Hard GCSE Factors, multiples and primes Questions

Find the HCF of $8$ and $12$

Start with the highest common factor.

Factors of $8$

List the factors of $8$.

Factors of $12$

List the factors of $12$.

Common factors

These appear in both lists.

HCF

The largest common factor is $4$.

What the HCF means

The HCF is the largest number that divides both $8$ and $12$.

When you use the HCF

Use the HCF for problems like the largest tile or longest equal piece.

Now find the LCM

Next find the lowest common multiple.

Multiples of $8$

List the multiples of $8$.

Multiples of $12$

List the multiples of $12$.

LCM

The smallest common multiple is $24$.

What the LCM means

The LCM is the smallest number that both $8$ and $12$ divide into.

When you use the LCM

Use the LCM for problems like bells ringing or lights flashing together again.

Compare their sizes

The HCF ($4$) is at most each number; the LCM ($24$) is at least each number.

State the difference

So the HCF ($4$) is for splitting into equal groups, and the LCM ($24$) is for events coinciding.

Decide HCF or LCM

LCM problems are about repeating events happening together again; HCF problems are about splitting into equal groups.

Choose an LCM context

Two lights flashing at different rates is a classic LCM setting.

Set the first rate

Light A flashes every $8$ seconds.

Set the second rate

Light B flashes every $12$ seconds.

State the starting point

They flash together at the start.

Write the question

Ask when they next flash at the same moment.

This needs the LCM

The joint flash happens at the lowest common multiple of $8$ and $12$.

Multiples of $8$

List the multiples of $8$.

Multiples of $12$

List the multiples of $12$.

LCM

$24$ is the smallest common multiple.

So the answer

They next flash together after $24$ seconds.

Check Light A

Light A has flashed $3$ times.

Check Light B

Light B has flashed $2$ times.

So the problem is solved by LCM

This is a valid real-life LCM problem with a clear answer.

State the problem and answer

So the designed problem gives $24$ seconds. (An HCF problem, such as cutting ribbons into equal pieces, would also be valid.)

The HCF is $8$

$8$ is the highest common factor of the two numbers.

So $8$ divides both numbers

Both numbers must be multiples of $8$.

The other number is a multiple of $8$

So the unknown number is a multiple of $8$.

Multiples of $8$ near the range

List the multiples of $8$ around $35$ to $45$.

Keep those between $35$ and $45$

Only $40$ is between $35$ and $45$ ($32$ is too small, $48$ too big).

So the other number is $40$

$40$ is the only candidate in the range.

Check the HCF of $24$ and $40$

We must confirm the HCF is really $8$, not more.

Factors of $24$

List the factors of $24$.

Factors of $40$

List the factors of $40$.

Common factors

These appear in both lists.

Highest common factor

The largest common factor is $8$, exactly as required.

So $40$ fits all the conditions

$40$ is a multiple of $8$, in range, and has HCF $8$ with $24$.

Why not $48$?

$48$ would give an HCF of $24$, not $8$, and is outside the range anyway.

So $40$ is the answer

$40$ is the only number that works.

State the answer

So the other number is $40$.

They ring together on common multiples

All three ring together at times that are multiples of $4$, $6$ and $10$.

Next together is the LCM

The next joint ring is the lowest common multiple of the three.

Start from multiples of $10$

The answer must be a multiple of $10$, so list those.

Keep the multiples of $4$

From those, keep the ones divisible by $4$.

Check which is a multiple of $6$

$20$ and $40$ are not multiples of $6$.

Check $60$

$60$ is a multiple of $6$.

So the LCM is $60$

$60$ is the smallest multiple of all three.

Check $4$

$60$ divides by $4$ exactly.

Check $6$

$60$ divides by $6$ exactly.

Check $10$

$60$ divides by $10$ exactly.

So all three divide $60$

$60$ works for all three bells.

Nothing smaller works

$30$ is a multiple of $6$ and $10$ but not $4$, so nothing below $60$ works.

Interpret

So they next ring together after $60$ minutes.

That is one hour

$60$ minutes is exactly $1$ hour.

State the answer

So the three bells next ring together after $60$ minutes.

Largest number dividing both is the HCF

The largest number that divides both $48$ and $72$ is their highest common factor.

Factors of $48$

List every factor of $48$.

Factors of $72$

List every factor of $72$.

Find the common factors

These appear in both lists.

HCF is the highest

The largest common factor is $24$.

Pick it

So the HCF is $24$.

Check $48$

$24$ divides $48$ exactly.

Check $72$

$24$ divides $72$ exactly.

Could anything bigger work?

The only bigger factor of $48$ is $48$ itself, but $48$ does not divide $72$.

Confirm $48$ fails

$72\div48$ is not a whole number, so $48$ does not divide both.

So $24$ is the largest

$24$ is the biggest number dividing both.

Interpret

$24$ divides both $48$ and $72$ exactly.

Restate

The largest number dividing both is $24$.

Sense check

Both numbers are whole-number multiples of $24$.

State the answer

So the largest number dividing both $48$ and $72$ is $24$.