 10} whose divisibility implies divisibility by all others.

Note that:
- $10 = 2 \times 5$, so divisibility by 10 implies divisibility by both 2 and 5.
- $8 = 2^3$, so divisibility by 8 implies divisibility by 4 and 2.
- If a number is divisible by both 8 and 10, then it is divisible by 8 (hence by 4 and 2) and by 10 (hence by 5 and 2). So divisibility by all five is guaranteed.

Check: $14560 \div 8 = 1820$ ✓ and $14560 \div 10 = 1456$ ✓.

Guna checked divisibility by 8 and 10. Divisibility by 8 ensures divisibility by 4 and 2; divisibility by 10 ensures divisibility by 5 and 2. Together, all five are covered.

For a number to be divisible by all of 2, 4, 5, 8, and 10, it must be divisible by LCM(2,4,5,8,10) = 40.

A number is divisible by 40 if it is divisible by both 8 and 5 (i.e., divisible by 8 and ends in 0).

- 572: Does not end in 0 → not divisible by 5 or 10. No.
- 2352: Does not end in 0 → not divisible by 5 or 10. No.
- 5600: Ends in 0 ✓. Last three digits: 600. $600 \div 8 = 75$ ✓. Yes.
- 6000: Ends in 0 ✓. Last three digits: 000. $0 \div 8 = 0$ ✓. Yes.
- 77622160: Ends in 0 ✓. Last three digits: 160. $160 \div 8 = 20$ ✓. Yes.

Numbers divisible by all of 2, 4, 5, 8 and 10: 5600, 6000, and 77622160.

$10000 = 2^4 \times 5^4$

We need to split this into two factors, neither ending in 0. A number ends in 0 only if it has both 2 and 5 as factors. So we must give all the 2s to one number and all the 5s to the other.

$$2^4 = 16 \quad \text{and} \quad 5^4 = 625$$

Check: $16 \times 625 = 10000$ ✓
Units digit of 16 is 6 (not 0) ✓
Units digit of 625 is 5 (not 0) ✓

The two numbers are 16 and 625.

Box 1: 5, 7, 12, 35
- 5 is special: it is the only single-digit number; also the only multiple of 5 that is prime.
- 7 is special: it is the only prime number that is not a factor of 35 among the primes listed; actually 7 is prime and a factor of 35.
- 12 is special: it is the only even number; it is the only composite number not involving 5 or 7; it is the only multiple of 4.
- 35 is special: it is the only two-digit composite number; it is the only multiple of both 5 and 7.

Box 2: 3, 8, 11, 24
- 3 is special: it is the only single-digit odd prime.
- 8 is special: it is the only power of 2 (and the only even number that is not a multiple of 3).
- 11 is special: it is the only two-digit prime number.
- 24 is special: it is the only composite number divisible by both 2 and 3; it is the only multiple of 8.

Box 3: 27, 3, 123, 31
- 3 is special: it is the only single-digit number; it is the only prime.
- 27 is special: it is the only perfect cube ($3^3$).
- 31 is special: it is the only two-digit prime.
- 123 is special: it is the only three-digit number; it is the only multiple of 41.

Box 4: 17, 27, 44, 65
- 17 is special: it is the only prime number.
- 27 is special: it is the only odd number that is a perfect cube ($3^3$).
- 44 is special: it is the only even number; it is the only multiple of 4.
- 65 is special: it is the only multiple of 5 and 13; it is the only number divisible by 5.

We need to fill a 3×3 grid with prime numbers such that:
- Row 1 product = 63, Row 2 product = 27, Row 3 product = 190
- Column 1 product = 45, Column 2 product = 42, Column 3 product = 171

Prime factorisations:
- $63 = 3^2 \times 7 = 3 \times 3 \times 7$
- $27 = 3^3 = 3 \times 3 \times 3$
- $190 = 2 \times 5 \times 19$
- $45 = 3^2 \times 5 = 3 \times 3 \times 5$
- $42 = 2 \times 3 \times 7$
- $171 = 3^2 \times 19 = 3 \times 3 \times 19$

Let the grid be:
$$\begin{pmatrix} a &amp; b &amp; c \\ d &amp; e &amp; f \\ g &amp; h &amp; i \end{pmatrix}$$

Row 1: $a \times b \times c = 63 = 3 \times 3 \times 7$
Row 2: $d \times e \times f = 27 = 3 \times 3 \times 3$
Row 3: $g \times h \times i = 190 = 2 \times 5 \times 19$

Column 1: $a \times d \times g = 45 = 3 \times 3 \times 5$
Column 2: $b \times e \times h = 42 = 2 \times 3 \times 7$
Column 3: $c \times f \times i = 171 = 3 \times 3 \times 19$

From Row 3: $g = 2, h = 5, i = 19$ (or permutations).
From Column 1: $a \times d \times 2 = 45$ → $a \times d = 22.5$ — not integer with $g=2$.

Try $g = 5, h = 2, i = 19$:
Column 1: $a \times d \times 5 = 45$ → $a \times d = 9 = 3 \times 3$
Column 2: $b \times e \times 2 = 42$ → $b \times e = 21 = 3 \times 7$
Column 3: $c \times f \times 19 = 171$ → $c \times f = 9 = 3 \times 3$

Row 1: $a \times b \times c = 63$; Row 2: $d \times e \times f = 27$.

Let $a = 3, d = 3$ (from $a \times d = 9$).
Let $b = 3, e = 7$ (from $b \times e = 21$).
Let $c = 3, f = 3$ (from $c \times f = 9$).

Check Row 1: $3 \times 3 \times 3 = 27 \neq 63$. ✗

Try $b = 7, e = 3$:
Row 1: $a \times 7 \times c = 63$ → $a \times c = 9 = 3 \times 3$, so $a = 3, c = 3$.
Row 2: $d \times 3 \times f = 27$ → $d \times f = 9 = 3 \times 3$, so $d = 3, f = 3$.
Check Column 1: $3 \times 3 \times 5 = 45$ ✓
Check Column 3: $3 \times 3 \times 19 = 171$ ✓

$$\boxed{\begin{pmatrix} 3 &amp; 7 &amp; 3 \\ 3 &amp; 3 &amp; 3 \\ 5 &amp; 2 &amp; 19 \end{pmatrix}}$$

Verification:
- Row 1: $3 \times 7 \times 3 = 63$ ✓
- Row 2: $3 \times 3 \times 3 = 27$ ✓
- Row 3: $5 \times 2 \times 19 = 190$ ✓
- Col 1: $3 \times 3 \times 5 = 45$ ✓
- Col 2: $7 \times 3 \times 2 = 42$ ✓
- Col 3: $3 \times 3 \times 19 = 171$ ✓

Prime factorisations:
- $343 = 7^3 = 7 \times 7 \times 7$
- $66 = 2 \times 3 \times 11$
- $44 = 2^2 \times 11 = 2 \times 2 \times 11$
- $28 = 2^2 \times 7 = 2 \times 2 \times 7$
- $154 = 2 \times 7 \times 11$
- $231 = 3 \times 7 \times 11$

Let the grid be:
$$\begin{pmatrix} a &amp; b &amp; c \\ d &amp; e &amp; f \\ g &amp; h &amp; i \end{pmatrix}$$

Row 1: $a \times b \times c = 343 = 7 \times 7 \times 7$ → $a = b = c = 7$

Column 1: $7 \times d \times g = 28$ → $d \times g = 4 = 2 \times 2$, so $d = 2, g = 2$.
Column 2: $7 \times e \times h = 154$ → $e \times h = 22 = 2 \times 11$.
Column 3: $7 \times f \times i = 231$ → $f \times i = 33 = 3 \times 11$.

Row 2: $d \times e \times f = 66 = 2 \times 3 \times 11$. With $d = 2$: $e \times f = 33 = 3 \times 11$.
Row 3: $g \times h \times i = 44 = 2 \times 2 \times 11$. With $g = 2$: $h \times i = 22 = 2 \times 11$.

From $e \times f = 33$ and $e \times h = 22$ and $f \times i = 33$ and $h \times i = 22$:

Let $e = 3, f = 11$: then $h = 22/3$ — not integer.
Let $e = 11, f = 3$: then $h = 22/11 = 2$ and $i = 33/3 = 11$.
Check Row 3: $2 \times 2 \times 11 = 44$ ✓

$$\boxed{\begin{pmatrix} 7 &amp; 7 &amp; 7 \\ 2 &amp; 11 &amp; 3 \\ 2 &amp; 2 &amp; 11 \end{pmatrix}}$$

Verification:
- Row 1: $7 \times 7 \times 7 = 343$ ✓
- Row 2: $2 \times 11 \times 3 = 66$ ✓
- Row 3: $2 \times 2 \times 11 = 44$ ✓
- Col 1: $7 \times 2 \times 2 = 28$ ✓
- Col 2: $7 \times 11 \times 2 = 154$ ✓
- Col 3: $7 \times 3 \times 11 = 231$ ✓