Alan

The upper limit can be written as y = 10, though the "y" is usually dropped for the upper limit.

With y - 2 = k what you do is ask what is the value of k when y = 3  (k = 3 - 2 = 1), and what is the value of k when y = 10 (k = 10 - 2 = 8).

$\sum_{k=2}^{k=9}\frac{1}{k}=\sum_{k=1}^{k=9}\frac{1}{k}-\frac{1}{1} =\sum_{k=1}^{k=8}\frac{1}{k}-\frac{1}{1}+\frac{1}{9}$

It's perhaps a little more elegant the following way:

$\sqrt{36x^{36}}=\sqrt{6^2x^{18\times2}}=\sqrt{6^2(x^{18})^2}=6x^{18}$

I think it means you just keep putting the number you calculate back into the function until the result is 1.  i.e. C(9) = 28, C(28) = 14, C(14) = 7, ...etc.

Here is a short computer program that does the calculations:

Like so:

Do you mean like this?  (A matrix approach seems like overkill here!)

You have the following two equations:

s = c                                       (s=nbr of sausage buns, c = nbr of custard buns)

c + 27 = 6(s - 68)

Can you take it from here?

@wisdom321:  Note that your equation, y^2 + 9y - 18 = 0, shoud be y^2 + 9y + 18 = 0

I set j = k-2, so k=3 becomes j = 1, and k = n becomes j = n-2.  Then:

The sum $\Sigma_{j=1}^{m}j$  is given by $m(m+1)/2$,  so when m = n-2, the sum is (n-2)(n-1)/2.

As 23 is a prime number we must have:

$(23x+a)(x+b)=23x^2+kx-5$

Multiply out the left-hand side

$23x^2+(a+23b)x+ab=23x^2+kx-5$

Compare coefficients on both sides.

Can you take it from here, remembering that the coefficients must be integers.