Chapter 3: Finding Common Ground

The tile is a square of side $$L$$ ft (whole number). To cover the floor without cutting any tile, $$L$$ must divide both $$12$$ and $$16$$ exactly.

So $$L$$ is a common factor of $$12$$ and $$16$$.

Factors of $$12$$: $$1, 2, 3, 4, 6, 12$$.
Factors of $$16$$: $$1, 2, 4, 8, 16$$.
Common factors: $$1, 2, 4$$.

To use the fewest tiles, the tile side must be as large as possible. So $$L$$ = the largest common factor $$ = 4$$ ft.

The tile size is $$4$$ ft $$\times\; 4$$ ft. Along the $$12$$ ft side she needs $$12 \div 4 = 3$$ tiles, and along the $$16$$ ft side she needs $$16 \div 4 = 4$$ tiles.

Total number of tiles $$= 3 \times 4 = 12$$.

Fractional lengths allowed: Suppose the side is $$L$$ (a positive number, not necessarily whole). We still need $$12/L$$ and $$16/L$$ to be whole numbers, say $$12/L = a$$ and $$16/L = b$$. Then $$L = 12/a = 16/b$$, which gives $$b/a = 16/12 = 4/3$$. The smallest whole numbers $$a$$ and $$b$$ with this ratio are $$a = 3$$ and $$b = 4$$, giving the same $$L = 4$$ ft and the same $$a\times b = 12$$ tiles.

So the answer does not change even if fractional lengths are allowed.

Let each bag weigh $$w$$ kg (a whole number). Since a bag contains rice from only one farm, $$w$$ must divide $$84$$ exactly and also divide $$108$$ exactly. So $$w$$ is a common factor of $$84$$ and $$108$$.

To use the fewest bags, $$w$$ must be the largest common factor, i.e., the HCF.

Prime factorisation:
$$84 = 2 \times 2 \times 3 \times 7$$
$$108 = 2 \times 2 \times 3 \times 3 \times 3$$

Common primes: two $$2$$s and one $$3$$.
$$\text{HCF} = 2 \times 2 \times 3 = 12$$.

So each bag should hold $$12$$ kg. Number of bags $$= 84/12 + 108/12 = 7 + 9 = 16$$ bags.

Yes. Jumpy starts at $$0$$ and makes jumps of equal size $$j$$. The numbers he can land on are $$0, j, 2j, 3j, \ldots$$ — the multiples of $$j$$.

For Jumpy to land on both treasure numbers, both must be multiples of $$j$$, i.e., $$j$$ must be a common factor of the two numbers.

The longest jump that still lands on both numbers is therefore the largest common factor, i.e., the HCF.

Yes. Listing every factor of each number is slow and error-prone — for large numbers we may miss some factors.

A simpler, more reliable process is to use prime factorisation:

1. Write each number as a product of primes.
2. Pick the primes that appear in both factorisations, taking each prime with the smallest power that appears.
3. The product of these picked primes is the HCF.

This method is systematic (we never miss a common factor) and works even for very large numbers.

Prime factorisation of $$840$$:

$$840 = 2 \times 2 \times 2 \times 3 \times 5 \times 7$$.

To be a factor of $$840$$, a number's prime factorisation must be built out of the primes above, and no prime may appear more times than it does in $$840$$.

Now $$28 = 2 \times 2 \times 7$$ uses two $$2$$s (allowed, since $$840$$ has three) and one $$7$$ (allowed, since $$840$$ has one). So $$28$$ divides $$840$$.

Check: $$840 \div 28 = 30$$, a whole number.

Since $$840 = 2 \times 2 \times 2 \times 3 \times 5 \times 7$$ and $$28 = 2 \times 2 \times 7$$, the primes left over in $$840$$ after taking out those of $$28$$ are one $$2$$, one $$3$$ and one $$5$$.

Their product: $$2 \times 3 \times 5 = 30$$.

Check: $$28 \times 30 = 840$$. ✓

Recall $$840 = 2 \times 2 \times 2 \times 3 \times 5 \times 7$$ — three $$2$$s, one $$3$$, one $$5$$, one $$7$$.

Is $$2 \times 7 = 14$$ a factor? $$14$$ uses one $$2$$ and one $$7$$. Both are present in $$840$$'s primes, in sufficient number. So yes, $$14$$ is a factor. (Check: $$840 \div 14 = 60$$.)

Is $$2 \times 2 \times 2 = 8$$ a factor? $$8$$ uses three $$2$$s. $$840$$ has exactly three $$2$$s. So yes, $$8$$ is a factor. (Check: $$840 \div 8 = 105$$.)

Is $$3 \times 3 \times 3 = 27$$ a factor? $$27$$ needs three $$3$$s, but $$840$$ has only one $$3$$. So no, $$27$$ is not a factor. (Check: $$840 \div 27 = 31.11\ldots$$, not a whole number.)

Prime factorise: $$225 = 9 \times 25 = 3 \times 3 \times 5 \times 5$$.

Every factor of $$225$$ is built by choosing $$0$$, $$1$$ or $$2$$ of the $$3$$s and $$0$$, $$1$$ or $$2$$ of the $$5$$s.

Listing them:

So the factors of $$225$$ are $$1, 3, 5, 9, 15, 25, 45, 75, 225$$ ($$9$$ factors in all).

We verify by direct division. $$225 \div 1 = 225$$, $$225 \div 3 = 75$$, $$225 \div 5 = 45$$, $$225 \div 9 = 25$$, $$225 \div 15 = 15$$, $$225 \div 25 = 9$$, $$225 \div 45 = 5$$, $$225 \div 75 = 3$$, $$225 \div 225 = 1$$ — every division gives a whole number.

Any number not in the list, for example $$2, 4, 6, 7, 8, 10, 11, \ldots$$, does not divide $$225$$ (because their prime factors don't appear in $$225 = 3 \times 3 \times 5 \times 5$$).

So all nine factors are captured, and no factor is missing.

Anshu's toran length is a multiple of $$6$$ (i.e., $$6, 12, 18, 24, 30, \ldots$$).
Guna's toran length is a multiple of $$8$$ (i.e., $$8, 16, 24, 32, \ldots$$).

They must pick a common length — a common multiple of $$6$$ and $$8$$. The smallest one is the LCM.

$$6 = 2 \times 3$$, $$8 = 2 \times 2 \times 2$$.
Taking each prime with its highest power: $$\text{LCM}(6, 8) = 2 \times 2 \times 2 \times 3 = 24$$.

Anshu will use $$24 \div 6 = 4$$ strips and Guna will use $$24 \div 8 = 3$$ strips.

Free gajak days: every $$7$$ days (Mondays), i.e., day $$7, 14, 21, 28, 35, 42, 49, 56, 63, 70, \ldots$$ from today.

Kabamai's visit days: every $$10$$ days, i.e., day $$10, 20, 30, 40, 50, 60, 70, \ldots$$ from today.

She gets free gajak on days that are common to both lists. The next such day is the LCM of $$7$$ and $$10$$.

$$7$$ is prime, $$10 = 2 \times 5$$ — they share no prime.
$$\text{LCM}(7, 10) = 7 \times 10 = 70$$.

Write each number as a product of its primes. Then, for every prime that appears in either factorisation, take it the maximum number of times it appears in a single factorisation. Multiply these together — that product is the LCM.

Example: find $$\text{LCM}(12, 18)$$.
$$12 = 2 \times 2 \times 3$$ (two $$2$$s, one $$3$$).
$$18 = 2 \times 3 \times 3$$ (one $$2$$, two $$3$$s).
Take the maximum of each prime: two $$2$$s and two $$3$$s.
$$\text{LCM} = 2 \times 2 \times 3 \times 3 = 36$$.

Why this works: the LCM must be divisible by each number, so it must contain at least as many copies of each prime as either number has. Taking the maximum gives the smallest such number.

The two numbers under discussion are $$14 = 2 \times 7$$ and $$35 = 5 \times 7$$.

Consider $$N = 2 \times 3 \times 5 \times 7 = 210$$.

Does $$14$$ divide $$210$$? $$14 = 2 \times 7$$, and $$210$$ contains one $$2$$ and one $$7$$. Yes. ($$210 \div 14 = 15$$.)

Does $$35$$ divide $$210$$? $$35 = 5 \times 7$$, and $$210$$ contains one $$5$$ and one $$7$$. Yes. ($$210 \div 35 = 6$$.)

So $$210$$ is a common multiple of $$14$$ and $$35$$. (It is not the lowest — the LCM is $$70$$ — but it is a common multiple, because it contains all the primes needed by both numbers.)

$$14 = 2 \times 7$$, $$35 = 5 \times 7$$.

Take each prime with its highest power in either number: one $$2$$, one $$5$$, one $$7$$.

$$\text{LCM}(14, 35) = 2 \times 5 \times 7 = 70$$.

Verify: $$70 \div 14 = 5$$ and $$70 \div 35 = 2$$. Both whole. And no smaller positive multiple of $$14$$ (namely $$14, 28, 42, 56$$) is a multiple of $$35$$, so $$70$$ is indeed the least.

The numbers under discussion are $$14 = 2 \times 7$$ and $$35 = 5 \times 7$$. The LCM must be divisible by each, so it must contain as many $$2$$s as the number that has the most $$2$$s.

$$14$$ has one $$2$$; $$35$$ has zero $$2$$s. Maximum $$= 1$$.

So the LCM contains exactly one $$2$$.

Look at the two numbers $$14 = 2 \times 7$$ and $$35 = 5 \times 7$$. Neither number has $$3$$ as a prime factor.

So the LCM needs zero $$3$$s — including a $$3$$ would only make it larger, not required for divisibility.

$$14 = 2 \times 7$$ has zero $$5$$s; $$35 = 5 \times 7$$ has one $$5$$. Maximum $$= 1$$.

So the LCM must contain exactly one $$5$$.

Some examples of pairs whose HCF is one of the two numbers:

• $$(5, 15)$$ — $$\text{HCF} = 5$$
• $$(6, 12)$$ — $$\text{HCF} = 6$$
• $$(7, 21)$$ — $$\text{HCF} = 7$$
• $$(9, 36)$$ — $$\text{HCF} = 9$$
• $$(11, 11)$$ — $$\text{HCF} = 11$$

In each example, the smaller number is itself the HCF. This happens exactly when the smaller number divides the larger number, because then the smaller number is a common factor of both, and no factor of a number can be larger than the number itself.

Description: Pairs $$(a, b)$$ where $$a$$ divides $$b$$ (i.e., $$b$$ is a multiple of $$a$$).

Explore:

In every case the HCF also doubles.

Why: Doubling multiplies each number by $$2$$, so it adds one extra $$2$$ to each of their prime factorisations. The HCF picks the common primes with the smallest count. Since each number now has (at least) one extra $$2$$, the count of common $$2$$s goes up by exactly $$1$$. Multiplying the old HCF by this extra $$2$$ gives an HCF that is doubled.

In symbols: if $$a$$ and $$b$$ have HCF $$h$$, then $$2a$$ and $$2b$$ have HCF $$2h$$.

Both numbers share the factor $$14$$. Beyond that, the leftover parts are $$6$$ and $$9$$; their HCF is $$3$$.

So $$\text{HCF}(14 \times 6,\; 14 \times 9) = 14 \times \text{HCF}(6, 9) = 14 \times 3 = 42$$.

Check with actual values: $$14 \times 6 = 84 = 2^2 \times 3 \times 7$$ and $$14 \times 9 = 126 = 2 \times 3^2 \times 7$$. Common primes: $$2 \times 3 \times 7 = 42$$. ✓

Recall the HCFs from question 25:

The HCF equals the common multiplier in cases (b) and (c) — exactly the cases where the two leftover parts are co-prime (share no factor other than $$1$$).

General rule: If two numbers are written as $$c \times p$$ and $$c \times q$$ with $$\text{HCF}(p, q) = 1$$, then $$\text{HCF}(cp, cq) = c$$. If $$p$$ and $$q$$ share a factor, the HCF is larger than $$c$$.

Write both numbers pulling out common pieces:

$$84 = (2 \times 2 \times 3) \times 7$$
$$180 = (2 \times 2 \times 3) \times 15$$

The common multiplier is $$c = 2 \times 2 \times 3 = 12$$. The leftover factors are $$7$$ and $$15$$.

Is $$\text{HCF}(7, 15) = 1$$? $$7$$ is prime and $$15 = 3 \times 5$$, so yes — $$7$$ and $$15$$ share no common prime.

By the rule from question 26, $$\text{HCF}(84, 180) = c \times \text{HCF}(7, 15) = 12 \times 1 = 12$$.

A common multiple of two numbers $$a$$ and $$b$$ must be divisible by both — so its prime factorisation must contain the prime factorisation of $$a$$ and the prime factorisation of $$b$$.

The least such number is obtained by taking each prime with the highest power occurring in either $$a$$ or $$b$$ (any less, and one of $$a, b$$ would fail to divide it; any more, and it would no longer be the smallest).

Example (300 and 150): $$300 = 2^2 \times 3 \times 5^2$$, $$150 = 2 \times 3 \times 5^2$$. Highest powers: $$2^2, 3, 5^2$$. So $$\text{LCM} = 2^2 \times 3 \times 5^2 = 300$$. Both $$300$$ and $$150$$ divide $$300$$, and no smaller positive multiple of $$300$$ exists. Since $$150$$ already divides $$300$$, the LCM is just $$300$$.

This is why the method — 'multiply each prime to its highest power in either factorisation' — always produces the LCM.

The product is always $$\ge$$ the LCM. They are equal only when the two numbers are co-prime.

Examples:

• $$4$$ and $$6$$: product $$= 24$$, LCM $$= 12$$. Product is greater.
• $$8$$ and $$9$$: product $$= 72$$, LCM $$= 72$$. Equal (co-prime).
• $$6$$ and $$10$$: product $$= 60$$, LCM $$= 30$$. Product is greater.

Why: the product $$a \times b$$ is a common multiple of $$a$$ and $$b$$ (since $$a$$ divides $$ab$$, and $$b$$ divides $$ab$$). Being a common multiple, it is at least as big as the least common multiple.

Reasoning: Let the two numbers be $$a$$ and $$b$$, and let $$P = a \times b$$.

Is $$P$$ a multiple of $$a$$? Yes — $$P = a \times b$$, so $$a$$ divides $$P$$.
Is $$P$$ a multiple of $$b$$? Yes — $$P = b \times a$$, so $$b$$ divides $$P$$.

So the product $$P$$ is a common multiple of $$a$$ and $$b$$.

By definition, the LCM is the least (smallest) common multiple. So the LCM cannot exceed $$P$$; the most it can be is $$P$$ itself.

Therefore $$\text{LCM}(a, b) \le a \times b$$, with equality exactly when $$a$$ and $$b$$ have no common prime factor (i.e., are co-prime).

Prime factorise:

$$105 = 3 \times 5 \times 7$$
$$95 = 5 \times 19$$

Take each prime to its highest power in either factorisation: $$3, 5, 7, 19$$.

$$\text{LCM}(105, 95) = 3 \times 5 \times 7 \times 19 = 15 \times 133 = 1995$$.

Yes. Collecting the results:

In every row, product $$\div$$ LCM $$= $$ HCF. So $$\text{HCF} \times \text{LCM} = $$ product of the two numbers.

Take two numbers $$a$$ and $$b$$. Look at any prime $$p$$ and suppose it appears $$x$$ times in $$a$$ and $$y$$ times in $$b$$ (either $$x$$ or $$y$$ can be $$0$$).

Then:

• $$p$$ appears $$\min(x, y)$$ times in the HCF (the smaller count).
• $$p$$ appears $$\max(x, y)$$ times in the LCM (the larger count).
• $$p$$ appears $$x + y$$ times in the product $$a \times b$$.

But for any two whole numbers, $$\min(x, y) + \max(x, y) = x + y$$. So the total count of $$p$$ in HCF $$\times$$ LCM is $$\min(x, y) + \max(x, y) = x + y$$, which is exactly the count of $$p$$ in $$a \times b$$.

Since this is true for every prime, HCF $$\times$$ LCM and $$a \times b$$ have the same prime factorisation and must be equal.

Concrete example: $$a = 45 = 3^2 \times 5$$, $$b = 105 = 3 \times 5 \times 7$$.
HCF $$= 3^1 \times 5^1 = 15$$. LCM $$= 3^2 \times 5^1 \times 7^1 = 315$$.
HCF $$\times$$ LCM $$= 3^{1+2} \times 5^{1+1} \times 7^{0+1} = 3^3 \times 5^2 \times 7 = 4725 = 45 \times 105$$. ✓

Test with $$2$$, $$4$$, $$6$$.

$$\text{HCF}(2, 4, 6) = 2$$.
$$\text{LCM}(2, 4, 6) = 12$$.
HCF $$\times$$ LCM $$= 2 \times 12 = 24$$.
Product $$= 2 \times 4 \times 6 = 48$$.

Since $$24 \ne 48$$, the property fails for three numbers.

Why the argument breaks: For a prime $$p$$ occurring $$x_1, x_2, x_3$$ times in the three numbers, the HCF has $$\min(x_1, x_2, x_3)$$ copies, the LCM has $$\max(x_1, x_2, x_3)$$ copies, and the product has $$x_1 + x_2 + x_3$$ copies. In general $$\min + \max \ne x_1 + x_2 + x_3$$ — it leaves out the middle value — so the identity does not hold with three (or more) numbers.

Try another example: $$4, 6, 9$$. $$\text{HCF} = 1$$, $$\text{LCM} = 36$$, HCF $$\times$$ LCM $$= 36$$; product $$= 4 \times 6 \times 9 = 216$$. Again unequal.

Anshu's claim is wrong. Bigger numbers do not always have longer prime factorisations.

Counterexample 1: $$90 = 2 \times 3 \times 3 \times 5$$ ($$4$$ prime factors), but the larger number $$97$$ is prime, so its factorisation is just $$97$$ ($$1$$ prime factor).

Counterexample 2: $$30 = 2 \times 3 \times 5$$ ($$3$$ prime factors), while the much larger $$1024 = 2^{10}$$ has $$10$$ prime factors — but any large prime like $$1009$$ still has only $$1$$.

The length of a prime factorisation depends on which primes divide the number and how many times, not on how big the number is. Any prime number, no matter how large, has factorisation length $$1$$.

List factors of each number.

Factors of $$45$$: $$1, 3, 5, 9, 15, 45$$.
Factors of $$75$$: $$1, 3, 5, 15, 25, 75$$.

Common factors: $$1, 3, 5, 15$$.

HCF is the largest common factor: $$\text{HCF}(45, 75) = 15$$.

Prime-factorisation check: $$45 = 3^2 \times 5$$, $$75 = 3 \times 5^2$$; common primes with smallest powers: $$3 \times 5 = 15$$. ✓

Prime factorise:

$$112 = 2^4 \times 7$$
$$84 = 2^2 \times 3 \times 7$$

Common primes taken with smallest powers: $$2^2 \times 7 = 28$$. So $$\text{HCF}(112, 84) = 28$$.

Common factors are exactly the factors of the HCF. Factors of $$28$$: $$1, 2, 4, 7, 14, 28$$.

Prime factorise:

$$96 = 2^5 \times 3$$
$$275 = 5^2 \times 11$$

The two prime factorisations share no prime, so the only common factor is $$1$$.

$$\text{HCF}(96, 275) = 1$$ (the numbers are co-prime).

Prime factorise:

$$30 = 2 \times 3 \times 5$$
$$72 = 2 \times 2 \times 2 \times 3 \times 3 = 2^3 \times 3^2$$

Common primes (with smallest count in either number): one $$2$$, one $$3$$.

$$\text{HCF}(30, 72) = 2 \times 3 = 6$$.

Prime factorise:

$$225 = 3 \times 3 \times 5 \times 5 = 3^2 \times 5^2$$
$$750 = 2 \times 3 \times 5 \times 5 \times 5 = 2 \times 3 \times 5^3$$

Common primes with smallest counts: one $$3$$, two $$5$$s (the count of $$5$$s in $$225$$).

$$\text{HCF}(225, 750) = 3 \times 5^2 = 75$$.

Prime factorise:

$$14 = 2 \times 7$$
$$35 = 5 \times 7$$

Take each prime with its highest count in either number: one $$2$$, one $$5$$, one $$7$$.

$$\text{LCM}(14, 35) = 2 \times 5 \times 7 = 70$$.

Prime factorise:

$$96 = 2^5 \times 3$$
$$360 = 2^3 \times 3^2 \times 5$$

Highest power of each prime in either number: $$2^5, 3^2, 5$$.

$$\text{LCM}(96, 360) = 2^5 \times 3^2 \times 5 = 32 \times 9 \times 5 = 1440$$.

No, one cannot conclude that. Comparing composite factors like $$\{6, 12\}$$ with $$\{8, 18\}$$ misses common factors hidden inside those composites.

For example, $$6$$ and $$8$$ share the factor $$2$$, and $$12$$ and $$18$$ share $$6$$. These shared factors show up only when we go all the way down to primes.

Do this properly with prime factorisation:

$$72 = 6 \times 12 = (2 \times 3) \times (2^2 \times 3) = 2^3 \times 3^2$$
$$144 = 8 \times 18 = 2^3 \times (2 \times 3^2) = 2^4 \times 3^2$$

Common primes: $$2^3$$ and $$3^2$$. So $$\text{HCF}(72, 144) = 2^3 \times 3^2 = 72$$.

Actually, $$72$$ divides $$144$$ ($$144 = 2 \times 72$$), so $$72$$ is a common factor. The composite factorisation was misleading — always break down to primes before comparing.

Read off the two rows carefully. The top row's colour pattern repeats every $$6$$ stars, with a blue star occurring once per cycle. The bottom row's pattern repeats every $$4$$ stars, again with one blue star per cycle. Both rows have their first blue star at the same starting column.

So in the top row, blue stars appear at columns $$4, 10, 16, 22, \ldots$$ (every $$6$$ steps).
In the bottom row, blue stars appear at columns $$4, 8, 12, 16, 20, \ldots$$ (every $$4$$ steps).

They meet again wherever a column is a common multiple of $$6$$ and $$4$$ apart from the first meeting. The next meeting is exactly $$\text{LCM}(6, 4)$$ columns after the first.

$$6 = 2 \times 3$$, $$4 = 2^2$$. $$\text{LCM}(6, 4) = 2^2 \times 3 = 12$$.

So the next meeting is $$12$$ columns after the first — i.e., at column $$4 + 12 = 16$$.

If HCF $$= 1$$, the two numbers $$a, b$$ are co-prime, and $$\text{HCF} \times \text{LCM} = a \times b$$ gives $$a \times b = 66$$.

Factor $$66 = 2 \times 3 \times 11$$. Split its primes into two co-prime groups:

• $$1 \times 66 = 66$$ — pair $$(1, 66)$$, HCF $$= 1$$, LCM $$= 66$$. ✓
• $$2 \times 33 = 66$$ — pair $$(2, 33)$$, and $$33 = 3 \times 11$$ shares no prime with $$2$$. HCF $$= 1$$, LCM $$= 66$$. ✓
• $$3 \times 22 = 66$$ — pair $$(3, 22)$$, $$22 = 2 \times 11$$, coprime with $$3$$. HCF $$= 1$$, LCM $$= 66$$. ✓
• $$6 \times 11 = 66$$ — pair $$(6, 11)$$, $$6 = 2 \times 3$$, coprime with $$11$$. HCF $$= 1$$, LCM $$= 66$$. ✓

Any of these pairs works. A common choice is $$\boxed{(6, 11)}$$.

Let $$N$$ be the number of cows. From the story:

• $$3$$ divides $$N$$ (equal split across $$3$$ gates).
• $$5$$ divides $$N$$ (equal split across $$5$$ gates).
• $$7$$ divides $$N$$ (equal split across $$7$$ gates).

So $$N$$ must be a common multiple of $$3, 5, 7$$. The smallest is

$$\text{LCM}(3, 5, 7) = 3 \times 5 \times 7 = 105$$ (since the numbers are pairwise co-prime).

The multiples of $$105$$ less than $$200$$ are just $$105$$ ($$210 > 200$$).

So the cowherd had $$105$$ cows.

Any number divisible by all of $$3, 4, 5, 7$$ is a multiple of $$\text{LCM}(3, 4, 5, 7)$$. Since these are pairwise co-prime (except $$4$$, which shares nothing with $$3, 5, 7$$),

$$\text{LCM}(3, 4, 5, 7) = 3 \times 4 \times 5 \times 7 = 420$$.

So the number must be $$420k$$ for some positive integer $$k$$.

We also need $$420k$$ to leave remainder $$10$$ when divided by $$11$$.

First find $$420 \bmod 11$$: $$11 \times 38 = 418$$, so $$420 - 418 = 2$$. Hence $$420 \equiv 2 \pmod{11}$$, and $$420k \equiv 2k \pmod{11}$$.

We need $$2k \equiv 10 \pmod{11}$$. Try small $$k$$: $$k = 5$$ gives $$2 \times 5 = 10 \equiv 10 \pmod{11}$$. ✓

Smallest number: $$420 \times 5 = 2100$$.

Verify: $$2100 \div 11 = 190$$ remainder $$10$$ (since $$11 \times 190 = 2090$$ and $$2100 - 2090 = 10$$). ✓

They leap at the same time: for every one leap the dog takes, the rabbit also takes one.

In one such pair of leaps, the dog covers $$9$$ ft while the rabbit only covers $$7$$ ft. So the dog closes the gap by $$9 - 7 = 2$$ ft per leap.

Initial gap $$= 150$$ ft. Number of leaps needed:

$$\dfrac{150}{2} = 75$$ leaps.

Verify: after $$75$$ leaps, the dog has moved $$75 \times 9 = 675$$ ft; the rabbit has moved $$75 \times 7 = 525$$ ft. Rabbit's total distance from dog's start $$= 150 + 525 = 675$$ ft. Both are at the same spot. ✓

The smallest common multiple is the LCM. Prime factorise each number:

$$1$$ — no primes.
$$2 = 2$$, $$3 = 3$$, $$4 = 2^2$$, $$5 = 5$$, $$6 = 2 \times 3$$, $$8 = 2^3$$, $$9 = 3^2$$, $$10 = 2 \times 5$$.

For each prime, take the highest power occurring in any of the numbers:

• $$2^3$$ (from $$8$$)
• $$3^2$$ (from $$9$$)
• $$5^1$$ (from $$5$$ or $$10$$)

$$\text{LCM} = 2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = 360$$.

Check that $$360$$ is indeed a multiple of every number in the list: $$360 = 1 \times 360 = 2 \times 180 = 3 \times 120 = 4 \times 90 = 5 \times 72 = 6 \times 60 = 8 \times 45 = 9 \times 40 = 10 \times 36$$. ✓

To add fractions with different denominators, use a common denominator. The most efficient choice is the LCM of all denominators.

Prime factorise each denominator:

$$15 = 3 \times 5$$
$$20 = 2^2 \times 5$$
$$36 = 2^2 \times 3^2$$
$$63 = 3^2 \times 7$$
$$21 = 3 \times 7$$

Highest power of each prime: $$2^2, 3^2, 5, 7$$.
$$\text{LCM} = 4 \times 9 \times 5 \times 7 = 1260$$.

Convert each fraction to denominator $$1260$$:

$$\dfrac{8}{15} = \dfrac{8 \times 84}{1260} = \dfrac{672}{1260}$$ (since $$1260 \div 15 = 84$$)
$$\dfrac{1}{20} = \dfrac{1 \times 63}{1260} = \dfrac{63}{1260}$$ (since $$1260 \div 20 = 63$$)
$$\dfrac{7}{36} = \dfrac{7 \times 35}{1260} = \dfrac{245}{1260}$$ (since $$1260 \div 36 = 35$$)
$$\dfrac{11}{63} = \dfrac{11 \times 20}{1260} = \dfrac{220}{1260}$$ (since $$1260 \div 63 = 20$$)
$$\dfrac{1}{21} = \dfrac{1 \times 60}{1260} = \dfrac{60}{1260}$$ (since $$1260 \div 21 = 60$$)

Add the numerators:

$$672 + 63 + 245 + 220 + 60 = 1260$$.

So the sum is $$\dfrac{1260}{1260} = 1$$.

The efficient trick: find the LCM of all denominators using prime factorisation once (rather than doing pairwise additions).