Finding Common Ground · Lesson 1 of 7
Common Factors and the HCF
“Discover how the largest shared factor solves equal-size grouping problems.”
• List factors and identify factors shared by two numbers. • Explain HCF as the greatest possible common whole-number group size. • Use a diagram to connect tile size and number of tiles. • Recognise HCF in packing and equal-jump situations.
The largest square that fits
Imagine covering a 12 ft by 16 ft rectangular floor with identical square tiles. A tile side must fit along both edges without leaving a strip uncovered. If its length must be a whole number of feet, it must divide both 12 and 16 exactly. Start with possible sides before deciding which is best.
The whole-number factors of 12 are 1, 2, 3, 4, 6 and 12; those of 16 are 1, 2, 4, 8 and 16. Their common factors are 1, 2 and 4. The biggest square side is therefore 4 ft. It takes 12 such tiles, while 2 ft tiles would take 48. Would allowing fractional sides produce a larger square here? No: if k equal tiles span 12 ft and m span 16 ft, then 12/k = 16/m, so 3m = 4k. The smallest positive whole counts are k = 3 and m = 4, still giving a 4 ft side.
The greatest positive whole number that divides each of the given positive whole numbers exactly. It is also called the Greatest Common Divisor (GCD).
Problem
A floor measures 12 ft by 16 ft. Find the largest whole-number square tile and the number required.
- 1.List common factors of 12 and 16: 1, 2, 4.
- 2.Choose the greatest, 4, for the tile side.
- 3.There are 12 ÷ 4 = 3 tiles across and 16 ÷ 4 = 4 along, so 3 × 4 = 12 tiles.
Equal bags and equal jumps
A grouping question often hides the same condition. If each bag has the same whole-number weight and rice from two farms is kept separate, that weight must divide both farm totals. A larger bag means fewer bags, so the greatest common factor gives the requested size.
Problem
Pack 84 kg and 108 kg separately into equal whole-kilogram bags, using as few bags as possible.
- 1.Common factors of 84 and 108 are 1, 2, 3, 4, 6 and 12.
- 2.Choose 12 kg per bag, the greatest common factor.
- 3.The farms fill 84 ÷ 12 = 7 and 108 ÷ 12 = 9 bags, or 16 bags altogether.
The Jump Jackpot picture gives another interpretation. Starting at 0, jumps of size d land on both 14 and 30 only if both numbers are multiples of d. The longest such jump is their HCF. A pair such as 7 and 11 has no shared jump longer than 1.
Problem
What is the longest whole-number jump from 0 that lands exactly on both 28 and 42?
- 1.List the common factors: 1, 2, 7 and 14.
- 2.The longest possible jump is 14.
- 3.Two jumps land on 28 and three land on 42.
Choosing a smaller common tile side gives more tiles, not fewer. Always connect the phrase “fewest equal groups” to the largest permitted size.
Quiz
Which number is a common factor of 12 and 16?
What is the HCF of 14 and 30?
Why does a whole-number tile side have to divide both floor dimensions?
How many 4 ft by 4 ft tiles cover a 12 ft by 16 ft floor?
Two quantities are 7 and 11. What is their HCF?
Practice Problems
- Find all common factors and the HCF of 18 and 30.
- A 20 cm by 28 cm card is covered by largest possible whole-centimetre square pieces. Find their side and number.
- Pack 48 kg and 72 kg of grain separately in equal whole-kilogram bags, minimising the bag count.
- From 0, find the longest jump size that lands on both 30 and 50. Explain the factor connection.
- Why is 6 a possible bag weight for 84 kg and 108 kg, but not the best weight when we want the fewest bags?
Key Takeaways
• A common factor divides every given number exactly. • HCF is the greatest common factor; GCD is another name for it. • A largest whole-number tile or bag size often signals HCF. • Count groups after finding their size: divide each quantity by the HCF. • A jump size that reaches two numbers is a common factor of them.
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Next · Lesson 2
Prime Factorisation and All the Factors