Chapter 3: Finding Common Ground (HCF and LCM)

The largest tile that fits both sides exactly gives the fewest tiles. With 4 ft tiles: $\frac{12}{4}\times \frac{16}{4}=3\times 4=$ 12 tiles.

Even if fractional sides were allowed, the tile side must still fit a whole number of times into both 12 and 16, i.e. $12÷s$ and $16÷s$ must be whole numbers. The largest such $s$ is still 4 ft (a bigger tile, like 4.5 ft or 6 ft, does not fit both sides), so the answer does not change.

The largest tile means the fewest tiles: 12 tiles of 4 ft; fractional sizes do not give a larger tile, so the answer is the same.

The weight must be a common factor of 84 and 108: 1, 2, 3, 4, 6, 12. The fewest bags come from the largest, 12 kg: $84÷12=7$ and $108÷12=9$, i.e. 16 bags.

12 kg bags (16 bags in all).

(a) 2 (b) 1 (c) 10 (d) 14

Yes, it is the HCF. Jumps of size $j$ from 0 land only on multiples of $j$, so $j$ must be a factor of both numbers; the longest such jump is the highest common factor.

(a) 2 (b) 1 (c) 10 (d) 14; the jump must divide both numbers, so the longest is their HCF.

$105=3\times 5\times 7$; $30=2\times 3\times 5$.

$1200=2\times 2\times 2\times 2\times 3\times 5\times 5$

The division method is easier: at each step we only divide by one small prime, so nothing is missed.

105 = 3 × 5 × 7; 30 = 2 × 3 × 5; 1200 = 2 × 2 × 2 × 2 × 3 × 5 × 5

• 14: yes, since 2 × 7 is a part of the factorisation: $840=(2\times 7)\times (2\times 2\times 3\times 5)=14\times 60$.
• 8: yes: $840=(2\times 2\times 2)\times (3\times 5\times 7)=8\times 105$.
• 27: no: 840 has only one 3 in its prime factorisation, so 3 × 3 × 3 cannot be taken from it.

14 and 8 are factors (they are parts of the prime factorisation); 27 is not (840 has only one 3).

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

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

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

(d) $360=2\times 2\times 2\times 3\times 3\times 5$: 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) $840=2\times 2\times 2\times 3\times 5\times 7$: 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

See the lists: 12, 8, 12, 24 and 32 factors respectively.

(a) 10 (b) 5 (c) 1 (d) 74 (e) 81

Take each common prime the least number of times it appears.

(a) $24=2^3\times 3$, $180=2^2\times 3^2\times 5$: HCF $=2\times 2\times 3=$ 12

(b) $42=2\times 3\times 7$, $75=3\times 5\times 5$, $24=2^3\times 3$: only 3 is common: 3

(c) $240=2^4\times 3\times 5$, $378=2\times 3^3\times 7$: HCF $=2\times 3=$ 6

(d) $400=2^4\times 5^2$, $2500=2^2\times 5^4$: HCF $=2^2\times 5^2=$ 100

(e) $300=2^2\times 3\times 5^2$, $800=2^5\times 5^2$: HCF $=2^2\times 5^2=$ 100

(a) 12 (b) 3 (c) 6 (d) 100 (e) 100

No. These are not prime factorisations: 6, 12, 8 and 18 can be broken further. In fact $72=2^3\times 3^2$ and $144=2^4\times 3^2$, so the HCF is $2^3\times 3^2=72$ (72 is a factor of 144). Common factors are only seen correctly from prime factorisations.

No; the factors 6, 12, 8, 18 are not primes. Their HCF is actually 72.

No. Common multiples of 6 and 8 are 24, 48, 72, 96, … (all multiples of 24) and go on forever, so there is no largest one.

No; the common multiples (24, 48, 72, …) never end.

(a) 12 (b) 77 (c) 210 (d) 165

Yes. "Idli-vada" is called at common multiples, and the first one is the lowest common multiple.

(a) 12 (b) 77 (c) 210 (d) 165; yes, the first common multiple is the LCM.

Yes: a multiple is the number × something, so its prime factors are those of the number plus the prime factors of that "something".

$2\times 3\times 5\times 7=210$ contains $2\times 7$ and $5\times 7$, so yes, it is a common multiple ($210=14\times 15=35\times 6$), but not the lowest (70 is).

More than five 2s still gives a common multiple, but it is at least twice as big as the one with five 2s, so it cannot be the lowest.

Yes; 210 is a common multiple of 14 and 35 (not the lowest); extra 2s only make the multiple larger.

Take each prime the greatest number of times it appears.

(a) $30=2\times 3\times 5$, $72=2^3\times 3^2$: LCM $=2^3\times 3^2\times 5=$ 360

(b) $36=2^2\times 3^2$, $54=2\times 3^3$: LCM $=2^2\times 3^3=$ 108

(c) $105=3\times 5\times 7$, $195=3\times 5\times 13$, $65=5\times 13$: LCM $=3\times 5\times 7\times 13=$ 1365

(d) $222=2\times 3\times 37$, $370=2\times 5\times 37$: LCM $=2\times 3\times 5\times 37=$ 1110

(a) 360 (b) 108 (c) 1365 (d) 1110

Examples: (6, 18), (5, 35), (12, 48), (9, 9). This happens exactly when one number is a factor of the other.

(a) The other number can be any multiple of m (2m, 3m, …), or any factor of m.

(b) Any multiple of $7k$ (such as $14k$, $21k$), or any factor of $7k$ (such as $k$ or 7).

When one number divides the other: with m, the other is a multiple or a factor of m; with 7k, e.g. 14k or k.

(a) HCF = 2. E.g. (10, 12) → 2. Both are even, and a common factor must also divide their difference 2.

(b) HCF = 1. E.g. (15, 17). A common factor divides the difference 2, but the numbers are odd, so it cannot be 2.

(c) HCF is even (at least 2), since 2 is a common factor. E.g. (12, 18) → 6.

(d) HCF = 1. A common factor would divide their difference, 1.

(e) HCF = 1 (that is the meaning of co-prime).

(a) 2 (b) 1 (c) an even number (d) 1 (e) 1; a common factor must divide the difference of the two numbers.

(a) (4, 20) → 20; (7, 14) → 14; (5, 5) → 5; (6, 36) → 36.

(b) The LCM is one of the numbers exactly when that number is a multiple of the other. In algebra: for $n$ and $kn$ (with $k$ a counting number), LCM $=kn$.

When one number is a multiple of the other: LCM of n and kn is kn.

(a) The LCM is also a multiple of 3 (e.g. 6 and 9 → 18).

(b) For $2n$ and $2n+2$: the LCM is half their product, $2n(n+1)$ (e.g. 10 and 12 → 60 = 120 ÷ 2), since their HCF is 2.

(c) The LCM is their product (e.g. 8 and 9 → 72), since their HCF is 1.

(d) The LCM is their product (e.g. 4 and 9 → 36).

(a) a multiple of 3 (b) half the product (c) the product (d) the product.

Doubling both numbers doubles the HCF (each gets one more 2, which joins the common part).

(a) $18\times \text{HCF}(10,15)=18\times 5=$ 90 (b) $10\times \text{HCF}(38,21)=10\times 1=$ 10 (c) $5\times 1=$ 5 (d) $12\times 4=$ 48

The HCF equals the common multiplier exactly when the other two multipliers are co-prime (38 and 21; 13 and 20).

The HCF doubles. (a) 90 (b) 10 (c) 5 (d) 48; the HCF is the common multiplier when the other multipliers are co-prime.

84, 180 → (÷2) 42, 90 → (÷2) 21, 45 → (÷3) 7, 15. At each step both numbers are divided by a common prime; we stop when the two numbers have no common factor.

• HCF = product of the divisors on the left: $2\times 2\times 3=$ 12.
• LCM = divisors × the last row: $12\times 7\times 15=$ 1260.

For 300, 150: HCF $=2\times 5\times 5\times 3=150$ and LCM $=150\times 2\times 1=300$. For 630, 770: HCF $=2\times 5\times 7=70$ and LCM $=70\times 9\times 11=6930$.

Why: the left column is exactly the common part of both prime factorisations (the HCF). Each number equals HCF × its last-row number, and the last-row numbers share no factor, so HCF × (both last-row numbers) contains each number's full factorisation and nothing extra: it is the LCM.

84, 180: HCF 12, LCM 1260. The left divisors are the common part (HCF); multiplying in the co-prime leftovers gives the smallest number containing both (LCM).

Dividing by a composite common factor such as 50 is the same as dividing by its primes ($2\times 5\times 5$) all at once, so the result is the same.

(a) 90, 150 → (÷30) 3, 5: HCF 30, LCM $=30\times 3\times 5=$ 450

(b) 84, 132 → (÷12) 7, 11: HCF 12, LCM $=12\times 7\times 11=$ 924

A composite divisor is just several prime steps together. (a) HCF 30, LCM 450 (b) HCF 12, LCM 924

The product is never smaller: the product $a\times b$ is itself a common multiple of $a$ and $b$, and the LCM is the smallest common multiple. So LCM ≤ product (they are equal when the numbers are co-prime).

LCM ≤ product, since the product is a common multiple; equal for co-prime numbers.

In each case the LCM is a factor of the product and the multiplier is the HCF.

Why: in the prime factorisations, the primes common to both numbers go once into the HCF and once into the LCM; every other prime goes into the LCM only. Altogether, HCF × LCM uses every prime of both numbers exactly as often as the product does.

For 3 numbers it fails, e.g. 4, 6, 8: HCF × LCM $=2\times 24=48$, but the product is 192.

The multiplier is the HCF each time (15, 11, 74): HCF × LCM = product for two numbers, but not for three (4, 6, 8 gives 48 ≠ 192).

The top row repeats every 6 stars and the bottom row every 4 stars; both have a blue star at position 4. Blue stars are together again after LCM(6, 4) = 12 positions: at positions 16, 28, 40, …

At position 16 (12 places after position 4), and then every 12 places.

(a) No: a multiple would have to contain 7 × 7 and 2, but 5 × 7 × 11 × 11 has only one 7 and no 2.

(b) No: it has 11 × 11, but the other number has only one 11.

(a) No (b) No

(a) $12\times 7\times 11=2\times 2\times 3\times 7\times 11$. HCF $=3\times 7$ (= 21); LCM $=2\times 2\times 3\times 3\times 5\times 7\times 7\times 11$ (= 97020).

(b) $45=3\times 3\times 5$, $36=2\times 2\times 3\times 3$. HCF $=3\times 3$ (= 9); LCM $=2\times 2\times 3\times 3\times 5$ (= 180).

(a) HCF 3 × 7, LCM 2 × 2 × 3 × 3 × 5 × 7 × 7 × 11 (b) HCF 3 × 3, LCM 2 × 2 × 3 × 3 × 5

Their product $=1\times 66=66$ and they are co-prime. $66=2\times 3\times 11$: possible pairs (6, 11), (2, 33), (3, 22), (1, 66).

E.g. 6 and 11 (also 2 and 33, 3 and 22, 1 and 66).

The number is a common multiple of 3, 5 and 7, i.e. a multiple of LCM = 105. The only one below 200 is 105 cows.

105 cows

The cube's side must divide 12, 18 and 36 exactly (common factors: 1, 2, 3, 6).

(b) 6 cm, (d) 3 cm, (e) 2 cm can be packed. 9 cm does not divide 12, and 4 cm does not divide 18.

(b), (d) and (e)

$306=2\times 3\times 3\times 17$, $36=2\times 2\times 3\times 3$: HCF $=2\times 3\times 3=$ 18, option (c).

(c) 18

It must be a multiple of LCM(3, 4, 5, 7) = 420. Check the multiples: 420 leaves 2, 840 leaves 4, 1260 leaves 6, 1680 leaves 8, 2100 leaves 10 when divided by 11 ($2100=11\times 190+10$).

2100

No one out for 6 and 9 → the number is a multiple of LCM(6, 9) = 18. Someone out for 10 → it is not a multiple of 10.

(a) 72 and (e) 36 fit. (90 is a multiple of 10; 45 and 3 are not multiples of 18.)

(a) 72 and (e) 36

Two different primes are co-prime, so their LCM is exactly $m\times n$, which is greater than both primes. Only (c) is correct.

(c) only (the LCM is m × n).

Each leap the dog gains $9-7=2$ ft. To close 150 ft: $150÷2=$ 75 leaps (the dog covers $75\times 9=675$ ft and the rabbit $150+75\times 7=675$ ft).

75 leaps

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

360

Use the LCM of the denominators as the common denominator: $15=3\times 5$, $20=2^2\times 5$, $36=2^2\times 3^2$, $63=3^2\times 7$, $21=3\times 7$, so LCM $=2^2\times 3^2\times 5\times 7=1260$.

$$\frac{672}{1260}+\frac{63}{1260}+\frac{245}{1260}+\frac{220}{1260}+\frac{60}{1260}=\frac{1260}{1260}=1$$

The sum is exactly 1!

1 (using the LCM 1260 as the common denominator)

At 1 digit per second without any break: 4,10,24,320 seconds $\approx 4,10,24,320÷86,400\approx 475$ days, i.e. well over a year of non-stop writing (several years if you write a few hours a day).

About 475 days of non-stop writing at one digit a second.

The ring around each number is cut into one part for each prime factor (counted with repeats), and each prime has its own colour: 2 → green, 3 → pink, 5 → orange, 7 → dark blue, primes above 7 → red, and 1 → grey. For example, $12=2\times 2\times 3$ has two green parts and one pink part; every prime number has a single-coloured ring.

Each ring has one part per prime factor (2 green, 3 pink, 5 orange, 7 blue, larger primes red); e.g. 105 = pink, orange, blue.