Investigation Guide 1: Bob's Buttons ... Investigation Guide 2: Bob's Buttons
Investigation Guide 1The 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, ...
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:
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:
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
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.
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, ...
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 2This 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) 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.
4.
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.
