1.2

Highest Common Factor (HCF)

AWhat is the HCF?

Let us list the factors of 18 and 24.

Factors of 18: 1, 2, 3, 6, 9, 18
Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24

The common factors of 18 and 24 are 1, 2, 3 and 6. That is, both 18 and 24 can be divided by 1, 2, 3 and 6 exactly. Among these common factors, the largest one is 6. Hence 6 is the highest common factor (HCF) of 18 and 24.

The highest common factor (HCF) of two or more whole numbers is the largest whole number that divides the given numbers exactly.

We also apply the concept of HCF when we express fractions in their simplest form directly. For example, to express $\dfrac{18}{24}$ in its simplest form, we divide the numerator and denominator by their HCF, 6, as shown.

$$\dfrac{18}{24} = \dfrac{18 \div 6}{24 \div 6} = \dfrac{3}{4}$$ 6 is the HCF of 18 and 24.

BMethods of Finding HCF

There are different methods of finding the HCF of two or more numbers. We can use prime factorisation to find their HCF in a more efficient way.

📖 WORKED EXAMPLE 4

Find the highest common factor of 225 and 750.

Solution

Method 1 Prime Factorisation

We express 225 and 750 each as a product of its prime factors.

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

We can visualise the process of taking common factors as shown:

225 = 3 × 3 × 5 × 5 750 = 2 × 3 × 5 × 5 × 5

HCF = 3 × 5 × 5

$\therefore \text{HCF} = 3 \times 5^2 = 75$

(3 is of a lower power than $3^2$, and $5^2$ is of a lower power than $5^3$.)

NOTE: The HCF of any three numbers is obtained by multiplying the lowest power of each common prime factor of the given numbers. In this case, the HCF is obtained by multiplying the lower power of each common prime factor — that is, 3 and $5^2$.

Method 2 Successive Short Division

We use short division to divide both numbers by their common prime factors successively until there are no common prime factors.

For example,

3 225 750 5 75 250 5 15 50 3 10 ← no common prime factor

$\therefore \text{HCF} = 3 \times 5 \times 5 = 75$

NOTE: The HCF is the product of all the common prime factors.

THINK²

For the fraction $\dfrac{225}{750}$, if we divide the numerator and the denominator by their HCF, 75, is the fraction in its simplest form? Explain.

EN Yes. $\dfrac{225 \div 75}{750 \div 75} = \dfrac{3}{10}$. The HCF of 3 and 10 is 1, so the numerator and denominator share no common factor greater than 1 — the fraction is in its simplest form.

⏱ 仔细阅读英文版，30 秒后可查看中文版

中文是。$\dfrac{225 \div 75}{750 \div 75} = \dfrac{3}{10}$。3 和 10 的 HCF 是 1，分子与分母没有大于 1 的公因数，所以这是最简形式。

看完答案，自评：

✏️ TRY IT YOURSELF 4

(a) Find the HCF of 252 and 360 using prime factorisation.
(b) Find the HCF of 882 and 945 using successive short division.

📖 WORKED EXAMPLE 5

Find the HCF of 84, 126 and 245.

Solution

84 = 2² × 3 × 7 126 = 2 × 3² × 7 245 = 5 × 7²

HCF = 7

7 is the only prime factor common to 84, 126 and 245.

$\therefore \text{HCF} = 7$

(By aligning the factors of the given numbers as shown, we can choose the lowest power of each common prime factor easily.)

THINK²

We found out that 7 is the HCF of 84, 126 and 245.
(a) Is 7 the HCF of any two of these three numbers?
(b) Is 7 greater or smaller than the HCF of those two numbers? Explain.

EN (a) No. $\text{HCF}(84,126)=42$, $\text{HCF}(84,245)=7$ and $\text{HCF}(126,245)=7$. So 7 is the HCF of two pairs out of the three, but not of the pair (84, 126).
(b) Adding a third number can only shrink the HCF, never grow it. So 7 is smaller than or equal to the HCF of any two of these numbers (and equal when the pair already shares only 7 as the largest common factor).

⏱ 仔细阅读英文版，30 秒后可查看中文版

中文 (a) 不全是。$\text{HCF}(84,126)=42$，$\text{HCF}(84,245)=7$，$\text{HCF}(126,245)=7$。三对中有两对的 HCF 是 7，但 (84, 126) 这对的 HCF 是 42，不是 7。
(b) 增加一个数只会让 HCF 不变或变小，不会变大。所以 7 小于等于任意两数的 HCF（当那两数本来最大公因数就是 7 时取等号）。

看完答案，自评：

✏️ TRY IT YOURSELF 5

(a) Find the HCF of 77, 110 and 242.
(b) Find the HCF of 21, 108 and 275.

CHCF in Word Problems

📖 WORKED EXAMPLE 6

A rectangular field measuring 20 m by 15 m is divided exactly into identical square plots.

(a) What is the largest possible length of a side of each square plot?
(b) Hence find the total number of square plots obtained.

Analysis

The length of a side of each square plot should divide both 20 and 15 exactly. Suppose we let the length of a side of each square plot be 10 m, then the shaded part cannot be divided into identical squares of side 10 m. Hence the length of a side of each square plot should be a common factor of 15 and 20.

When we find the largest possible length of a side of each plot, we are actually finding the highest common factor (HCF) of 20 and 15.

Solution

(a) $20 = 2^2 \times 5$
    $15 = 3 \times 5$
    The HCF of 20 and 15 is 5.
    The largest possible length of a side of each square plot is 5 m.

(b) Number of squares along the length $= 20 \div 5 = 4$
Number of squares along the breadth $= 15 \div 5 = 3$
Hence the number of square plots obtained $= 4 \times 3 = \mathbf{12}$.

Alternatively, the number of square plots obtained
$= \text{Area of the rectangle} \div \text{Area of one square}$
$= (20 \times 15) \div (5 \times 5) = \mathbf{12}$.

✏️ TRY IT YOURSELF 6

A cuboid measuring 56 cm by 40 cm by 24 cm is cut exactly into identical cubes.

(a) Find the largest possible length of a side of each cube.
(b) Hence find the total number of cubes obtained.

📖 WORKED EXAMPLE 7

Jane is packing 252 apples and 360 oranges into identical baskets. She wants to distribute each type of fruit equally among the baskets with no fruit left.

(a) Find the largest number of baskets that Jane can pack.
(b) What is the number of apples in each basket?

Analysis

For a start, to help us understand and solve the problem, we can think of a simpler problem such as packing 20 apples and 15 oranges instead.

When there is 1 basket, we can distribute the apples and oranges equally — 20 apples and 15 oranges in 1 basket.
When there are 2 baskets, we can distribute 20 apples equally between the baskets but we cannot distribute the 15 oranges equally.
When there are 3 baskets, we cannot distribute the 20 apples equally among the baskets but we can distribute the 15 oranges equally.

Hence the number of baskets that can be packed is a common factor of the number of apples and the number of oranges. The greatest number of baskets that Jane can pack is actually the HCF of the numbers of apples and oranges.

Solution

(a) We find the prime factorisation of each number.
    $252 = 2^2 \times 3^2 \times 7$
    $360 = 2^3 \times 3^2 \times 5$
    $\text{HCF of }252 \text{ and } 360 = 2^2 \times 3^2 = 36$
    The greatest number of baskets that Jane can pack is 36.

(b) $252 \div 36 = 7$
The number of apples in each basket is 7.

THINK²

If Jane also packs 72 pears equally into the baskets, will the answer in Worked Example 7(a) change? Explain briefly.

EN Yes, it changes. The greatest number of baskets must now divide 252, 360 and 72 exactly. So we need $\text{HCF}(252, 360, 72)$. $72 = 2^3 \times 3^2$, so $\text{HCF}(252, 360, 72) = 2^2 \times 3^2 = 36$ — but that's a coincidence here. The answer is still 36 because 72 already shares $2^2 \times 3^2$ with the previous HCF. (General rule: adding a third number can only keep the HCF the same or shrink it.)

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中文 会变（也可能不变）。现在最多的篮子数要同时整除 252、360 和 72，所以要算 $\text{HCF}(252, 360, 72)$。$72 = 2^3 \times 3^2$，所以 $\text{HCF}(252,360,72) = 2^2 \times 3^2 = 36$ — 恰好和之前一样。这是因为 72 已经包含了原 HCF 中的 $2^2 \times 3^2$。（一般来说，加入第三个数只会让 HCF 不变或更小。）

看完答案，自评：

✏️ TRY IT YOURSELF 7

Members from three performing arts groups get together to organise an outdoor activity. The Dance Society has 48 members, the Band has 84 members and the Choir has 144 members. All these members are divided equally to form teams for an adventure quest.

(a) What is the greatest number of teams that can be formed?
(b) How many members from each performing arts group are there in a team?
Dance: Band: Choir:

📝 PRACTICE EXERCISE 1.2

BASIC MASTERY

Find the HCF of each pair of numbers.

(a) 8 and 12 →

(b) 45 and 72 →

(c) 74 and 99 →

(d) 120 and 225 →

(e) 108 and 240 →

(f) 385 and 396 →
Find the HCF of each group of numbers.

(a) 28, 63 and 91 →

(b) 60, 75 and 100 →

(c) 14, 36 and 175 →

(d) 70, 210 and 350 →

INTERMEDIATE

Find the largest whole number that is a factor of both 54 and 180.
Find the largest whole number that will divide 77, 110 and 242 exactly.
Two numbers written as the product of their prime factors are $2^3 \times 3^3 \times 5^2 \times 7$ and $2 \times 3^4 \times 7^3 \times 13$. Find the largest whole number that will divide both numbers exactly.
There are two metal bars of length 72 cm and 96 cm. Both bars are cut exactly into pieces of equal length. Find the largest possible length of each piece.
A rectangular piece of tin plate measuring 360 cm by 280 cm is cut exactly into identical squares of the largest possible length. Find

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

(b) the number of square tin plates obtained.
Amirah needs to pack 132 pieces of white chocolate and 154 pieces of dark chocolate into identical boxes. She distributes each type of chocolate equally among the boxes. Find

(a) the largest number of boxes that can be packed,

(b) the number of pieces of each type of chocolate in a box.
white: dark:
The total sales of a particular model of calculator on three days are $180, $414 and $306 respectively.

(a) Find the greatest possible price of the calculator. $

(b) Hence find the total number of calculators sold on the three days.
A charity distributes 420 packets of rice, 168 bottles of oil and 504 oranges equally among some senior citizens.

(a) Find the greatest possible number of senior citizens who can benefit. Hence find how many packets of rice each senior citizen will get.
citizens: rice/person:

(b) Mr Tan donates 250 tins of milk powder for distribution to all the senior citizens. Is it possible to distribute the tins of milk powder equally among them? If not, how many more tins are needed?
more tins needed:

ADVANCED

[OPEN] State two composite numbers such that the HCF of the two numbers is 1.

EN Many valid answers — pick two composite numbers whose prime factorisations share no common prime. Examples: 4 and 9 ($2^2$ and $3^2$); 8 and 9 ($2^3$ and $3^2$); 9 and 25; 4 and 25; 16 and 27.

⏱ 仔细阅读英文版，30 秒后可查看中文版

中文答案不唯一——挑两个质因数完全不重合的合数。例如： 4 和 9（$2^2$ 和 $3^2$）；8 和 9；9 和 25；4 和 25；16 和 27 等。

自评：
[OPEN] Find two different numbers less than 100 such that the HCF of the two numbers is 18.

EN We need two multiples of 18 (less than 100) whose HCF is exactly 18 — i.e. dividing them by 18 must give two coprime numbers. Multiples of 18 below 100: 18, 36, 54, 72, 90. Examples: 18 and 90 (18 = 18·1, 90 = 18·5; HCF(1,5)=1 ✓); 54 and 72 (54 = 18·3, 72 = 18·4; HCF(3,4)=1 ✓); 18 and 54 ✓; 18 and 36 ✓; 72 and 90 ✓.

⏱ 仔细阅读英文版，30 秒后可查看中文版

中文需要找两个 100 以内的 18 的倍数，且两数除以 18 后互质（HCF=1）。100 以内的 18 的倍数：18, 36, 54, 72, 90。可选： 18 与 90（18=18·1, 90=18·5; HCF(1,5)=1 ✓）；54 与 72（HCF(3,4)=1 ✓）； 18 与 54；18 与 36；72 与 90 等。

自评：
A number has exactly eight factors. Two of its factors are 4 and 6. What is the number?
[OPEN] Find three numbers such that the HCF of each pair of these numbers is greater than 1 but the HCF of all three numbers is 1.
Hint: For instance, the numbers 6, 10 and 15 satisfy the conditions.

EN Pick three numbers, each being the product of two distinct primes, with the three primes shared "in pairs": 6, 10, 15 (= 2·3, 2·5, 3·5 — pairs share 2, 3 or 5; all-three share none). Other examples: 14, 22, 77 (2·7, 2·11, 7·11) ✓; 15, 21, 35 (3·5, 3·7, 5·7) ✓.

⏱ 仔细阅读英文版，30 秒后可查看中文版

中文选三个数，每个数都是两个不同质数的积，三个质数"两两共享"： 6, 10, 15（= 2·3, 2·5, 3·5 — 任意两数共享一个质因数；三个数没有共同的质因数）。其他例子：14, 22, 77（2·7, 2·11, 7·11）✓；15, 21, 35（3·5, 3·7, 5·7）✓。

自评：
[OPEN] The price (in dollars) of a model car in a shop is a whole number greater than 1. The sales from the model cars on two days are $1518 and $2254. Find two possible total number of model cars sold over the two days.

EN The price must be a common factor > 1 of 1518 and 2254. $1518 = 2 \cdot 3 \cdot 11 \cdot 23$ and $2254 = 2 \cdot 7^2 \cdot 23$, so $\text{HCF} = 46$ and the common factors > 1 are 2, 23, 46. Total cars sold:
· Price = $2: $1518/2 + 2254/2 = 759 + 1127 = \mathbf{1886}$ cars
· Price = $23: $66 + 98 = \mathbf{164}$ cars
· Price = $46: $33 + 49 = \mathbf{82}$ cars
Any two of these are valid answers.

⏱ 仔细阅读英文版，30 秒后可查看中文版

中文价格必须是 1518 和 2254 大于 1 的公因数。$1518 = 2 \cdot 3 \cdot 11 \cdot 23$，$2254 = 2 \cdot 7^2 \cdot 23$，HCF = 46，大于 1 的公因数为 2、23、46。两天售出总数：
· 价格 = $2: 1518/2 + 2254/2 = 759 + 1127 = \mathbf{1886}$ 辆
· 价格 = $23: 66 + 98 = \mathbf{164}$ 辆
· 价格 = $46: 33 + 49 = \mathbf{82}$ 辆
任写两个即可。

自评：
The product of two numbers is 8820. The HCF of the two numbers is 42. Find the greater number.
Find the greatest number that will divide 171, 255 and 304, so as to leave the same remainder in each case.

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 LCM 之前会重新弹出，确保你掌握 1.2 的核心。