International Baccalaureate · MYP 1

Number Properties

⌚ ~106 min ▢ 11 blocks ✓ 50 exercises ➣ Prerequisites : Whole Numbers and Operations, Factors, multiples and primes, Squares, square roots and exponents

I Zero and One

Zero $(0)$ and one $(1)$ are numbers with specific properties.

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Proposition — Additive Identity Property

Adding $0$ to any number results in the number itself.
For any number $a$, $$0 + a = a \quad \text{and} \quad a + 0 = a.$$

Example

$7 + 0 = 7$

Proposition — Multiplicative Identity Property

Multiplying any number by $1$ results in the number itself.
For any number $a$, $$1 \times a = a \quad \text{and} \quad a \times 1 = a.$$

Example

$1 \times 7 = 7$

Multiplying by $0$:
- Any number multiplied by $0$ always results in $0$.
- For example, $5 \times 0 = 0 + 0 + 0 + 0 + 0 = 0$.
Dividing by $0$:
- Dividing a non-zero number by $0$ leads to a contradiction.
- For example, if we assume $a = 5 \div 0$, then $a \times 0 = 5$. But this is impossible because $a \times 0$ is always $0$, not $5$.
- The case $0 \div 0$ is different: there is no contradiction, since $a \times 0 = 0$ holds for every number $a$ — but then no number stands out, there is no unique quotient. Either way, dividing by $0$ makes no sense.

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Proposition — Multiplying by $0$

For any number $a$, $$a \times 0 = 0 \quad \text{and} \quad 0 \times a = 0.$$

Proposition — Dividing by $0$

For any number $a$, the division $a \div 0$ is undefined.

Skills to practice

Applying the Properties of 0 and 1

II The Tests for 4 and 6

Proposition — Divisibility Tests for 3 and 9

A whole number is divisible by $3$ exactly when the sum of its digits is divisible by $3$.
A whole number is divisible by $9$ exactly when the sum of its digits is divisible by $9$.

Proposition — Divisibility Criterion for 4

A whole number is divisible by $4$ exactly when the number formed by its last two digits is divisible by $4$ (for a number with a single digit, this is the number itself).

Example

Determine whether $917$ is divisible by $4$, then whether $1\,316$ is.

Answer

$917$ is not divisible by $4$ because the number formed by its last two digits, $17$, is not divisible by $4$ ($17 = 4 \times 4 + 1$). $1\,316$ is divisible by $4$ because the number formed by its last two digits, $16$, is divisible by $4$ ($16 = 4 \times 4$). Check: $1\,316 = 4 \times 329$.

Proposition — Divisibility Criterion for 6

A whole number is divisible by $6$ exactly when it is divisible both by $2$ and by $3$. Indeed $6 = 2 \times 3$: every multiple of $6$ is even and a multiple of $3$. The converse is accepted without proof: a number that is both even and a multiple of $3$ is always a multiple of $6$.

Example

Determine whether $8\,514$ is divisible by $6$.

Answer

Both conditions must be checked.

By $2$: the last digit is $4$, which is even — so $8\,514$ is divisible by $2$.
By $3$: the sum of the digits is $8 + 5 + 1 + 4 = 18$, a multiple of $3$ — so $8\,514$ is divisible by $3$.

Both hold, so $8\,514$ is divisible by $6$. Check: $8\,514 = 6 \times 1\,419$.

Example

Determine whether $2\,346$ is divisible by $6$, then whether $2\,344$ is.

Answer

For $2\,346$: the last digit $6$ is even, and $2 + 3 + 4 + 6 = 15$ is a multiple of $3$. Both conditions hold, so $2\,346$ is divisible by $6$ ($2\,346 = 6 \times 391$). For $2\,344$: the last digit $4$ is even, but $2 + 3 + 4 + 4 = 13$ is not a multiple of $3$. One condition failing is enough: $2\,344$ is not divisible by $6$.

Skills to practice

Determining Divisibility by 4
Determining Divisibility by 6
Applying the Tests for 4 and 6

III Factors and Factor Pairs

Definition — Divisibility Relationships

We say that a non-zero natural number $b$ divides a natural number $a$ if $a$ can be obtained by multiplying $b$ by a natural number $k$: $$ a = k \times b $$ In other words, the number $a$ appears in the multiplication table of $b$.
We can also use the following formulations:

$b$ is a divisor of $a$
$b$ is a factor of $a$
$a$ is divisible by $b$
$a$ is a multiple of $b$

The numbers $0$ and $1$ play special roles in divisibility:

$1$ divides every whole number: $n = n \times 1$;
every non-zero whole number divides $0$: $0 = n \times 0$;
this is why the definition requires $b \neq 0$: dividing by $0$ is never allowed.

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Method — Finding All the Factors by Pairs

Look for factors in pairs: for a non-zero whole number $n$, each factor $d$ comes with its partner $n \div d$.
Start from $1$ and test $2, 3, 4, \dots$ in order, writing each pair found.
Stop when the pairs cross: once the candidate $d$ satisfies $d \times d > n$, every positive factor has been found — beyond that point, each factor has already turned up as the partner of a smaller one. If a pair has two equal members ($d \times d = n$), that factor counts only once.

Example

Find all the factors of $60$.

Answer

Test the candidates in order, in pairs: $$1 \times 60 \qquad 2 \times 30 \qquad 3 \times 20 \qquad 4 \times 15 \qquad 5 \times 12 \qquad 6 \times 10$$ The candidate $7$ does not divide $60$ ($60 = 7 \times 8 + 4$). The next candidate is $8$, and $8 \times 8 = 64 > 60$: stop, the pairs have crossed. $$\text{Factors of } 60: \; 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 \quad (\text{12 factors}).$$

Skills to practice

Recognising All the Factors of a Number
Listing All the Factors by Pairs

IV Testing Whether a Number Is Prime

Definition — Prime Number

A prime number is a whole number greater than $1$ that has exactly two different divisors: $1$ and itself.
A whole number greater than $1$ that is not prime is called a composite number.

Method — Testing whether a Number is Prime

To decide whether a whole number $n$ greater than $1$ is prime:

Check the quick criteria: if $n$ is even, or if its digit sum is a multiple of $3$, or if it ends in $0$ or $5$, then $n$ is composite (unless $n$ is $2$, $3$ or $5$ itself).
Otherwise, test the next prime numbers one after the other ($7$, $11$, $13$, …): divide $n$ by each one.
Stop as soon as the prime $p$ to test satisfies $p \times p > n$: if no prime before that point divides $n$, then $n$ is prime. Indeed, if $n$ is composite it can be written $n = a \times b$ with $1 < a \leq b$, so $a \times a \leq n$. And the smallest divisor of $n$ above $1$ is necessarily prime: were it not, it would itself have a smaller divisor, which would divide $n$. So a composite number always has a prime factor $p$ with $p \times p \leq n$.

Example

Is $101$ a prime number?

Answer

$101$ is odd, its digit sum $1 + 0 + 1 = 2$ is not a multiple of $3$, and it does not end in $0$ or $5$: neither $2$, $3$ nor $5$ divides it. Testing the next primes: $101 = 7 \times 14 + 3$, so $7$ does not divide it. The next prime is $11$, and $11 \times 11 = 121 > 101$: we stop. No prime up to the stopping point divides $101$: the number $101$ is prime.

Skills to practice

Deciding Whether a Number Is Prime

V The Highest Common Factor

Definition — Common Divisor

A number is a common divisor of two numbers when it divides both. To find them all: write the divisors of the first number, write those of the second, and keep the ones that appear in both lists.

Definition — Highest Common Factor (HCF)

The highest common factor (HCF) of two (or more) whole numbers, not all zero, is the largest positive whole number that is a factor of each of them (it divides each of them exactly, with remainder $0$). It is also called the greatest common divisor (GCD).
We usually write the HCF of $a$ and $b$ as $\operatorname{HCF}(a, b)$.

Example

Find the HCF of $18$ and $24$.

Answer

The factors of $18$ are: $1, 2, 3, \textcolor{olive}{6}, 9, 18$.
The factors of $24$ are: $1, 2, 3, 4, \textcolor{olive}{6}, 8, 12, 24$.
The highest common factor is $\textcolor{olive}{6}$, so $\operatorname{HCF}(18, 24) = 6$.

Skills to practice

Finding the Highest Common Factor by Listing
Solving Real-Life Problems with the Highest Common Factor

VI Going further

Skills to practice

Investigating Patterns in Divisibility
Solving Problems with Factors and Multiples
Quiz