Integers · Study Notes · Edexcel IGCSE 4MA1 Higher

Foundation for the whole Number topic. Master types of integers, factors, multiples, primes, HCF and LCM — plus the divisibility rules and word-problem cues that examiners reward.

At a glance

Integers $\mathbb{Z} = \{ \dots, -2, -1, 0, 1, 2, \dots \}$ — whole numbers, positive, negative or zero.
Prime = exactly TWO factors. $1$ is NOT prime; $2$ IS prime (only even prime).
Prime factorisation — every integer $> 1$ has a unique product of primes. Always give in INDEX form when asked.
HCF = lower powers of common primes. LCM = higher powers of ALL primes appearing.
Identity: $a \times b = \text{HCF}(a,b) \times \text{LCM}(a,b)$ — useful when 3 of 4 are known.
Word problems: 'when do they meet again?' = LCM. 'biggest equal share?' = HCF.
Divisibility: $3$ and $9$ via digit sum; $4$ via last two digits; $5$ via last digit; $11$ via alternating sum.

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

1.1 — Identify and use natural numbers, integers, even and odd numbers, prime numbers, square and cube numbers.
1.1 — Use the terms factor, multiple, common factor, common multiple.
1.1 — Find the HCF and LCM of two or more numbers.
1.1 — Write integers as a product of prime factors using index notation.
1.1 — Use BIDMAS/BODMAS for order of operations including negative numbers.

Types of integers — what each one is

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Natural, integer, prime, composite, square, cube — Edexcel exam vocabulary.

Edexcel 4MA1 expects you to use precise vocabulary. Mark schemes credit specific keywords.

Natural numbers ( $\mathbb{N}$ ). $1, 2, 3, 4, \ldots$ — positive integers used for counting. $0$ is NOT a natural number under the Edexcel convention.

Integers ( $\mathbb{Z}$ ). $\ldots, -3, -2, -1, 0, 1, 2, 3, \ldots$ — every whole number, positive, negative, or zero.

Even and odd integers. Even = divisible by $2$ ; odd = not divisible by $2$ . Note $0$ is even.

Prime numbers. Integers $> 1$ with EXACTLY two positive factors ( $1$ and itself). The first ten primes: $2, 3, 5, 7, 11, 13, 17, 19, 23, 29$ . $2$ is the only even prime. $1$ is NOT prime.

Composite numbers. Positive integers $> 1$ that have more than two factors. Equivalently: positive integers $> 1$ that aren't prime.

Square numbers. $n^{2}$ : $1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, \ldots$ . Memorise to $15^{2} = 225$ .

Cube numbers. $n^{3}$ : $1, 8, 27, 64, 125, 216, 343, 512, 729, 1000$ . Memorise to $10^{3}$ .

Worked qualitative. Why is $1$ not a prime number?

A prime needs EXACTLY two distinct factors.
$1$ has only ONE factor (itself).
Excluding $1$ from primes also makes the Fundamental Theorem of Arithmetic clean (every integer has a unique prime factorisation).

Edexcel tip. State 'integer' explicitly when asked. Many candidates write 'whole number' instead — Edexcel mark schemes accept both, but 'integer' is the technical term and is always rewarded.

Natural: $1, 2, 3, \ldots$ (no $0$ ).
Integer: any whole number including $0$ and negatives.
Prime: exactly 2 factors. $1$ excluded.
Square + cube — memorise common ones.
$0$ is even.

Prime factorisation — the foundation for HCF and LCM

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Every integer $> 1$ has a unique product of primes. Use a factor tree. Always give in INDEX form.

Why prime factorisation? Two numbers' prime factorisations make HCF and LCM trivial to compute. Almost every Number 1 question expects you to factorise first.

Method 1 — Factor tree.

Start with the number. Divide by the smallest prime that fits. Continue branching until all leaves are prime.

Example: $360$

Read off all the primes: $360 = 2 \times 2 \times 2 \times 3 \times 3 \times 5$ .

In INDEX form: $360 = 2^{3} \times 3^{2} \times 5$ .

Green leaves are primes; collect them into index form.

Method 2 — Repeated division.

Same answer.

Worked qualitative. What's special about index form vs the long product?

Long product: $2 \times 2 \times 2 \times 3 \times 3 \times 5$ .
Index form: $2^{3} \times 3^{2} \times 5^{1}$ .
Edexcel mark schemes have separate marks for METHOD (factor tree shown) and ANSWER in INDEX FORM.
Stopping at the long product = lose the A1.

Edexcel tip. When the question says "in index form" and you don't, you lose the accuracy mark. Always group repeats AND drop the $^1$ on lone primes (write $5$ , not $5^{1}$ ).

Factor tree or repeated division.
Always give answer in INDEX form.
Group repeats: $2 \times 2 \times 2 = 2^{3}$ .
Lone primes: drop the $^1$ .
M1 for method, A1 for index form.

HCF and LCM — the rules and identity

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HCF: lower powers of common primes. LCM: higher powers of all primes. Memorise the product identity.

Once you've prime-factorised both numbers, computing HCF and LCM is mechanical.

HCF (Highest Common Factor) rule. Take each prime that appears in BOTH factorisations, raised to the LOWER power.

LCM (Lowest Common Multiple) rule. Take EVERY prime that appears in EITHER factorisation, raised to the HIGHER power.

Worked example. Find HCF and LCM of $A = 2^{4} \times 3^{2} \times 5$ and $B = 2^{2} \times 3^{3} \times 7$ .

Prime	$A$	$B$	HCF (lower)	LCM (higher)
$2$	$2^{4}$	$2^{2}$	$2^{2}$	$2^{4}$
$3$	$3^{2}$	$3^{3}$	$3^{2}$	$3^{3}$
$5$	$5^{1}$	—	—	$5^{1}$
$7$	—	$7^{1}$	—	$7^{1}$

$\text{HCF} = 2^{2} \times 3^{2} = 4 \times 9 = 36$ .
$\text{LCM} = 2^{4} \times 3^{3} \times 5 \times 7 = 16 \times 27 \times 5 \times 7 = 15{,}120$ .

HCF uses only the shared overlap; LCM uses every prime in the diagram.

The product identity (memorise — NOT on Edexcel formula sheet):

$a \times b = \text{HCF}(a, b) \times \text{LCM}(a, b)$

Useful when given 3 of the 4 quantities.

Worked qualitative. When do you use HCF vs LCM in word problems?

Phrase	Use
"When will they next coincide?"	LCM (find when both repeat together)
"Greatest length / largest box / biggest equal share"	HCF (find biggest divisor of both)
"How many groups of equal size?"	HCF
"How long until they all flash together?"	LCM

Edexcel tip. Show the prime factorisations FIRST, then circle/highlight common primes. This earns method marks even if final arithmetic is wrong.

HCF: common primes, LOWER power.
LCM: all primes, HIGHER power.
Identity: $ab = \text{HCF} \times \text{LCM}$ .
When-they-meet-again → LCM.
Biggest equal share → HCF.

Divisibility tests — quick checks

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Memorise these rules. They save time on Paper 1H non-calculator questions.

Edexcel Paper 1H has no calculator. These rules let you test factors instantly.

Divisor	Test
$2$	Last digit even ( $0, 2, 4, 6, 8$ )
$3$	Digit sum divisible by $3$
$4$	Last two digits form a multiple of $4$
$5$	Last digit is $0$ or $5$
$6$	Divisible by both $2$ AND $3$
$8$	Last three digits form a multiple of $8$
$9$	Digit sum divisible by $9$
$10$	Last digit is $0$
$11$	Alternating digit sum is divisible by $11$

Example: is $7{,}524$ divisible by $9$ ?

Digit sum: $7 + 5 + 2 + 4 = 18$ .
$18 \div 9 = 2$ — yes.

Example: is $7{,}524$ divisible by $11$ ?

Alternating sum: $7 - 5 + 2 - 4 = 0$ .
$0$ is a multiple of $11$ — yes.

Worked qualitative. A 5-digit number $4\,a\,2\,7\,5$ is divisible by $9$ . Find $a$ .

Digit sum: $4 + a + 2 + 7 + 5 = 18 + a$ .
For divisibility by $9$ : $18 + a$ must be a multiple of $9$ .
Since $0 \le a \le 9$ : $a = 0$ gives $18$ , $a = 9$ gives $27$ .
Both work — answer: $a = 0$ or $a = 9$ .

Edexcel tip. State the rule before applying it. Mark schemes for digit-puzzles award M1 for the rule and A1 for the answer.

$3, 9$ — digit sum.
$4$ — last two digits.
$8$ — last three digits.
$11$ — alternating sum.
State the rule for the M1 mark.

Working with negative integers

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Even powers stay positive. Subtracting a negative is adding. BIDMAS rules unchanged.

Negative integers appear throughout 4MA1 — coordinates, temperatures, finance.

The four arithmetic rules.

Operation	Result
$+ \times +$	$+$
$- \times -$	$+$
$+ \times -$	$-$
$- \times +$	$-$

Same rules apply to division.

Powers of negatives.

$(-2)^{2} = +4$ (even exponent → positive).
$(-2)^{3} = -8$ (odd exponent → negative).
General: $(-x)^{n} = +x^n$ if $n$ even, $-x^n$ if $n$ odd.

Watch the brackets.

$(-3)^{2} = 9$ .
$-3^{2} = -9$ (just the $3$ is squared).

Subtracting a negative.

Replace 'subtract negative' with 'add positive':

$9 - (-20) = 9 + 20 = 29$ .

BIDMAS still applies:

Brackets
Indices (powers/roots)
Division/Multiplication (left to right)
Addition/Subtraction (left to right)

Worked qualitative. Calculate $(-2)^{4} + (-3) \times (-5) - (-7)$ .

Powers first: $(-2)^{4} = 16$ .
Multiplication next: $(-3) \times (-5) = 15$ .
Substitute: $16 + 15 - (-7) = 16 + 15 + 7 = 38$ .

Edexcel tip. Underline negative numbers in the question to remind yourself they're there. The double-negative trap is the single most common arithmetic error in Paper 1H.

$- \times - = +$ .
Even power of $-x$ is positive.
$(-3)^2 = 9$ , but $-3^2 = -9$ .
Subtract negative = add positive.
BIDMAS applies as usual.

Quick recap

Integers: whole numbers including 0 and negatives.
Primes: exactly 2 factors; $1$ excluded.
Prime factorisation — always in INDEX form.
HCF: lower powers of common primes. LCM: higher powers of all.
Identity: $a \times b = \text{HCF} \times \text{LCM}$ .
Divisibility: $3, 9$ digit sum; $4$ last two; $11$ alternating.
Negatives: $- \times - = +$ ; brackets matter for powers.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Integer — any whole number, positive, negative or zero.
Prime — integer $> 1$ with exactly two factors.
HCF — lower powers of common primes.
LCM — higher powers of all primes appearing.
Identity: $a \times b = \text{HCF}(a,b) \times \text{LCM}(a,b)$ .
Divisibility by 3, 9: digit sum.
Divisibility by 4: last two digits multiple of 4.

How it’s examined

Integers appear on EVERY 4MA1 paper (Paper 1H + Paper 2H). Typical formats: prime factorise (2-3 marks), HCF/LCM (3-4 marks), digit puzzle (2-3 marks), word problem with HCF or LCM (3-5 marks). Examiner reports flag (1) candidates not giving prime factorisation in index form, (2) using HCF instead of LCM in 'when do they meet?' problems, and (3) the double-negative arithmetic trap on Paper 1H.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/1H May/Jun 2023, 4MA1/2H May/Jun 2024, 4MA1/1H Oct/Nov 2023. Last reviewed 2026-05-07.

Step-by-step solutions to past-paper-style questions on integers, written exactly the way a tutor would explain them at the board.

1Write a number as a product of prime factors in index form
Foundation• Adapted from 4MA1/1H May/Jun 2023 Q3• prime factorisation, index form
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Question
Write $504$ as a product of its prime factors. Give your answer in index form.
Step-by-step solution
1. Step 1
  Use a factor tree, dividing by the smallest prime each time. Start with $2$ : $504 \div 2 = 252$ .
2. Step 2
  Continue: $252 \div 2 = 126$ , $126 \div 2 = 63$ .
3. Step 3
  $63$ is odd, move to the next prime: $63 \div 3 = 21$ , $21 \div 3 = 7$ .
4. Step 4
  $7$ is itself prime, so stop. Collect the primes:
  $504 = 2 \times 2 \times 2 \times 3 \times 3 \times 7$
5. Step 5
  Group repeats into index form.
  $504 = 2^{3} \times 3^{2} \times 7$
Answer
$504 = 2^{3} \times 3^{2} \times 7$
Examiner tip
Edexcel mark schemes split this into M1 (correct method shown — factor tree or repeated division) and A1 (correct prime factor product in INDEX form). Stopping at $2 \times 2 \times 2 \times 3 \times 3 \times 7$ loses the A1.
2Find the HCF and LCM from prime factorisations
Higher• Adapted from 4MA1/2H May/Jun 2024 Q5• HCF, LCM, prime factorisation
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Question
Two numbers $A$ and $B$ are written as products of their prime factors:\n $A = 2^{3} \times 3 \times 5^{2}$ \n $B = 2^{2} \times 3^{2} \times 7$ \nFind (a) the highest common factor (HCF) of $A$ and $B$ and (b) the lowest common multiple (LCM) of $A$ and $B$ .
Step-by-step solution
1. Step 1
  HCF rule: take each prime that appears in BOTH numbers, raised to the LOWER power.
2. Step 2
  Common primes: $2$ (lower power $2^{2}$ ) and $3$ (lower power $3^{1}$ ). The primes $5$ and $7$ aren't in both — exclude them.
  $\text{HCF} = 2^{2} \times 3 = 4 \times 3 = 12$
3. Step 3
  LCM rule: take EVERY prime that appears in either number, raised to the HIGHER power.
4. Step 4
  Primes used: $2^{3}$ , $3^{2}$ , $5^{2}$ , $7$ .
  $\text{LCM} = 2^{3} \times 3^{2} \times 5^{2} \times 7 = 8 \times 9 \times 25 \times 7 = 12600$
Answer
(a) $\text{HCF} = 12$ (b) $\text{LCM} = 12{,}600$
Examiner tip
The 4MA1 mark scheme awards 2 marks for each part: M1 for the correct rule applied (showing the lower/higher powers) and A1 for the answer. State the powers explicitly — examiners cannot award the M1 from a bare numerical answer.
3Use the HCF × LCM identity
Higher• HCF, LCM
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Question
Two positive integers have HCF $= 8$ and LCM $= 240$ . One of the numbers is $48$ . Find the other.
Step-by-step solution
1. Step 1
  Use the identity for two integers $a$ and $b$ :
  $a \times b = \text{HCF}(a,b) \times \text{LCM}(a,b)$
2. Step 2
  Substitute: $48 \times b = 8 \times 240$ .
  $48b = 1920$
3. Step 3
  Solve for $b$ :
  $b = \frac{1920}{48} = 40$
Answer
The other number is $40$ .
Examiner tip
This identity is examined regularly on Paper 2H. Quote it explicitly to score the method mark — the formula is NOT given on the Edexcel formula sheet.
4Classify numbers as prime, square or cube
Foundation• Adapted from 4MA1/1H Oct/Nov 2023 Q2• prime, square, cube
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Question
From the list $\{ 1,\ 9,\ 17,\ 21,\ 27,\ 49,\ 64 \}$ write down (a) the prime numbers, (b) the cube numbers, (c) the numbers that are both square AND cube.
Step-by-step solution
1. Step 1
  Primes: integers $> 1$ with exactly two factors. Check each: $9 = 3 \times 3$ (not prime), $17$ (only $1, 17$ — prime), $21 = 3 \times 7$ (not prime), $27 = 3^{3}$ (not prime), $49 = 7^{2}$ (not prime), $64 = 2^{6}$ (not prime). Note $1$ is NOT prime.
2. Step 2
  Cubes: $n^{3}$ for some positive integer $n$ . From the list: $1 = 1^{3}$ , $27 = 3^{3}$ , $64 = 4^{3}$ .
3. Step 3
  Both square AND cube: must be of the form $n^{6}$ . From the list, $1 = 1^{6}$ and $64 = 2^{6}$ .
Answer
(a) $17$ (b) $1, 27, 64$ (c) $1$ and $64$
Examiner tip
$1$ is a square AND a cube AND a sixth power. Edexcel reports note candidates routinely forget to include $1$ and lose the mark.
5Use divisibility rules to find a missing digit
Higher• Adapted from 4MA1/1H Jan 2024 Q4• divisibility, digit puzzles
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Question
The five-digit number $7\,2\,a\,1\,8$ is divisible by $9$ . Find the value of the digit $a$ .
Step-by-step solution
1. Step 1
  Divisibility rule for $9$ : the digit sum must be a multiple of $9$ .
2. Step 2
  Digit sum: $7 + 2 + a + 1 + 8 = 18 + a$ .
3. Step 3
  $18 + a$ must be a multiple of $9$ . Since $a$ is a single digit ( $0 \le a \le 9$ ), the only options are $a = 0$ (giving $18$ ) or $a = 9$ (giving $27$ ).
4. Step 4
  If the question expects a unique answer, $a = 0$ is the smallest. (Both are mathematically valid — list both unless the question constrains.)
Answer
$a = 0$ or $a = 9$
Examiner tip
Always state the divisibility rule before computing. Mark schemes for these problems award M1 for the correct rule (digit sum) and A1 for both/single value of $a$ .
6Order of operations with negative integers
Foundation• negative numbers, BIDMAS
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Question
Calculate $(-3)^{2} - 4 \times (-5) + (-2)^{3}$ .
Step-by-step solution
1. Step 1
  Apply BIDMAS. Powers first: $(-3)^{2} = 9$ (positive — even power), and $(-2)^{3} = -8$ (negative — odd power).
2. Step 2
  Multiplication next: $4 \times (-5) = -20$ .
3. Step 3
  The expression becomes $9 - (-20) + (-8) = 9 + 20 - 8$ .
4. Step 4
  Simplify: $9 + 20 - 8 = 21$ .
Answer
$21$
Examiner tip
Even powers of negatives give positives; odd powers stay negative. The double-negative trap ( $-(-20) = +20$ ) is one of the most common Paper 1H errors flagged in Edexcel reports.
7Real-life LCM word problem
Higher• Adapted from 4MA1/2H May/Jun 2024 Q11• LCM, word problem
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Question
Two lighthouses flash at regular intervals. Lighthouse A flashes every $24$ seconds. Lighthouse B flashes every $36$ seconds. They flash together at $9{:}00$ am. At what time will they next flash together?
Step-by-step solution
1. Step 1
  When two events repeat at fixed intervals, they re-align at the LCM of the two intervals.
2. Step 2
  Prime factorise: $24 = 2^{3} \times 3$ , $36 = 2^{2} \times 3^{2}$ .
3. Step 3
  $\text{LCM} = 2^{3} \times 3^{2} = 8 \times 9 = 72$ seconds.
4. Step 4
  $72$ seconds $= 1$ minute $12$ seconds. So they next flash together at $9{:}01{:}12$ am.
Answer
$9{:}01{:}12$ am ( $72$ seconds after $9{:}00$ ).
Examiner tip
Recognising LCM (not HCF) is the key insight. Edexcel reports note that ~40% of candidates use HCF in this type of question — losing all the marks. RULE: 'when do they meet?' = LCM; 'how can we share equally?' = HCF.
8Height above and depth below sea level [1 mark]
Getting started• Paper 1H/2H style — 1 mark• directed numbers, context
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Question
The top of a hill is $215$ m above sea level. The bottom of a mine shaft is $380$ m below sea level.

Work out the vertical distance between the top of the hill and the bottom of the mine shaft. [1]
Step-by-step solution
1. Step 1
  Write the heights as directed numbers: above sea level is positive, below is negative. The distance between them is the higher value minus the lower value.
2. Step 2
  B1 for the distance.
  $215 - (-380) = 215 + 380 = 595$
Answer
$595$ m
Examiner tip
Examiner tip: Sea level is the zero on this number line. One height is above zero and the other below, so the two distances from zero are added.

Common mistake: Subtracting the sizes, $380 - 215 = 165$ m, as if both were on the same side of sea level.
9A common multiple within a range [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• lcm, multiples
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Question
Find a number between $100$ and $150$ that is a multiple of $6$ and a multiple of $14$ . Show your working clearly. [3]
Step-by-step solution
1. Step 1
  A number that is a multiple of both is a multiple of their LCM. M1 for both numbers as products of prime factors.
  $6 = 2 \times 3, \qquad 14 = 2 \times 7$
2. Step 2
  A1 for the LCM.
  $\text{LCM} = 2 \times 3 \times 7 = 42$
3. Step 3
  The multiples of $42$ are $42, 84, 126, 168, \ldots$ Only one lies between $100$ and $150$ . A1 for $126$ .
  $42 \times 3 = 126$
Answer
$126$
Examiner tip
Examiner tip: Check the answer against both conditions: $126 \div 6 = 21$ and $126 \div 14 = 9$ , both exact.

Common mistake: Using $6 \times 14 = 84$ as the only common multiple and then answering $168$ (the next multiple of $84$ ), which is outside the range. The LCM is $42$ , not $84$ , because $6$ and $14$ share a factor of $2$ .
10Smallest multiplier for a square that is also a multiple of 7 [5 marks]
Stretch• Paper 1H/2H style — 5 marks• prime factors, square numbers
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Question
$k$ is a positive integer.

$90k$ is a square number and $90k$ is a multiple of $7$ .

Find the smallest possible value of $k$ . Show your working clearly. [5]
Step-by-step solution
1. Step 1
  M1 for a correct method to write $90$ as a product of primes.
  $90 \div 2 = 45, \quad 45 \div 3 = 15, \quad 15 \div 3 = 5$
2. Step 2
  A1 for the product in index form.
  $90 = 2 \times 3^{2} \times 5$
3. Step 3
  A square number has every prime to an even power. $3^{2}$ is fine, but $2$ and $5$ each need one more factor. M1 for identifying that $k$ must contain $2 \times 5$ .
4. Step 4
  $90k$ must also contain the prime $7$ . In a square, $7$ must appear to an even power, so the smallest is $7^{2}$ . M1 for including $7^{2}$ in $k$ .
  $k = 2 \times 5 \times 7^{2}$
5. Step 5
  A1 for $k = 490$ .
  $90 \times 490 = 2^{2} \times 3^{2} \times 5^{2} \times 7^{2} = 44\,100 = 210^{2}$
Answer
$k = 490$
Examiner tip
Examiner tip: Build the target first: the smallest square that is a multiple of both $90$ and $7$ is $2^{2} \times 3^{2} \times 5^{2} \times 7^{2}$ . Then $k$ is whatever is missing from $90$ .

Common mistake: Using $k = 2 \times 5 \times 7 = 70$ . That makes $90k$ a multiple of $7$ , but $6300$ is not a square because $7$ appears only once.

Question 1
Paper 1H/2H style — 1 mark1 mark
Write down the value of the digit $7$ in the number $3\,074\,219$ . [1]
Model answer
$70\,000$ (seven ten-thousands)
Why this scores
Mark scheme:
- B1 for $70\,000$ or seventy thousand oe
Why this scores full marks: The candidate gives the value of the digit, not its place name alone and not the digit itself. Counting columns from the right — units, tens, hundreds, thousands, ten-thousands — puts the $7$ in the ten-thousands column.
Question 2
Paper 1H/2H style — 2 marks2 marks
(a) Write down two integers whose product is $-12$ and whose sum is $1$ . [1]

(b) Write down two integers whose product is $20$ and whose sum is $-9$ . [1]
Model answer
(a) $4$ and $-3$

Check: $4 \times (-3) = -12$ , $\ 4 + (-3) = 1$

(b) $-4$ and $-5$

Check: $(-4) \times (-5) = 20$ , $\ -4 + (-5) = -9$
Why this scores
Mark scheme:
- B1 for $4$ and $-3$
- B1 for $-4$ and $-5$
Why this scores full marks: The candidate uses the signs to narrow the search. A negative product needs one positive and one negative integer. A positive product with a negative sum needs both integers negative. Each pair is then checked against both conditions.
Question 3
Paper 1H/2H style — 2 marks2 marks
(a) Write down the even multiple of $9$ that lies between $50$ and $70$ . [1]

(b) Write down all the factors of $45$ . [1]
Model answer
(a) Multiples of $9$ between $50$ and $70$ : $54, 63$ . The even one is $54$ .

(b) $1 \times 45, \ 3 \times 15, \ 5 \times 9$

Factors: $1, 3, 5, 9, 15, 45$
Why this scores
Mark scheme:
- B1 for $54$
- B1 for $1, 3, 5, 9, 15, 45$ and no others
Why this scores full marks: The factors are found in pairs, working up from $1$ until the pairs meet, so none is missed. Weaker scripts leave out $1$ and $45$ , or include a multiple such as $90$ .
Question 4
Paper 1H/2H style — 3 marks3 marks
(a) Find the highest common factor (HCF) of $126$ and $294$ . Show your working clearly. [2]

(b) Hence write $\dfrac{126}{294}$ in its simplest form. [1]
Model answer
(a) $126 = 2 \times 3^{2} \times 7$

$294 = 2 \times 3 \times 7^{2}$

$\text{HCF} = 2 \times 3 \times 7 = 42$

(b) $\frac{126}{294} = \frac{126 \div 42}{294 \div 42} = \frac{3}{7}$
Why this scores
Mark scheme:
- M1 for both numbers as products of prime factors, or a list of factors of each including $42$
- A1 for $42$ cao
- B1 for $\frac{3}{7}$ , ft their HCF only if it gives a fraction in its lowest terms
Why this scores full marks: The prime factor forms make the HCF visible, and the word Hence is obeyed — dividing top and bottom by the HCF reaches the simplest form in one step. A correct HCF with no working scores nothing in (a).
Question 5
Paper 1H/2H style — 3 marks3 marks
Two gear wheels turn together. Wheel A has $18$ teeth. Wheel B has $45$ teeth.

A tooth on wheel A and a tooth on wheel B are marked, and the marked teeth are touching.

Work out the least number of complete turns each wheel makes before the marked teeth touch again. [3]
Model answer
$18 = 2 \times 3^{2}, \qquad 45 = 3^{2} \times 5$

$\text{LCM} = 2 \times 3^{2} \times 5 = 90 \text{ teeth}$

Wheel A: $90 \div 18 = 5$ turns

Wheel B: $90 \div 45 = 2$ turns
Why this scores
Mark scheme:
- M1 for $18$ and $45$ as products of prime factors, or multiples of each listed to $90$
- A1 for $90$
- A1 for wheel A $5$ turns and wheel B $2$ turns
Why this scores full marks: The candidate sees that the marked teeth meet again when the same number of teeth has passed on each wheel — a common multiple of $18$ and $45$ — and then converts teeth into turns for each wheel. Stopping at $90$ answers a different question.
Question 6
Paper 1H/2H style — 3 marks3 marks
A submarine is at a depth of $240$ m below sea level. It rises at $15$ m per minute for $8$ minutes. It then dives a further $95$ m.

Work out the final position of the submarine. [3]
Model answer
Start: $-240$ m

Rise: $15 \times 8 = 120$ m

$-240 + 120 = -120$

$-120 - 95 = -215$

The submarine is $215$ m below sea level.
Why this scores
Mark scheme:
- M1 for $15 \times 8$ or $120$ seen
- M1 for $-240 + 120 - 95$ oe
- A1 for $215$ m below sea level or $-215$ m
Why this scores full marks: Positions are written as directed numbers with sea level as $0$ , so a rise is added and a dive is subtracted. The final answer states both the size and the direction.
Question 7
Paper 1H/2H style — 4 marks4 marks
$N = 2^{3} \times 3^{4} \times 7$

(a) Explain why $N$ is a multiple of $56$ . [1]

(b) Find the smallest integer greater than $1$ that is not a factor of $N$ . [1]

(c) Write $6N$ as a product of prime factors in index form. [1]

(d) Write $N^{2}$ as a product of prime factors in index form. [1]
Model answer
(a) $56 = 2^{3} \times 7$ , and $N = (2^{3} \times 7) \times 3^{4}$ , so $N = 56 \times 81$ .

(b) $2, 3, 4$ are factors of $N$ . $5$ is prime and does not appear in $N$ , so $5$ is not a factor.

Answer: $5$

(c) $6N = 2 \times 3 \times 2^{3} \times 3^{4} \times 7 = 2^{4} \times 3^{5} \times 7$

(d) $N^{2} = 2^{6} \times 3^{8} \times 7^{2}$
Why this scores
Mark scheme:
- B1 for $56 = 2^{3} \times 7$ and a statement that these factors are in $N$ , or $N = 56 \times 81$
- B1 for $5$
- B1 for $2^{4} \times 3^{5} \times 7$
- B1 for $2^{6} \times 3^{8} \times 7^{2}$
Why this scores full marks: Every part is answered from the prime factor form without evaluating $N$ . In (d) the candidate doubles every index — weaker scripts square only some of the powers, or add $2$ to each index instead of doubling it.
Question 8
Paper 1H/2H style — 4 marks4 marks
(a) Write $7056$ as a product of its prime factors. Give your answer in index form. Show your working clearly. [2]

(b) Hence find $\sqrt{7056}$ . [1]

(c) Hence write $\sqrt{7056}$ as a product of its prime factors. [1]
Model answer
(a) $7056 \div 2 = 3528$ , $\ 3528 \div 2 = 1764$ , $\ 1764 \div 2 = 882$ , $\ 882 \div 2 = 441$

$441 \div 3 = 147$ , $\ 147 \div 3 = 49$ , $\ 49 \div 7 = 7$

$7056 = 2^{4} \times 3^{2} \times 7^{2}$

(b) Halve each index: $\sqrt{7056} = 2^{2} \times 3 \times 7 = 84$

(c) $84 = 2^{2} \times 3 \times 7$
Why this scores
Mark scheme:
- M1 for at least two correct steps of repeated division or a factor tree
- A1 for $2^{4} \times 3^{2} \times 7^{2}$
- B1 for $84$ from the prime factor form (a calculator value with no link to (a) does not follow Hence)
- B1 for $2^{2} \times 3 \times 7$
Why this scores full marks: Every index in (a) is even, which signals a square number. The square root halves each index, and the candidate shows that step so the Hence is honoured.
Question 9
Paper 1H/2H style — 4 marks4 marks
Find the largest integer that divides into $249$ and into $405$ , leaving a remainder of $9$ each time. Show your working clearly. [4]
Model answer
Remove the remainder from each number:

$249 - 9 = 240, \qquad 405 - 9 = 396$

$240 = 2^{4} \times 3 \times 5$

$396 = 2^{2} \times 3^{2} \times 11$

$\text{HCF} = 2^{2} \times 3 = 12$

Check: $249 = 20 \times 12 + 9$ and $405 = 33 \times 12 + 9$ . Also $12 > 9$ , so a remainder of $9$ is possible.

Answer: $12$
Why this scores
Mark scheme:
- M1 for $240$ and $396$ seen
- M1 for $240$ and $396$ both written as products of prime factors, or lists of factors of each
- M1 for $2^{2} \times 3$ oe, or a common factor of $240$ and $396$ greater than $9$ identified
- A1 for $12$ cao
Why this scores full marks: The key step is subtracting the remainder, which turns the problem into an ordinary HCF. The candidate then checks both remainders and that the divisor is larger than $9$ — a divisor of $9$ or less could never leave a remainder of $9$ .
Question 10
Paper 1H/2H style — 5 marks5 marks
$A = 2^{m} \times 3 \times 5^{2}$ and $B = 2^{4} \times 3^{n} \times 7$ , where $m$ and $n$ are positive integers.

The HCF of $A$ and $B$ is $24$ . The LCM of $A$ and $B$ is $25\,200$ .

(a) Find the value of $m$ and the value of $n$ . [4]

(b) Work out the value of $A$ and the value of $B$ . [1]
Model answer
(a) $24 = 2^{3} \times 3$

$25\,200 = 2^{4} \times 3^{2} \times 5^{2} \times 7$

HCF uses the lower power of $2$ : $2^{3}$ is lower than $2^{4}$ , so $m = 3$ .

LCM uses the higher power of $3$ : $3^{2}$ is higher than $3^{1}$ in $A$ , so $n = 2$ .

$m = 3, \ n = 2$

(b) $A = 2^{3} \times 3 \times 5^{2} = 600$

$B = 2^{4} \times 3^{2} \times 7 = 1008$
Why this scores
Mark scheme:
- M1 for $24 = 2^{3} \times 3$
- M1 for $25\,200 = 2^{4} \times 3^{2} \times 5^{2} \times 7$
- A1 for $m = 3$
- A1 for $n = 2$
- B1 for $A = 600$ and $B = 1008$ , ft their $m$ and $n$
Why this scores full marks: The candidate uses the HCF to fix the power of $2$ (it cannot come from the LCM, which only shows $2^{4}$ from $B$ ) and the LCM to fix the power of $3$ . A quick check confirms the HCF of $600$ and $1008$ is $24$ .

4MA1 Integers

At a glance

What you’ll learn

Types of integers — what each one is

Prime factorisation — the foundation for HCF and LCM

HCF and LCM — the rules and identity

Divisibility tests — quick checks

Working with negative integers

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Write a number as a product of prime factors in index form

2Find the HCF and LCM from prime factorisations

3Use the HCF × LCM identity

4Classify numbers as prime, square or cube

5Use divisibility rules to find a missing digit

6Order of operations with negative integers

7Real-life LCM word problem

8Height above and depth below sea level [1 mark]

9A common multiple within a range [3 marks]

10Smallest multiplier for a square that is also a multiple of 7 [5 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10