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Bob's Buttons

Lesson10
Thank you for visiting the Classroom Contributions page. This is where we collect stories about this lesson from the Maths300 community. We are grateful to the teachers, schools and students below for taking time to share their excitement.

If you have been using this lesson in your classroom, we would be delighted to hear your story. Please send your comments, student comments, photographs (with permission if people can be identified) or sample student work to our web site contact person.

Buckley Park Secondary College

Andrew Dunstall, Year 7
I thought you might like to see some student work done on one of my favourites, Bob's Buttons. Juliana's work is typical of a number of my students' work, but has the advantage of being an MS-Word document.

I have also attached the sheets I wrote to accompany Bob's Buttons. I basically do the first lesson as described in the Maths300 introduction - kids forming into groups and remainders - then, I did this last year, this was the group size and remainder, what was the class size?. Then:

  • Sheet 1 in pairs with counters, followed by
  • Sheet 2 in pairs, usually without counters, and finally
  • Sheet 3 completed individually mainly as a homework assignment.

I like Bob's Buttons because:

  1. At first glance it seems really simple, then it becomes infuriatingly tricky.
  2. All kids can engage it.
  3. There is something about the thinking required to solve each problem that is different to the usual thinking we do in maths class. Not very well expressed but if you do the activity you will get what I mean.
  4. The activity is not algebra, not arithmetic, I not sure what it is, but it does make them think and its definitely maths. Seriously, the patterns that emerge from the table are really powerful and the way kids express the patterns they see often demonstrates some real thinking.
  5. The tables on my sheets and the conversation around them, I think, demonstrate a very logical, step by step, approach to discovery in maths and science.

Alby Gymnasium, Sweden

Mahmoud M. Sarabi, Year 8
When we set Option 3 of the software to test Groups of 4 with remainder 2 and Groups of 2 with remainder 1, it was quite stunning. It just kept on checking endlessly without finding a solution. We had to ask why.

An algebraic explanation is:

  • Groups of 4 with 2 remainder can be represented by G14 + 2.
  • Groups of 2 with 1 remainder can be represented by G22 + 1.
  • The software is searching for a situation where:
    G14 + 2 = G22 + 1
  • But G1 = 2G2, so
    8G2 + 2 = 2G2 + 1
  • which implies:
    6G2 = -1
But there are no whole number solutions of this for G2, so the software will never find a solution.

Murrayville Community College

Tracey Willersdorf, Year 7
We also found Option 3 of the software using Groups of 4 with remainder 2 and Groups of 2 with remainder 1 to be fascinating. I appreciate Mahmoud's algebraic explanation because it may have broad application, but in this case, a simpler explanation for some students may be that a solution can never be found because:
  • Groups of 4 with remainder 2 will always give even numbers, and
  • Groups of 2 with remainder 1 will always give odd numbers.

McAuley School, England

V. Lynch
A solution can be obtained using the Chinese Remainder Theorem and teachers might like to explore this with senior students.

Let x be the number of buttons that satisfies both conditions simultaneously. Then:

x congruent a [mod n] AND x congruent b [mod m], where congruent means 'is congruent to'.

Find:
the inverse of n [mod m] so that nn-1 congruent 1 [mod m] and
the inverse of m [mod n] so that mm-1 congruent 1 [mod n].

Then:
x congruent amm-1 + bnn-1 [mod mn]

Check:
Using mod n arithmetic, x congruent a1 + b0 = a
Using mod m arithmetic, x congruent a0 + b1 = b

In the particular case of the original problem, a = 2, n = 4 and b = 1, m = 5.
The inverse of 4 [mod 5] is 4 because 4 x 4 = 16 congruent 1, so n-1 = 4.
The inverse of 5 [mod 4] is 1 because 5 x 1 = 5 congruent 1, so m-1 = 1.

Modulo 4 & 5

Substituting in the general rule gives:
x congruent 251 + 144 [mod 20]
   congruent 10 + 16 [mod 20]
   congruent 6 [mod 20]

Which implies that:
x congruent 20k + 6 where k is integral, which is the known result for the original problem.

 

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