## Counting Pennies, Generating Functions and STEP

STEP I 1997, Question 1
Show that you can make up 10 pence in eleven ways using 10p, 5p, 2p and 1p coins.

In how many ways can you make up 20 pence using 20p,10p 5p, 2p and 1p coins?

There is an interesting way of approaching this question using generating functions.

A generating function is of the form:

$a_{0}+a_{1}t+a_{2}t^{2}+a_{3}t^{3}+\dots$

where the coefficients of $t$ count various problems and issues of convergence are unimportant.

Show that you can make up 10 pence in eleven ways using 10p, 5p, 2p and 1p coins.

This will be the co-efficient of $t^{10}$ in,

$(1+t+t^{2}+t^{3}+\dots+t^{10})(1+t^{2}+t^{4}+\dots+t^{10})(1+t^{5}+t^{10})(1+t^{10})$

by way of explanation:

$\overset{\textnormal{the 1p s}}{\overbrace{(1+t+t^{2}+t^{3}+\dots+t^{10})}}\overset{\textnormal{the 2p s}}{\overbrace{(1+t^{2}+t^{4}+\dots+t^{10})}}\overset{\textnormal{the 5p s}}{\overbrace{(1+t^{5}+t^{10})}}\overset{\textnormal{the 10p s}}{\overbrace{(1+t^{10})}}$

The candidate still needs to expand the function to get the answer, but having changed the problem to one of algebra, they might enjoy the benefits of practice and familiarity in the manual computation.

For those of at liberty to explore the question at leisure, a Computer Algebra System (CAS) is useful. In fact, the website Wolfram Alpha will expand the brackets for us, computing the answers to all similar counting problems in the process. The Wolfram Alpha query box understands the latex code if this is just pasted in, but only if it is presented in an explicit form, that is without the dots.  In order to simplify the question for the CAS notice that each bracket is a geometric progression, which can be summed using standard A level theory reducing the function to:

$\dfrac{(1-t^{11})}{(1-t)}\dfrac{(1-t^{12})}{(1-t^{2})}\dfrac{(1-t^{15})}{(1-t^{5})}\dfrac{(1+t^{10})}{1}$

Use of Wolfram Alpha then gives the following:

which contains our answer to the first part: 11.

Equipped with some methods and machinery we can now proceed to the second part of the question.  In fact, we can adapt the method and functions to count any similar loose change problem.

In how many ways can you make up 20 pence using 20p,10p 5p, 2p and 1p coins?

This will be the coefficient of $t^{20}$ in,

$(1+t+t^{2}+t^{3}+\dots+t^{20})(1+t^{2}+t^{4}+\dots+t^{20})(1+t^{5}+t^{10}+\dots+t^{20})(1+t^{10}+t^{20})(1+t^{20})$

$\overset{\textnormal{the 1 p s}}{\overbrace{(1+t+t^{2}+t^{3}+\dots+t^{20})}}\overset{\textnormal{the 2 p s}}{\overbrace{(1+t^{2}+t^{4}+\dots+t^{20})}}\overset{\textnormal{the 5 p s}}{\overbrace{(1+t^{5}+t^{10}+\dots+t^{20})}}\overset{\textnormal{the 10p s}}{\overbrace{(1+t^{10}+t^{20})}}\overset{\textnormal{the 20p s}}{\overbrace{(1+t^{20})}}$

which is equal to

$\dfrac{(1-t^{21})}{(1-t)}\dfrac{(1-t^{22})}{(1-t^{2})}\dfrac{(1-t^{25})}{(1-t^{5})}\dfrac{(1+t^{10}+t^{20})(1+t^{20})}{1}$

and which Wolfram Alpha gives and expansion of:

## Carr’s Synopsis and Sixth Term Entrance Papers

Students in search of extension work in maths, particularly as far as Sixth Term Entrance Papers (STEP) are concerned, should peruse Carr’s Synopsis of results in Elementary Pure Mathematics.

This book is famous for being the volume used by the autodidact Ramanujan to teach himself mathematics.  The book is merely a list of results and student how works through them, proving each one, will develop a strong ability in the subject.  This life line into higher mathematics could not be produced in the same form in today’s educational and publishing context but, perhaps, Carr deserves a place as one of the subject’s great educators.

Higher mathematics is about generalisation and abstraction.  Because of this, school mathematics students can find university text books unreadable.  In such a context how can harder, higher level questions be set using the frame work of sixth form mathematics? The answer to this question is to wind the clock back to the 1880s and see how the subject was constructed then.  Carr does this for us; many useful stages in school mathematics are listed here with extensions and generalisation written in generally sixth form recognisable form.

Take question 1 from this year’s (2016) STEP I exam.  Pupils will only have to be familiar with the first six results to tackle this quickly.

The question play is all about the divisibility of,

$x^{2n+1}+1$

which result 6, above, tells us is

$(x+1)(x^{2n}-x^{2n-1}+\dots+1)$.

Question game play leads us to factorise large numbers (without use of calculator):

$\dfrac{300^{3}+1}{301}=89911\times 90091$

and

$\dfrac{7^{49}+1}{7^{7}+1}=[(7^{7}+1)^{3}-7^{4}(7^{14}+7^{7}+1)][(7^{7}+1)^{3}+7^{4}(7^{14}+7^{7}+1)]$

in a difference of two squares (result 1) construction.

Full solution is attached.

step-i-2016-q1

I think that mathematically interested sixth formers, wanting to study the subject at University, should have some background knowledge.  Many would have heard about the Riemann Hypothesis and a link to prime numbers; this subject is beyond the scope of this blog but school pupils, with knowledge of prime numbers and geometric progressions, can see how the Zeta Function relates to prime numbers in the above bit of mathematics.  I like it anyway.