Highest common factor

Key wordsLearning outcomes

Rounding in context

Problemsolving

Numbersless than zero

Commonmultiples

Prime numbers

Rounding in context

A railway carriage can seat 40 passengers.

The 16:30 train from Glasgow to Ayr normally leaves Central Station with around 330 passengers on board.

What is the least number of carriages the train should have so that every passenger has a seat?

The number of carriages needed = 330 ÷ 40 = 33 ÷ 4 = = 8·25

So, 9 carriages will be needed.

Because of the context, we round up.

8 14

Problem solving

This plane can carry 167 passengers.

On one flight, seats 1 to 36, 38 to 95 and 104 to 167 were occupied.

How many passengers were on board?

By counting all the unoccupied seats: Seat 37: 1Seat 96 to 103: 103 – 96 + 1 = 8Total unoccupied: 1 + 8 = 9So total occupied: 167 – 9 = 158 passengers.

There are two different ways to solve this problem.

By counting all the occupied seats: 1 to 36: 36 38 to 95: 95 – 38 + 1 = 58104 to 167: 167 – 104 + 1 = 64Totalling: 36 + 58 + 64 = 158 passengers.

When objects are numbered, we can determine how many

objects there are by subtractingthe low number from the

high number then adding 1.

When objects are numbered, we can determine how many

objects there are by subtractingthe low number from the

high number then adding 1.

The table shows the monthly profit made over a five - month period, by an ice-cream manufacturer.

Find the overall profit or loss for the five months.

Find the sum of the losses (i.e. the negative profits):-240 + (-70) + (-80) = -390 = -£390

and the sum of the profits:350 + 480 = £830

The overall profit or loss : £830 + (- £390) = £830 - £390 = £440

i.e. a profit of £440.

Month May June J uly August September

Profit (£) –240 –70 350 480 –80

Numbers less than zero

A red bus leaves the bus station every eight minutes, a green bus leaves every twelve minutes, and a blue bus every eighteen minutes.

If all three colours of bus leave together at 8.06 a.m., when is the next time all three colours of bus leave together?

We need to find the lowest common multiple (L.C.M.) of 8, 12 and 18.

Start with the multiples of the largest number:18: 18, 36, 54, 72, 90, 108, …12: 12, 24, 36, 48, 60, 72, 84, 96, 108, … 8: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, …

72 is the lowest common multiple.

So 72 minutes after 8.06 a.m. is the next time the three buses leave together.That is 1 hour 12 minutes after 8.06 a.m.i.e. at 9.18 a.m.

Common multiples

Find the highest common factor (H.C.F.) of (a) 36 and 90.(b) 6x and 2x2

(a) Start with the bigger number:

The factors of 90: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90.The factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36.

The common factors are 1, 2, 3, 6, 9, 18.So the highest common factor (H.C.F.) is 18.

(b) The factors of 6x: 1, 2, 3, 6, x, 2x, 3x, 6x The factors of 2x2: 1, 2, x, 2x, x2, 2x2

The common factors are 1, 2, x, 2x

So the highest common factor (H.C.F.) is 2x.

Highest common factor

The age of Sara’s grandfather is a prime number. When he was a year younger, 6 was a factor of his age.When he is a year older, 8 will be a factor of his age.

Find the age of Sara’s grandfather. A prime number is a

whole number that has exactly two

factors, itself and 1.

When 6 was a factor, his age was ... 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96, 102, …

When 8 will be a factor, his age will be … 40, 48, 56, 64, 72, 80, 88, 96, 104, …

Look for numbers that are 2 apart …

These point to three possibilities to consider: 55, 79 and 103.

However, grandfather’s age is a prime number, so he cannot be 55. It is most likely that he is 79, since 103 is a very old age.

(55 is not a prime number, it is divisible by 5. 79 and 103 are prime, since each number is divisible only by 1 and itself).

Prime numbers

Keywords Note

Negative number … A number less than zero.

Multiple … Multiples of 6 are 0, 6, 12, 18, 24, … because each is formed by multiplying a whole number by 6.

Common multiple … 12 is a common multiple of 4 and 6 being a multiple of 4 (0, 4, 8, 12, 16, …)and a multiple of 6 (0, 6, 12, 18, …).

Factor … The factors of 16 are 1, 2, 4, 8 and 16 because they divide 16 without remainder.

Common factor … A common factor of two numbers will divide both numbers without remainder.

Prime number … A prime number has exactly two factors, itself and 1.The first five prime numbers are 2, 3, 5, 7, 11.

Learning outcomes

• I can round a number taking into account the context of the problem.

• I can use a variety of methods to solve number problems in familiar contexts.

• I can recall number facts quickly and use them accurately when making calculations.

• I can use my understanding of numbers less than zero to solve simple problems in context.

• I can determine common multiples and common factors and can apply my understanding to solve problems.

• I know how to identify if a number is prime.

• I can communicate my reasoning when solving problems.