PrepShorts · Teaching notes · Class 6 Mathematics · Chapter 5, Prime TimePrepShorts

Chapter 5 · Prime Time

Prime factorisation, and why there is only one of them per number

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What to assume they know

What they should be able to do

  • Explain why one factorisation of a number does not reveal all its factors
  • Break a composite number down repeatedly until only primes remain
  • State the stopping rule and say why a prime cannot be broken further
  • Use the words prime factorisation and prime factors correctly, and distinguish them
  • State what the book says about 1, and about a prime's own factorisation
  • Show that different routes into the same number end at the same collection of primes
  • Explain why the order of the primes is not part of the answer, and write a factorisation in increasing order by convention
  • Build a number's prime factorisation upward from the factorisations of two of its factors

Where it usually goes wrong

  • "I found a factorisation, so I know the factors." The exact error the section opens with. One factor pair tells you two factors and says nothing about the rest. Play the wrong reasoning out at full length before correcting it, or the correction lands on nobody.
  • "More factorisations means more certainty." Four ways of splitting 80 still prove nothing, because you cannot tell you have finished. Only breaking all the way down terminates.
  • "Different routes give different answers." The whole point of the 36 diagram. What can differ is the order the primes come out in and the amount of work; the collection itself cannot.
  • "2 × 2 × 2 × 7 has four prime factors." It has four factors written down and two prime factors, 2 and 7. The book makes this distinction explicitly for 56 and students routinely lose it.
  • "1 is a prime factor of everything, since everything is divisible by 1." The book blocks this on p.118 by saying 1 has no prime factorisation. If 1 were admitted, no factorisation would ever be finished.
  • "Increasing order is a rule, and writing 3 × 2 × 5 is wrong." It is a convention for tidiness. The order carries no information — that is section 9's entire argument — so a differently ordered answer is a correct answer.
  • "A prime cannot have a prime factorisation, since it cannot be broken." It has one, and it is itself. This is stated on p.118.

Questions to check understanding

  • Break a given two-, three- or four-digit number down into its primes
  • Name the prime factors of a number, as distinct from writing its factorisation
  • Reconstruct a number from a stated collection of prime factors
  • Factorise a product without multiplying it out first
  • Find the smallest number whose factorisation uses a stated number of different primes
  • Given three primes and their product, identify the primes
  • Explain why two different routes into the same number must agree
  • The appendix bound with this chapter answers the p.120 practice set that this topic ends on across footer pages 7 and 8; §5.4's answers as a whole run footer pages 7 to 9

Examples worth working on the board

  • The opening exchange (§5.4, p.117). A teacher asks whether 56 and 63 are co-prime. One child splits them as 56 = 14 × 4 and 63 = 21 × 3, sees no repeat among 14, 4, 21 and 3, and concludes they are co-prime. Another child splits them as 56 = 7 × 8 and 63 = 9 × 7, sees the 7, and concludes they are not. The second child is right. This is the motivating failure for the whole section and it deserves real attention — the first child's arithmetic is not wrong.
  • The second attempt (p.117). 80 = 40 × 2 = 20 × 4 = 10 × 8 = 16 × 5, with a trailing row of question marks; 63 = 9 × 7 = 3 × 21, likewise. Even with four factorisations of 80 in hand, no conclusion is safe, because the list is not known to be finished.
  • Breaking 56 down (p.118). The printed route: 56 = 4 × 14; then 14 = 2 × 7, giving 56 = 4 × 2 × 7; then 4 = 2 × 2, giving 56 = 2 × 2 × 2 × 7. The two primes appearing are 2 and 7. Note that 2 appears three times but is one prime factor, which is the distinction section 6 is built on.
  • The two edge cases (p.118). The book states that 1 has no prime factorisation at all, and that a prime's factorisation is just itself. Checked against the printed page. Both are needed and neither is a technicality: without the first, a factorisation could be padded with 1s for ever.
  • 63 by two routes (p.118). Written once as 3 × 3 × 7 and once as 3 × 7 × 3. The same two primes appear, with 3 twice and 7 once in both.
  • The four-route diagram for 36 (p.118). Checked against the printed page. A circle holding 36 branches to four boxes: 2×18, 3×12, 4×9 and 6×6. Each then descends. The first column runs 2×18 → 2×2×9 → 2×2×3×3. The second runs 3×12 → 3×3×4 → 3×3×2×2 → 2×2×3×3. The third runs 4×9 → 2×2×9 → 2×2×3×3. The fourth runs 6×6 → 2×3×6 → 2×3×2×3 → 2×3×2×3. Every route ends with two 2s and two 3s. See Notes on the fourth column's last box.
  • The block of thirty (p.119). Checked against the printed page. A cuboid of small cubes, three cubes wide, five deep and two tall, with the top face yellow, the front face red and the right face green. The reader is asked to use it to explain why 30 comes out as the same product whichever way the three numbers are multiplied. The three visible faces carry 3 × 5, 3 × 2 and 5 × 2 cubes respectively; counting the whole block layer by layer in three different directions is the argument, and the figure should count it all three ways.
  • The convention (p.119). Primes are normally written smallest first. The book's two printed illustrations are 225 as 3 × 3 × 5 × 5 and 30 as 2 × 3 × 5. It also names commutativity and associativity here and defers them.
  • Building upward (pp.119–120). 72 is first written as 12 × 6. The factorisations of the two pieces are put together, giving 72 as 2 × 2 × 3 × 2 × 3, which is where p.119 stops; p.120 then tidies it into increasing order and asks the reader to tally each prime in the answer against its tally across the two pieces, so that the two tallies can be seen to agree.
  • The practice set (p.120). Prime factorisations wanted for 64, 104, 105, 243, 320, 141, 1728, 729, 1024, 1331 and 1000. Then: the number whose factorisation is one 2, two 3s and one 11; three primes below 30 whose product is 1955; three products to factorise without multiplying out first, namely 56 × 25, 108 × 75 and 1000 × 81; and the smallest numbers whose factorisations use three, and then four, different primes.

Figures to have open

  • A branching diagram from a single number to four separate routes down to the primes. This is the topic's central figure and must reproduce the four routes the book prints for 36, so that the student's page and the figure agree.
  • The three-dimensional block of unit cubes, 3 by 5 by 2, with distinguishable faces. Redraw; the printed art on p.119 is the book's own. It must be countable in three directions.
  • A two-line merge diagram for section 11: the factorisations of two factors above, the merged and sorted result below. Standard schematic.
  • No photograph or textbook data table is needed.

Where this sits in the book

The book

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