In Australian Rules football a score can be worth either 6 points or 1 point. The cheerleaders wave their pompoms each time their team scores, with the number of waves equaling the number of points that their team has scored thus far. So if a score of 1 point was followed by a score of 6 points, there would be one occasion for waving the pompoms once, and a second occasion, on which the pompoms would be waved 7 times (the total score at that point).

In a recent match, the team didn't do so well: the total number of waves that the cheerleaders gave was fewer than 50.

If I told you what that total number of waves was, you'd be able to deduce that the number of occasions on which they waved could be any of three different possibilities. Even if I then told you on how many occasions they waved, you could still find three different orders of 6's and 1's scored that would have led to that total number of waves.

On how many occasions did they wave, and what was the total number of waves?

OK, here we go.

Suppose n is the number of occasions they wave, N is the total number of waves, m is the number of 6-point scores and a(k) are the positions of 6-point scores counting from the end. Let’s represent 6 point scores as 1+5, therefore there are n 1-point scores and m 5-point scores like this:

1111
55

In this case n=4, m=2, a(k)=3,4.

One points will give a final cumulative score of 1+2+…+n=n(n+1)/2.
Five points will give a final cumulative score of 5*∑a(k), therefore

N=n(n+1)/2+5*∑a(k)

Since N<50, possible values for n are 2 to 9, and the values of n(n+1)/2 are: 3,6,10,15,21,28,36,45. N and n(n+1)/2 have the same reminder when divided by 5, so if we are given N and we can narrow n down to 3 cases, that means in 3 cases the reminder of n(n+1)/2 by 5 is the same, which is true for two sets: n=3,6,8 and n=4,5,9.

For the first set the common value among possible values of N is 36, however in each case there is only one sequence with N=36, so this case does not satisfy the second condition.

For the second set the common value among possible values of N is 45. In case of n=9 there is only one possible sequence with N=45. In case of n=4, N=45, then ∑a(k)=7. From numbers 1,2,3,4 we can form only two sums equal to 7, so this case again does not satisfy the second condition.

If n=5, N=45, then ∑a(k)=6. From numbers 1,2,3,4,5 we can form three sums equal to 6: 1+2+3, 1+5, 2+4. This satisfies the second condition of the problem. The possible sequences are:

11666
61116
16161


February 17, 2007, 8:01 pm
Edited on February 18, 2007, 2:05 am

Posted by Art M on 2007-02-17 19:58:40