1.3

Lowest Common Multiple (LCM)

AWhat is the LCM?

Let us list the first 10 multiples of 4 and 6.

Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40
Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60

We see that 12, 24 and 36 are common multiples of both 4 and 6. Among these common multiples, the smallest one is 12. Hence 12 is the lowest common multiple (LCM) of 4 and 6.

The lowest common multiple (LCM) of two or more whole numbers is the smallest whole number that is a common multiple of the given numbers.

Now, suppose we need to find the LCM of 24 and 90, we can do so by listing some multiples of 24 and 90.

Multiples of 24: 24, 48, 72, 96, 120, 144, 168, 192, 216, 240, 264, 288, 312, 336, 360, ...
Multiples of 90: 90, 180, 270, 360, ...

However, it is tedious to list the multiples to find the LCM of 24 and 90, which is 360. In fact, we can use other methods to find the LCM of these two numbers in a more efficient way as illustrated in Worked Example 8.

BMethods of Finding LCM

📖 WORKED EXAMPLE 8

Find the lowest common multiple of 24 and 90.

Solution

Method 1 Prime Factorisation

We express 24 and 90 each as a product of its prime factors.

$24 = 2^3 \times 3$
$90 = 2 \times 3^2 \times 5$

We can visualise the process of identifying all the different factors for the LCM as shown:

24 = 2 × 2 × 2 × 3 90 = 2 × 3 × 3 × 5

LCM = 2 × 2 × 2 × 3 × 3 × 5

$\therefore \text{LCM} = 2^3 \times 3^2 \times 5 = 360$

($2^3$ is of a higher power than 2, and $3^2$ is of a higher power than 3.)

NOTE: The LCM is obtained by multiplying the higher power of each prime factor ($2^3$, $3^2$ and 5) of the given numbers. If there are three given numbers, then the LCM is obtained by multiplying the highest power of each prime factor of the numbers.

Method 2 Successive Short Division

Carry out successive short division by common prime factors.

2 | 24 90 3 | 12 45 4 15 ← no common prime factors

LCM = product of the common prime factors and the remaining factors
= 2 × 3 × 4 × 15 = 360

✏️ TRY IT YOURSELF 8

(a) Find the LCM of 40 and 150 using prime factorisation.
(b) Find the LCM of 63 and 147 using successive short division.

CLCM of Three Numbers

📖 WORKED EXAMPLE 9

Find the LCM of 20, 24 and 70.

Solution

Method 1 Prime Factorisation

20 = 2² × 5 24 = 2³ × 3 70 = 2 × 5 × 7

LCM = 2³ × 3 × 5 × 7

$\therefore \text{LCM} = 2^3 \times 3 \times 5 \times 7 = 840$

(By aligning the factors as shown, we choose the highest power of each prime factor.)

Method 2 Successive Short Division

We do successive short divisions by using the common prime factors of at least two numbers. If a number is not divisible by the factor, copy the number to the next row.

2 | 20 24 70 2 | 10 12 35 ← 35 not divisible by 2; copy to next row 5 | 5 6 35 ← 6 not divisible by 5; copy to next row 1 6 7 ← stop, no common prime factor among any two

LCM = 2 × 2 × 5 × 6 × 7 = 840

THINK²

We found out that 840 is the LCM of 20, 24 and 70.
(a) Is 840 the LCM of any two of these three numbers?
(b) Is 840 greater or smaller than the LCM of those two numbers? Explain.

EN (a) No. $\text{LCM}(20,24)=120$, $\text{LCM}(20,70)=140$, $\text{LCM}(24,70)=840$. Only the pair (24, 70) has LCM equal to 840.
(b) Adding a third number can only make the LCM greater than or equal to the LCM of any two of the original numbers. So 840 ≥ LCM of any pair, with equality when the third number is already a factor of that pair's LCM.

⏱ 仔细阅读英文版，30 秒后可查看中文版

中文 (a) 不全是。$\text{LCM}(20,24)=120$，$\text{LCM}(20,70)=140$，$\text{LCM}(24,70)=840$。只有 (24, 70) 这对的 LCM 等于 840。
(b) 增加第三个数只会让 LCM 不变或变大。所以 840 大于等于任意两数的 LCM（当第三个数已经是那对的 LCM 的因数时取等号）。

看完答案，自评：

✏️ TRY IT YOURSELF 9

(a) Find the LCM of 54, 84 and 110.
(b) Given three numbers, $2 \times 3 \times 7^2$, $2^2 \times 11$ and $3$, find the LCM in index notation.

DLCM of Coprime Numbers

📖 WORKED EXAMPLE 10

Find the LCM of 26 and 99.

Solution

$26 = 2 \times 13$
$99 = 3^2 \times 11$
$\therefore \text{LCM} = 2 \times 3^2 \times 11 \times 13 = \mathbf{2574}$

✏️ TRY IT YOURSELF 10

(a) Find the LCM of 34 and 57.
(b) Given that the product of two whole numbers is also their LCM, find three possible sums of the two numbers if the LCM is 30.

EN "Product = LCM" means the two numbers are coprime (HCF = 1). Coprime pairs whose product = 30: (1, 30), (2, 15), (3, 10), (5, 6). Sums: $1 + 30 = \mathbf{31}$, $2 + 15 = \mathbf{17}$, $3 + 10 = \mathbf{13}$, $5 + 6 = \mathbf{11}$. Any three of these sums.

⏱ 仔细阅读英文版，30 秒后可查看中文版

中文 "乘积 = LCM" 意味着两数互质（HCF = 1）。乘积为 30 的互质对：(1, 30), (2, 15), (3, 10), (5, 6)。和： $1+30=\mathbf{31}$，$2+15=\mathbf{17}$，$3+10=\mathbf{13}$，$5+6=\mathbf{11}$。任写三个即可。

自评：

ELCM in Word Problems

📖 WORKED EXAMPLE 11

The circumference of the front wheel and that of the two identical rear wheels of a tricycle are 60 cm and 45 cm respectively. When Bob starts riding the tricycle in a straight line, point P (on the front wheel) and point Q (on the rear wheel) touch the ground.

(a) What is the distance travelled before P and Q next touch the ground at the same time?
(b) Find the number of revolutions that the front and rear wheels will each have made by then.

Analysis

Point P touches the ground again when the front wheel makes one revolution, that is, when the tricycle moves by 60 cm. Point Q touches the ground again when the rear wheel makes one revolution, that is, when the tricycle moves by 45 cm.

Therefore, the distance travelled, in centimetres, before P and Q next touch the ground at the same time is the smallest common multiple of the circumferences (i.e. 45 and 60) of the two wheels.

Solution

(a) $60 = 2^2 \times 3 \times 5$
    $45 = 3^2 \times 5$
    $\therefore \text{LCM} = 2^2 \times 3^2 \times 5 = \mathbf{180}$
    The distance travelled is 180 cm.

(b) $180 \div 60 = 3$ → front wheel makes 3 revolutions.
$180 \div 45 = 4$ → rear wheel makes 4 revolutions.

We can illustrate the horizontal distances covered:

Front wheel P

60 cm

P₃

Rear wheel Q

45 cm

Q₄

✏️ TRY IT YOURSELF 11

The figure shows a gear system. The big wheel has 20 teeth, the small wheel has 16 teeth. Tooth X on the big wheel and tooth Y on the small wheel are engaged at the start.

(a) Find the number of tooth contacts the two wheels will make before X and Y are engaged again.
(b) Find the number of revolutions each wheel will have made by then.
big wheel: small wheel:

📖 WORKED EXAMPLE 12

The dimensions of a rectangular tile are 40 cm by 35 cm. John arranges some of these tiles to form a square. What is the smallest possible area of the square?

Analysis

Use a diagram to represent the square. The length of the pink side is a multiple of 40. The length of the blue side is a multiple of 35. Since the lengths of the sides of a square are the same, the length should be a common multiple of 40 and 35.

Finding the smallest possible area of the square is equivalent to finding the area of the square with the shortest possible sides. Therefore, we need to find the lowest common multiple of 40 and 35.

Solution

$40 = 2^3 \times 5$
$35 = 5 \times 7$
Shortest possible length of side = LCM = $2^3 \times 5 \times 7 = \mathbf{280}$
Hence smallest possible area = $280 \times 280 = \mathbf{78\,400 \text{ cm}^2}$

THINK²

If John uses twice as many tiles in Worked Example 12, will he still be able to form a square?

EN No, not necessarily a square. The original square uses $(280/40) \times (280/35) = 7 \times 8 = 56$ tiles. Doubling: 112 tiles. Possible new shapes: $7 \times 16$ (rectangle 280 × 560), $14 \times 8$ (560 × 280), $14 \times 16$ (560 × 560 — yes a square, but that's 4× the tiles, not 2×). With exactly 112 tiles arranged side-by-side, the dimensions can't both be the same → it's a rectangle, not a square.

⏱ 仔细阅读英文版，30 秒后可查看中文版

中文 不一定能。原本的正方形用 $(280/40)\times(280/35)=7\times 8=56$ 块。翻倍后 112 块。可能的新形状：$7\times 16$（长方形 280×560）、$14\times 8$（560×280）、$14\times 16$（560×560 是正方形，但要 4 倍而不是 2 倍）。112 块按整块拼，长宽不可能完全相等 → 只能是长方形，不是正方形。

看完答案，自评：

✏️ TRY IT YOURSELF 12

The rectangular surface of each desk measures 63 cm by 36 cm. Some of these desks are arranged side by side to form a large square table. Find

(a) the shortest possible length of a side of the square table.
(b) the total number of desks needed.

📝 PRACTICE EXERCISE 1.3

BASIC MASTERY

Find the LCM of each pair of numbers.

(a) 12 and 15 →

(b) 6 and 28 →

(c) 23 and 32 →

(d) 60 and 75 →

(e) 59 and 118 →

(f) 65 and 91 →
Find the LCM of each group of numbers.

(a) 9, 12 and 30 →

(b) 13, 14 and 15 →

(c) 28, 42 and 105 →

(d) 22, 132 and 253 →

INTERMEDIATE

Two numbers expressed in index notation are $2^4 \times 3^5 \times 5^3 \times 7^2$ and $2^3 \times 3^6 \times 5 \times 7^8$. Find

(a) the HCF of these two numbers, in index notation:

(b) the LCM of these two numbers, in index notation:
Doris has piano lessons once every 6 days, swimming lessons once every 4 days, and ballet lessons once every 8 days. If she has all three lessons on 1 April, on which date in April will she next have all three lessons again?
A fairy light flashes once every 10 seconds, another once every 15 seconds. They are flashing together now. How long will it take for them to flash together again?
Two drones take off to fly along a circuit. They take 48 seconds and 56 seconds, respectively, to complete one circuit. They begin at different heights but at the same point, at the same time and fly in the same direction.

(a) How long does it take for them to meet at the starting point again?

(b) How many circuits each? Drone A: Drone B:
The thickness of a Science book is 20 mm and that of a Mathematics book is 28 mm. Books of each type are stacked in separate piles.

(a) What should the minimum height of each pile be so that both piles are of the same height?

(b) Number of books in each pile? Science: Math:
Each desk in a classroom has a rectangular desktop measuring 56 cm by 42 cm. Some of these desks are arranged side by side to form a large square table for a class activity.

Find

(a) the shortest length of a side of the square,

(b) the number of rows and columns of desks used to form the large square table.
rows: columns: （按 56 cm 边为列方向，42 cm 边为行方向）
Buses A, B and C arrive at a bus interchange at intervals of 20 minutes, 30 minutes and 45 minutes respectively. If all of them meet at the interchange at 08 30, find the time they will next meet at the interchange.

ADVANCED

[OPEN] Find three pairs of numbers such that the LCM of each pair of numbers is 24.

EN Many valid pairs — any pair (a, b) with LCM(a, b) = 24. Examples: (1, 24), (2, 24), (3, 8), (3, 24), (4, 24), (6, 8), (8, 12), (8, 24), (12, 24). Pick any three.

⏱ 仔细阅读英文版，30 秒后可查看中文版

中文答案不唯一——任何 LCM = 24 的数对都行。例如： (1, 24), (2, 24), (3, 8), (3, 24), (4, 24), (6, 8), (8, 12), (8, 24), (12, 24)。任写三对。

自评：
[OPEN] Find two possible pairs of numbers such that the HCF and LCM of each pair are 21 and 630 respectively.

EN Use the identity $\text{HCF} \times \text{LCM} = a \times b$. So $a \times b = 21 \times 630 = 13230$. Write $a = 21m, b = 21n$ where $\gcd(m,n) = 1$ and $m \cdot n = 30$. Coprime factor pairs of 30: (1,30), (2,15), (3,10), (5,6). So: (21, 630), (42, 315), (63, 210), (105, 126). Pick any two.

⏱ 仔细阅读英文版，30 秒后可查看中文版

中文利用 $\text{HCF} \times \text{LCM} = a \times b$，所以 $a \times b = 21 \times 630 = 13230$。设 $a = 21m, b = 21n$（$m, n$ 互质），$mn = 30$。互质对：(1,30), (2,15), (3,10), (5,6)。所以： (21, 630), (42, 315), (63, 210), (105, 126)。任写两对。

自评：
Find the greatest 3-digit number that is divisible by 15, 20 and 24.
Find the smallest positive number that should be added to 1628 so that the sum is exactly divisible by 4, 5 and 6.
There are two whole numbers m and n. Their HCF is p and LCM is q. What is the relationship between $m \times n$ and $p \times q$?

EN $m \times n = p \times q$. Why: write $m = p \cdot a$, $n = p \cdot b$ where $\gcd(a, b) = 1$. Then $\text{LCM}(m, n) = p \cdot a \cdot b$, so $q = p \cdot a \cdot b$. Therefore $m \cdot n = p \cdot a \cdot p \cdot b = p \cdot (p \cdot a \cdot b) = p \cdot q$.

⏱ 仔细阅读英文版，30 秒后可查看中文版

中文 $m \times n = p \times q$。理由：设 $m = p \cdot a$，$n = p \cdot b$，其中 $\gcd(a, b) = 1$。那么 $\text{LCM}(m, n) = p \cdot a \cdot b$，即 $q = p \cdot a \cdot b$。所以 $m \cdot n = p \cdot a \cdot p \cdot b = p \cdot (p \cdot a \cdot b) = p \cdot q$。
"两数之积 = HCF × LCM" 是一个非常有用的恒等式。

自评：
There are n apples. If divided equally among 3 students, there will be 2 left. If divided equally among 5 students, there will be 4 left. What is the smallest possible value of n? Show your working.
Hint: What is the relationship between 2 and 3? Between 4 and 5? What if we had $(n+1)$ apples?

EN Notice $2 = 3 - 1$ and $4 = 5 - 1$. So both remainders are "one less than the divisor". This means $(n+1)$ would be divisible by 3 (giving 0 remainder) AND by 5 (giving 0 remainder). So $n + 1$ is a common multiple of 3 and 5. The smallest is $\text{LCM}(3, 5) = 15$. Hence $n + 1 = 15$ and $\mathbf{n = 14}$. Verify: $14 \div 3 = 4$ r 2 ✓, $14 \div 5 = 2$ r 4 ✓.

⏱ 仔细阅读英文版，30 秒后可查看中文版

中文注意 $2 = 3 - 1$，$4 = 5 - 1$——两个余数都"比除数少 1"。这意味着 $(n+1)$ 既能被 3 整除（余 0），也能被 5 整除（余 0）。所以 $n+1$ 是 3 和 5 的公倍数。最小的是 $\text{LCM}(3, 5) = 15$。所以 $n+1 = 15$，$n = 14$。验证：$14 \div 3 = 4$ 余 $2$ ✓，$14 \div 5 = 2$ 余 $4$ ✓。

自评：

章末概念检查 · Concept Checkpoints

5 道封闭题，自动判分。这些题会进入跨章节复习池——后面学 Square Roots 之前会重新弹出，确保你掌握 1.3 的核心。