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Why do this problem?

At the heart of this problem is using a representation which helps to simplify the situation. For example, in the first part, using the fact that any number is either odd or even reveals the structure. The context and argument can be based on modulo arithmetic. Working from specific cases in order to generalise is a mathematical technique that can also be highlighted.


Possible approach


This printable worksheet may be useful: Differences.

You might like to look at the problem Make 37 first.
 
Ask the group to give you three numbers and write them on the board, in a second column write their three differences and in the third column the product.
Repeat several times until you are ready to discuss any patterns and relationships the group have noticed. Particularly that the product is always even.
  • Can they find three numbers where this is not the case?
  • Why not?
Discuss the partitioning of integers into odds and even numbers. For example:
All numbers are either even (E)
XX...XXXXXXXX
XX...XXXXXXXX
or odd (O)
XX...XXX
XX...XXXX
  • For the product not to be even what can you say about the differences?
  • What does this mean about the original three numbers?

Now move on to the second part of the problem. Some help with describing the three types of number related to multiples of three might be needed (see the notes to the problem Take Three from Five ).
 

Key questions

  • Why are there only three and six differences in the lists? (because $(a-b)$ is numerically equal to $(b-a) etc$)
  • How can we describe all numbers in terms of muliples of $3$, or $4$ or $5$ ...?
  • For the product not to be a multiple of three what can you say about the differences?

 

Possible support

Make 37 might be good to try first.

 

Possible extension

How many integers do you need to ensure that the product of all the differences is divisible by $5$?

Some students may go on to investigate this context more thoroughly, including posing and pursuing their own questions. For example: What about divisibility by $4$ and $6$, and then more generally?

Odd Stones and Take Three from Five might provide suitable follow-up problems.


 

You may also like

Adding All Nine

Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some other possibilities for yourself!

Counting Factors

Is there an efficient way to work out how many factors a large number has?

Repeaters

Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.

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The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice.

NRICH is part of the family of activities in the Millennium Mathematics Project.

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