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14 Divisors

Stage: 3 Challenge Level: Challenge Level:2 Challenge Level:2

Why do this problem ?

This problem allows students to explore prime factorisation.

Possible approach

Ask for the factors of $12$. Ask for a number with more factors, and list them. Ask for a number with fewer factors. Ask for a bigger number with fewer factors, etc, etc, until the group has had enough revision of techniques to list factors systematically omitting none.
Challenge pupils to see who'll be first to find a number with exactly fourteen factors - but ensure they know that intermediate results should also be clear in their books, in case they're useful later.
Part-way through the lesson, get a progress report, ask pupils to describe useful approaches (without giving too much away).
Once a solution has been found, refine the challenge - to find the SMALLEST number with fourteen factors, and to provide a convincing argument that it is the smallest.

Key questions

Are there any more factors? How do you know?
"Big numbers have lots of factors" - discuss.
"That's odd, they all have an even numbers of factors - or do they?"

Possible extension

Find more numbers with fourteen factors, until you can describe/define ALL the numbers with fourteen factors.
Looking at the structure of prime factorisiation, and working with the primes and their powers as values to be varied may help pupils appreciate the role of prime factors and their powers.

Experiment until you can make an educated guess of - the modal number of factors for the numbers $1$-$10$, $1$-$100$, $501$-$600$, $a$-$b$.

Go and research the distribution of primes.

Possible support

Use ten rather than fourteen.
How many rectangles can be made with an area of $24$? - draw or cut out all the (integer) possibilities.
Shall we consider $4 \times 6$ the same as or different to $6 \times4$?
What other areas could we try? Everyone tries out one or two different areas - trying to get ALL the rectangles.
I want to find an area for which there are five different rectangles (or ten if we decide that $4 \times6$ is different to $6 \times4$).