Prime factorisation and divisibility

What is it about?

A prime number is a natural number with exactly two divisors: $1$ and itself. $2, 3, 5, 7, 11, 13, \ldots$ are primes. $1$ is not a prime, because it has only one divisor. $2$ is the only even prime. Numbers with more than two divisors are called composite.

Prime factorisation: Every natural number greater than $1$ can be written as a product of primes. In a factor tree you keep splitting until only primes are left:

360 = 18 \cdot 20 = 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 5 = 2^3 \cdot 3^2 \cdot 5

However you start the tree, you always end with the same prime factors. The prime factorisation is unique (apart from the order).

Divisibility rules help you find divisors quickly:

by $2$ , $5$ or $10$ : look at the last digit.
by $4$ or $25$ : look at the number formed by the last two digits.
by $3$ or $9$ : check the digit sum, e.g. $3 + 6 + 0 = 9$ .
by $6$ : the number is divisible by $2$ and by $3$ .

Sieve of Eratosthenes: Cross out $1$ . The smallest number not crossed out is a prime; circle it and cross out all its multiples. For a prime $p$ , the first multiple that gets newly crossed out is $p \cdot p$ , e.g. $49$ for $7$ – smaller multiples are already crossed out. Up to $100$ the primes $2, 3, 5$ and $7$ are enough, because $11 \cdot 11 = 121$ is already greater than $100$ .

gcd and lcm: The greatest common divisor is the product of the common prime factors. The least common multiple is the product of all prime factors, where common ones are counted only once. With $24 = 2 \cdot 2 \cdot 2 \cdot 3$ and $36 = 2 \cdot 2 \cdot 3 \cdot 3$ :

\gcd(24, 36) = 2 \cdot 2 \cdot 3 = 12 \qquad \operatorname{lcm}(24, 36) = 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 = 72

Try it

Split 360 step by step: tap the dashed numbers or press Split.
Press Another tree several times. What changes, and what always stays the same?
Always split off the smallest prime factor. What does the tree look like now?
Split 512 and write the result as a power.
Tap a number in the tree: next to it (below it on a phone) you see which numbers divide it. Try 783 – the digit sum tells you a lot.
Run the sieve up to 100 with Next prime. Tap a crossed-out number: which prime crossed it out?
Arrange the numbers in 6 columns. In which columns are the primes from $5$ on? Why?
Find gcd and lcm of 24 and 36 with Sort the factors. What happens with 8 and 15?

Tasks

Task 1: Prime factorisation

Load task Split $84$ , $126$ and $500$ into prime factors and write them with powers.

Show solution

$84 = 2^2 \cdot 3 \cdot 7$ , $126 = 2 \cdot 3^2 \cdot 7$ and $500 = 2^2 \cdot 5^3$ . Show 126 · Show 500

Task 2: Divisibility rules

Check without a calculator whether $852$ is divisible by $2$ , $3$ , $4$ , $6$ and $9$ .

Show solution

The last digit $2$ is even: divisible by $2$ . The digit sum $8 + 5 + 2 = 15$ is divisible by $3$ , but not by $9$ . $52$ is divisible by $4$ , so $852$ is too. Divisible by $2$ and by $3$ means divisible by $6$ . $852$ is only not divisible by $9$ . Check

Task 3: Finding primes

Which primes lie between $50$ and $70$ ?

Show solution

$53$ , $59$ , $61$ and $67$ . All other numbers in between are divisible by $2$ , $3$ , $5$ or $7$ , e.g. $51 = 3 \cdot 17$ and $57 = 3 \cdot 19$ . Show the sieve

Task 4: gcd and lcm

Load task Find the gcd and lcm of $60$ and $84$ using their prime factors.

Show solution

$60 = 2 \cdot 2 \cdot 3 \cdot 5$ and $84 = 2 \cdot 2 \cdot 3 \cdot 7$ . The common factors are $2 \cdot 2 \cdot 3$ , so $\gcd(60, 84) = 12$ . All factors: $\operatorname{lcm}(60, 84) = 2 \cdot 2 \cdot 3 \cdot 5 \cdot 7 = 420$ .

Task 5: Buses

Two buses leave at the same time at $8$ o’clock. Bus A leaves every $12$ minutes, bus B every $18$ minutes. When do they next leave at the same time?

Show solution

You need the least common multiple: $\operatorname{lcm}(12, 18) = 36$ . The buses leave together again at $8{:}36$ . Show in the simulation

Task 6: Why is 7 enough?

Explain why you can stop after the multiples of $7$ when sieving up to $100$ .

Show solution

A composite number up to $100$ is a product $a \cdot b$ with $a \le b$ . If $a$ were greater than $10$ , then $a \cdot b > 10 \cdot 10 = 100$ . So every composite number up to $100$ has a divisor up to $10$ and therefore a prime factor $2$ , $3$ , $5$ or $7$ – it has already been crossed out.

Ideas for the classroom

Use: About grade 5 (prime factorisation and divisibility rules). gcd and lcm are needed later for simplifying fractions and finding common denominators (year 6).
Three views: factor tree with divisibility rules, sieve of Eratosthenes and gcd/lcm in a set diagram. Each prime keeps the same colour in all views.
Step by step: If “Show the result right away” is off (show=0), tree, sieve and set diagram start empty and are built up with the buttons or by tapping.
Uniqueness: “Another tree” splits with random divisors. That the same prime factors always appear is a good reason to ask why (without proof in year 5).
Common stumbling blocks:
- “ $1$ is a prime.”
- “Odd numbers are primes.” – $9$ , $15$ , $51 = 3 \cdot 17$ and $91 = 7 \cdot 13$ are not.
- The digit-sum rule is applied to $4$ .
- gcd and lcm are mixed up; the set diagram shows: the intersection gives the gcd, everything together the lcm.