Bob's Buttons ... Answers & Discussion

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Teachers notes below are numbered to match these Investigation Guides:

Investigation Guide 1: Bob's Buttons ... Investigation Guide 2: Bob's Buttons

 

Investigation Guide 1

The worksheet places the mathematics in the context of sharing buttons between friends, which is consistent with the original problem from the task centre collection.

1. The strategy of systematically setting out all the possibilities in a table shows that every extra 20 buttons more than the lowest solution will also be an answer, ie: 6, 26, 46, 66, ...
4s remainder 2:
{2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, 46, ...}
5s remainder 1:
{1, 6, 11, 16, 21, 26, 31, 36, 41, 46, ... }

Note that 4 x 5 = 20 and that 4 and 5 are the group sizes in the problem. So, once you have one solution, eg: 6 buttons, each extra 20 can be put into exact groups of 4 or of 5, hence the remainders will stay the same.

So the answers are all the numbers in the set (6 + 20n) where n is a whole number. Therefore 316, 220, 715 and 3,476 are not in this group, but 846 is.

2. For groups of (4 rem 3) and (5 rem 3), the same strategy produces:
4s remainder 3:
{3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, ...}
5s remainder 3:
{3, 8, 13, 18, 23, 28, 33, 38, 43, 48, 53, ... }

The smallest solution is 23 and every 20 thereafter.

3. For other combinations of remainders (still using groups of 4 and 5), the results for the smallest solution could be set out in a grid:

Table: Smallest solution for groups of 4 & 5 and all possible remainders. Remainders for groups of 4 on the left side. Remainders for groups of 5 across the top.
0 1 2 3 4
0          
1          
2   6      
3       23  

It could be a class project to share the labour and fill in the table. Then the table could be scrutinised for any patterns.

Answer Table

0 1 2 3 4
0 20 16 12 8 24
1 5 21 17 13 9
2 10 6 22 18 14
3 15 11 7 23 19

There are patterns in the table which, once one entry is known, will allow all others to be found. However, at this stage we do not know a simple way of calculating the smallest number directly.

4. For groups of 3 and 4, the answer will be the smallest number and then extra groups of 12.

  • For 3 & 5, it will be the smallest answer plus groups of 15.
  • For 2 & 7, it will be the smallest answer plus groups of 14.
  • For 5 & 7, it will be the smallest answer plus groups of 35.
  • For groups of a and b, (Note: a and b are not allowed to share common factors), it will be the smallest answer plus extra groups of a x b.

5. For groups of (3 rem 2), (4 rem 1) and (5 rem 2), a similar listing strategy to above soon finds the smallest answer at 17 and every 60 thereafter, ie: 17, 77, 137, 197, ...
3s rem 2:
{2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68, 71, 74, 77, ...}
4s rem 1:
{1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, 45, 49, 53, 57, 61, 65, 69, 73, 77, 81, 85, ...}
5s rem 2:
{2, 7, 12, 17, 22, 27, 32, 37, 42, 47, 52, 57, 62, 67, 72, 77, ...}

Note that 60 = 3 x 4 x 5, which is the product of the three group sizes. This is because if 17 is the smallest answer, then the extra buttons to the next answer must be able to be put into exact groups of 3, 4 and 5 and the lowest number for which this is possible is 60.

Investigation Guide 2

This problem is extraordinarily deep. While it starts as a task of finding just one number, Investigation Guide 1 shows how it can be extended to finding other solutions of the same problem, leading to generalisations about lowest common multiples, common factors etc.
Investigation Guide 2 follows a slightly different path. Up to this point students will have found the smallest solution for a given problem by trial and error. For example, if groups of 5 leave 3 over, and also groups of 4 leave 2 over, then systematic listing or just casting around, leads to a solution in which 3 groups of 5 and 3 over are the same as 4 groups of 4 and 2 over, ie:
3 x 5 + 3 = 4 x 4 + 2 = 18 If we now ask:
How can this number (18) be obtained from the four numbers in the problem?
and
Can this method be generalised?
we open a large and very deep mine, full of rich veins. The potential would stretch a good Year 12 student, and possibly even some teachers. It has certainly stretched us!

The worksheet tries to make this a little accessible by following what is often a good problem solving strategy of breaking the problem into parts, and seeking patterns.

1. If the group sizes are coprime (no common factor) then there will always be a solution. If they have a common factor, then there will be a solution only if the difference between the remainders is 0 or a multiple of the common factor. In the example given, the common factor is 3. The difference between the remainders is 1, so there is no solution. One way to see this visually is to mark the possible numbers in each pile along a number line in two different colours. This quickly demonstrates why the numbers stay 'out of phase'.

2.
a) 8
b) Example: groups of 4 give remainder 1, groups of 3 give remainder 0. c) If the group with none over is the larger group size, then the solution is the product of the remainder and the larger group size. In the example, 2 x 4 = 8.

  • If the group with no remainder is the smaller group size, then the solution is remainder + larger group size x (difference between smaller group size and remainder).
    Solution to example in (b) is 1 + 4 x (3 - 1) = 9.
  • Symbolically: if a > c, then any problem may be written as:
    ma + b = nc + d, where m, n, a, b, c, d are integers and a - c = 1.
    If b = 0, then the solution is ad.
    If d = 0, the solution is b + (c - b)a.
3.
a) 16
b) Example: groups of 8 give remainder 2, groups of 6 give remainder 0.
c) First step: divide all the numbers by the common factor. This then reduces the problem to one of the two types in Question 2. The example reduces to:
groups of 4, rem 0 = groups of 3, rem 2.
The answer obtained must then be multiplied by the common factor.
  • If the group with none over is the larger group size, then the solution is the product of the remainder and the larger group size.
    In the example, 2 x 4 = 8. Then multiply by the common factor 2 to get 16.
  • If the group with no remainder is the smaller group size, then the solution is remainder + larger group size x (difference between smaller group size and remainder).
    The problem in example in (b) reduces to:
    groups of 4 rem 1 = groups of 3 rem 0.
    The solution is 1 + 4 x (3 - 1) = 9. Then multiply by the common factor to get 18.
  • Symbolically: if a > c, then any problem may be written as:
    ma + b = nc + d, where m, n, a, b, c, d are integers.
    If the common factor of a and c is e, then:
    If b = 0, then the solution is ade.
    If d = 0, the solution is [b + (c - b)a]e.

4.
a) 21
b) 7
c) 8
d) 14
e) We are looking for a multiple of 7 that is greater than a multiple of 5 by the remainder.

  • If the remainder is equal to the difference between 7 and 5, then the solution is 7. If it is twice that difference the answer is twice 7, and so on.
  • If the remainder is between these multiples we need to find a formula that gives the number of 7s needed. You add 5 to the remainder, divide by 2 , and then multiply by 7.
  • Symbolically: if a > c, then any problem may be written as:
    ma + b = nc + d, where m, n, a, b, c, d are integers.
    If b = 0, and d = m(a - c), then the solution is: mad (parts b and d)
    If b = 0, and d = m(a - c) + 1, then the solution is: a(d + c)/2 (parts a and c).
  • Please note that although further generalisations are not requested and solutions are not offered, students could be encouraged to solve further what becomes a very complex problem.

5. You may continue to use the methods already discovered if you subtract the amount of the smaller remainder from both remainders first, solve the problem then add that amount back on again at the end.
Example: In groups of 8, rem 1 = groups of 6 rem 5:

  • Reduce both remainders by 1, to get:
    groups of 8, rem 0 = groups of 6, rem 4
    The problem has a solution because the remainder is a multiple of the difference between the group sizes (Qu. 1).
  • Divide by the common factor of 2, to get:
    groups of 4, rem 0 = groups of 3, rem 2 (Qu. 2).
    The answer is the product of 4 and 2 = 8.
    Check: 2 groups of 4, rem 0 = 2 groups of 3 rem 2 = 8
  • Multiply by the common factor, and then add the amount by which you reduced the smaller remainder: 8 x 2 + 1 = 17.
    Check in the original problem:
    2 groups of 8, rem 1 = 17 = 2 groups of 6 rem 5.

 

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