For square tiles to fit both dimensions exactly, the tile’s side must be a common factor of 12 and 16.

\( \text{Factors of }12 = 1,2,3,4,6,12 \qquad \text{Factors of }16 = 1,2,4,8,16 \)

Common factors: \(1, 2, 4\). To use the fewest tiles, each tile should cover the largest possible area — so we pick the largest common factor, the HCF.

Number of tiles: Along the breadth: \(12 \div 4 = 3\) tiles. Along the length: \(16 \div 4 = 4\) tiles.

Why the largest tile minimises the count: a bigger tile covers more floor area per piece, so fewer pieces are needed to cover the same total area.

If the tile’s side could be a fraction of a foot: Suppose the side is \(x\), with \(12 = ax\) and \(16 = bx\) for positive integers \(a, b\). Then \( \dfrac{16}{12} = \dfrac{b}{a} = \dfrac{4}{3}\), and the smallest whole numbers satisfying \(b:a = 4:3\) are \(a = 3, b = 4\), giving \(x = 12 \div 3 = 4\). No larger valid \(x\) exists even among fractions.

The bag weight must divide both 84 and 108 exactly — it must be a common factor. Using few bags means choosing the largest common factor.

\(84 = 2^2 \times 3 \times 7\) and \(108 = 2^2 \times 3^3\). Common primes: two 2’s and one 3.

Number of bags: \(84 \div 12 = 7\) bags from the first farm, \(108 \div 12 = 9\) bags from the second — 16 bags in total, the fewest possible.

To land exactly on both numbers, the jump size must divide both — so it’s a common factor, and the longest jump is the HCF.

(a) \(14 = 2 \times 7,\ \ 30 = 2 \times 3 \times 5\) → common factor \(2\) → HCF = 2

(b) \(7\) and \(11\) are both prime and different → no common factor except 1 → HCF = 1

(c) \(30 = 2 \times 3 \times 5,\ \ 50 = 2 \times 5^2\) → common \(2 \times 5\) → HCF = 10

(d) \(28 = 2^2 \times 7,\ \ 42 = 2 \times 3 \times 7\) → common \(2 \times 7\) → HCF = 14

Yes. Since Jumpy starts at 0, every number he lands on is a multiple of the jump size. To land exactly on both treasure numbers, the jump size must divide both of them — that is, it must be a common factor of the two numbers. To make the jump as long as possible, Jumpy must pick the greatest such common factor — which is exactly the definition of the HCF.

(a) 90 = \(2 \times 3^2 \times 5\) → Factors: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90

(b) 105 = \(3 \times 5 \times 7\) → Factors: 1, 3, 5, 7, 15, 21, 35, 105

(c) 132 = \(2^2 \times 3 \times 11\) → Factors: 1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, 132

(d) 360 = \(2^3 \times 3^2 \times 5\) → Factors: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360

(e) 840 = \(2^3 \times 3 \times 5 \times 7\) → Factors: 1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 15, 20, 21, 24, 28, 30, 35, 40, 42, 56, 60, 70, 84, 105, 120, 140, 168, 210, 280, 420, 840

Anshu’s claim: False. Counterexample — \(96 = 2^5 \times 3\) has a longer factorisation (6 factors written out) than \(121 = 11 \times 11\) (2 factors written out), yet \(121 > 96\). A single counterexample is enough to disprove a conjecture.

(a) \(50 = 2\times5^2,\ \ 60 = 2^2\times3\times5\). Common: \(2 \times 5\). Common factors: 1, 2, 5, 10. HCF = 10

(b) \(140 = 2^2\times5\times7,\ \ 275 = 5^2\times11\). Common: \(5\). Common factors: 1, 5. HCF = 5

(c) \(77 = 7\times11,\ \ 725 = 5^2\times29\). No common prime. Common factor: 1. HCF = 1

(d) \(370 = 2\times5\times37,\ \ 592 = 2^4\times37\). Common: \(2 \times 37\). Common factors: 1, 2, 37, 74. HCF = 74

(e) \(81 = 3^4,\ \ 243 = 3^5\). Common: \(3^4\). Common factors: 1, 3, 9, 27, 81. HCF = 81

Is 14 a factor? Yes — \(14 = 2 \times 7\) uses one 2 and one 7, both available in \(840\)’s factorisation. \(840 \div 14 = 60\).

Is 8 a factor? Yes — \(8 = 2^3\), and \(840\) contains exactly \(2^3\). \(840 \div 8 = 105\).

Is 27 a factor? No — \(27 = 3^3\), but \(840\) contains only one factor of 3 (\(3^1\)), not three. Since we can’t find three 3’s inside \(840\)’s factorisation, 27 does not divide it exactly.

(a) \(24=2^3\times3,\ 180=2^2\times3^2\times5\) → HCF = \(2^2\times3=\)12

(b) \(42=2\times3\times7,\ 75=3\times5^2,\ 24=2^3\times3\) → only prime common to all three is 3 → HCF = 3

(c) \(240=2^4\times3\times5,\ 378=2\times3^3\times7\) → HCF = \(2\times3=\)6

(d) \(400=2^4\times5^2,\ 2500=2^2\times5^4\) → HCF = \(2^2\times5^2=\)100

(e) \(300=2^2\times3\times5^2,\ 800=2^5\times5^2\) → HCF = \(2^2\times5^2=\)100

Q2 — 72 & 144: No, we cannot conclude that. \(6\) and \(12\) (or \(8\) and \(18\)) are not prime factorisations, so this split hides the real common structure. In fact \(72 = 2^3\times3^2\) and \(144 = 2^4\times3^2\), and since 72 divides 144 exactly, their true HCF is 72 — far more than “1”. Only a full prime factorisation reveals the genuine common factors.

Anshu’s toran lengths: multiples of 6 → 6, 12, 18, 24, 30, 36…
Guna’s toran lengths: multiples of 8 → 8, 16, 24, 32, 40…

The length must be a common multiple of 6 and 8, and the shortest toran uses the lowest common multiple.

\(6 = 2\times3,\quad 8 = 2^3\). LCM needs the highest power of each prime: \(2^3 \times 3 = 24\).

Mondays fall on multiples of 7: 7, 14, 21, 28, 35, 42, 49, 56, 63, 70…
Kabamai’s visits fall on multiples of 10: 10, 20, 30, 40, 50, 60, 70…

She gets free sweets again on the first day common to both lists — the LCM.

\(7\) and \(10\) share no common prime factor (co-prime), so LCM \(= 7 \times 10 = 70\).

(a) \(4=2^2,\ 6=2\times3\) → LCM = \(2^2\times3 = \)12

(b) \(7, 11\) are co-prime → LCM = \(7\times11 = \)77

(c) \(14=2\times7,\ 30=2\times3\times5\) → LCM = \(2\times3\times5\times7 = \)210

(d) \(15=3\times5,\ 55=5\times11\) → LCM = \(3\times5\times11 = \)165

Is it always the LCM? Yes — “idli-vada” is called on numbers that are multiples of both, i.e. common multiples, and the first time it happens is at the smallest such number: the LCM.

(a) \(30=2\times3\times5,\ 72=2^3\times3^2\) → LCM = \(2^3\times3^2\times5 = \)360

(b) \(36=2^2\times3^2,\ 54=2\times3^3\) → LCM = \(2^2\times3^3 = \)108

(c) \(105=3\times5\times7,\ 195=3\times5\times13,\ 65=5\times13\) → LCM = \(3\times5\times7\times13 = \)1365

(d) \(222=2\times3\times37,\ 370=2\times5\times37\) → LCM = \(2\times3\times5\times37 = \)1110

\(2\times3\times5\times7 = 210\). It contains \(2\times7\) (=14) as a subpart, and \(5\times7\) (=35) as a subpart — so it is divisible by both 14 and 35.

This happens exactly when one number is a factor of the other, i.e. the second number is a multiple of the first.

(a) If the number is \(m\), the other number can be \(km\) for any positive integer \(k\) — then HCF\((m, km) = m\).

(b) If the number is \(7k\), the other number can be \(j \times 7k\) for any positive integer \(j\) — then HCF\((7k,\, 7kj) = 7k\).

1(a) Two consecutive even numbers (e.g. 4,6 / 8,10): HCF is always 2. Reasoning: writing them as \(2k\) and \(2k+2 = 2(k+1)\), their HCF is \(2 \times \text{HCF}(k, k+1) = 2 \times 1 = 2\), since consecutive integers are always co-prime.

1(b) Two consecutive odd numbers (e.g. 5,7 / 9,11): HCF is always 1. Any common factor must divide their difference (2); since both are odd, they share no factor of 2, so only 1 remains.

1(c) Two even numbers (general): HCF is always even — it always contains at least one factor of 2, though it can be larger depending on the numbers.

1(d) Two consecutive numbers (e.g. 7,8): HCF is always 1 — consecutive integers share no common factor greater than 1.

1(e) Two co-prime numbers: HCF is always 1 — this is the very definition of co-prime numbers.

2. More pairs: (4,24), (6,24), (8,24), (12,24). General pattern: whenever one number \(m\) divides another number \(n=km\), then \(\text{LCM}(m, km) = km\) — the larger multiple itself.

3(a) Two multiples of 3 (e.g. 6,9 → 18; 12,15 → 60): the LCM is always itself a multiple of 3, since 3 is a common prime factor carried into the LCM.

3(b) Two consecutive even numbers (e.g. 4,6 → LCM 12; 8,10 → LCM 40): since HCF is always 2, LCM = (product of the two numbers) ÷ 2.

3(c) Two consecutive numbers (e.g. 7,8 → LCM 56): since HCF = 1, LCM = product of the two numbers.

3(d) Two co-prime numbers: since HCF = 1, LCM = product of the two numbers — same reasoning as (c).

The HCF also doubles. Doubling both numbers adds one extra factor of 2 to each of their prime factorisations, and this extra 2 carries straight into the largest common subpart.

Example: \(270 = 2\times3^3\times5,\ \ 50 = 2\times5^2\) → HCF = \(2\times5=10\).
Doubled: \(540 = 2^2\times3^3\times5,\ \ 100 = 2^2\times5^2\) → HCF = \(2^2\times5 = 20 = 2 \times 10\). ✓

When both numbers share an obvious common multiplier \(c\) (numbers \(c\!\times\!p\) and \(c\!\times\!q\)), their HCF = \(c \times \text{HCF}(p,q)\).

(a) HCF(10,15) = 5 → HCF = \(18\times5=\)90

(b) HCF(38,21) = 1 (38 = 2×19, 21 = 3×7, no common prime) → HCF = \(10\times1=\)10

(c) HCF(13,20) = 1 (13 is prime, doesn’t divide 20) → HCF = \(5\times1=\)5

(d) HCF(16,20) = 4 → HCF = \(12\times4=\)48

At each step, divide both numbers by a common prime factor and write the quotients below; stop when the quotients share no common prime factor. The HCF is the product of the divisors used; the LCM is the product of the divisors and the final quotients.

We don’t have to remove primes one at a time — any common factor we can spot works, as long as we finish once the remaining numbers are co-prime.

45, 105: \(45=3^2\times5,\ 105=3\times5\times7\). HCF \(=3\times5=15\). LCM \(=3^2\times5\times7=315\). Product \(=45\times105=4725=15\times315.\) ✓ multiplier = HCF = 15.

275, 352: \(275=5^2\times11,\ 352=2^5\times11\). HCF \(=11\). LCM \(=2^5\times5^2\times11=8800\). Product \(=275\times352=96800=11\times8800.\) ✓ multiplier = HCF = 11.

222, 370: \(222=2\times3\times37,\ 370=2\times5\times37\). HCF \(=2\times37=74\). LCM \(=2\times3\times5\times37=1110\). Product \(=222\times370=82140=74\times1110.\) ✓ multiplier = HCF = 74.

Why? Every prime factor in either number is either common to both (it goes into the HCF, contributing once, and into the LCM, contributing once) or unique to one number (it goes into the LCM only, contributing once, but appears once in the product from that one number). Multiplying HCF × LCM reassembles exactly the primes that make up the product — each common prime is “used” once by HCF and once by LCM, matching its double appearance across both original numbers, while each unique prime appears once via the LCM, matching its single appearance in the product.

Does it hold for three numbers? No. Counterexample: 2, 3, 4 → HCF = 1, LCM = 12, but \(1 \times 12 = 12 \ne 2\times3\times4 = 24\).

The product of two numbers is always a common multiple of both (each number obviously divides the product). Since the LCM is the smallest common multiple, the LCM can never exceed the product.

Reading the repeating block in each row: the top row‘s colour pattern repeats every 5 stars (with blue as the 5th colour in the cycle), and the bottom row‘s pattern repeats every 4 stars (with blue as the 4th colour in its cycle).

Blue stars occur at positions that are multiples of 5 in row 1, and multiples of 4 in row 2. Both rows show blue together at a common multiple of 4 and 5 — the smallest such position is the LCM.

\(4 = 2^2,\quad 5 = 5\) → co-prime → LCM \(= 4 \times 5 = 20\)

Let \(A = 5\times7\times11\times11 = 5\times7\times11^2\) and \(B = 5\times7\times7\times11\times2 = 2\times5\times7^2\times11\).

(a) Is A a multiple of B? A would need to contain every prime factor of B. But A has only one factor of 7 (B needs two), and A has no factor of 2 at all.

(b) Is A a factor of B? A would need every one of its prime factors to fit inside B. A has two 11’s, but B has only one 11.

(a) First number \(= 3^2\times5\times7^2\). Second number \(=12\times7\times11 = 2^2\times3\times7\times11\).

Common primes: 3 (min power 1) and 7 (min power 1).

(b) \(45 = 3^2\times5,\quad 36 = 2^2\times3^2\).

Since HCF = 1, the numbers share no common prime factor, so their LCM equals their product. We need two co-prime numbers whose product is 66 \(= 2\times3\times11\).

Split the prime factors into two co-prime groups, e.g. \(\{2\times3\}=6\) and \(\{11\}=11\).

The total number of cows must divide exactly by 3, by 5, and by 7 — so it must be a common multiple of 3, 5 and 7.

\(3, 5, 7\) are all prime and distinct, so LCM \(= 3\times5\times7 = 105\).

Multiples of 105 under 200: only 105 itself (\(105\times2=210 > 200\)).

A cube of side \(s\) fills the box exactly only if \(s\) divides all three dimensions — i.e. \(s\) must be a common factor of 12, 18 and 36, so \(s\) must divide their HCF.

\(12=2^2\times3,\ 18=2\times3^2,\ 36=2^2\times3^2\) → HCF \(= 2\times3 = 6\). So \(s\) must be a factor of 6: only 1, 2, 3, 6 qualify.

This is asking for HCF(306, 36). \(306 = 2\times3^2\times17,\quad 36 = 2^2\times3^2\).

Common: \(2\times3^2 = 18\).

The number must be a multiple of LCM(3,4,5,7). Since these are pairwise co-prime (except sharing no common factors), LCM \(= 3\times4\times5\times7 = 420\).

So the number is \(N = 420k\), and we need \(N \equiv 10 \pmod{11}\).

\(420 = 11\times38 + 2\), so \(420 \equiv 2 \pmod{11}\). We need \(2k \equiv 10 \pmod{11}\), i.e. \(k \equiv 5 \pmod{11}\) (since 2 × 6 ≡ 1 mod 11, multiply both sides by 6: \(k \equiv 60 \equiv 5 \pmod{11}\)).

Smallest positive \(k = 5\) → \(N = 420\times5 = 2100\).

Check: \(2100 \div 11 = 190\) remainder \(10\). ✓

“No one gets out” means the total splits evenly into groups of that size — the total must be divisible by both 6 and 9, i.e. a multiple of LCM(6,9) = 18. “Someone got out” for 10 means the total is not divisible by 10.

Check each option: 72 → ÷6=12 ✓, ÷9=8 ✓, ÷10=7.2 ✗(not exact) → fits!
90 → ÷6=15 ✓, ÷9=10 ✓, ÷10=9 ✓(exact) → doesn’t fit (10 should NOT divide evenly).
45 → ÷6 not exact → doesn’t fit.
3 → ÷6 not exact → doesn’t fit.
36 → ÷6=6 ✓, ÷9=4 ✓, ÷10=3.6 ✗(not exact) → fits!

Two different primes share no common factor (HCF = 1), so LCM\((m,n) = m \times n\) exactly.

Since \(m, n > 1\), the product \(m\times n\) is always strictly greater than both \(m\) and \(n\) individually, and it is neither less than nor greater than itself.

With each simultaneous leap, the dog gains on the rabbit by \(9 – 7 = 2\) feet.

Number of leaps needed to close the 150 ft gap: \(150 \div 2 = 75\).

We need the LCM of \(1,2,3,4,5,6,8,9,10\). Taking the highest power of each prime that appears:

\(2^3\) (from 8), \(3^2\) (from 9), \(5\) (from 5 or 10).

LCM \(= 2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = 360\)

To add fractions efficiently, use the LCM of the denominators as the common denominator — this keeps numbers as small as possible (rather than multiplying all denominators together).

\(15=3\times5,\ \ 20=2^2\times5,\ \ 36=2^2\times3^2,\ \ 63=3^2\times7,\ \ 21=3\times7\)

LCM \(= 2^2\times3^2\times5\times7 = 1260\)

Convert each fraction to a denominator of 1260:

\(\dfrac{8}{15}=\dfrac{672}{1260},\quad \dfrac{1}{20}=\dfrac{63}{1260},\quad \dfrac{7}{36}=\dfrac{245}{1260},\quad \dfrac{11}{63}=\dfrac{220}{1260},\quad \dfrac{1}{21}=\dfrac{60}{1260}\)

\[ \frac{672+63+245+220+60}{1260} = \frac{1260}{1260} \]