STEP Maths I,II,III 1987 Solutions

Dystopia

This question also had a fairly automatic (albeit somewhat time-consuming) solution. Would this method of turning two linear simultaneous differential equations into a single second-order differential equation have been taught at the time? It's in both of the mechanics books I've been using (one old, one new).

I don't remember seeing it (and 1987 is only 2 years after I did A-levels). Pretty standard to see in a maths methods book though.

Reply 184

lordcrusade9

1

when you click on the link of the papers, you get junk sites. why is that? the same with mathsexams.tk

Reply 185

STEP III

question 1)

Take n to be 2r + 1

Clearly (n-1)! = (2r)!

http://mathworld.wolfram.com/Factorial.html

(2r)! = 2^r \prod^{r}_{k=1} T_{2k-1}

with

T_n

the nth triangle number = n + (n-1) + (n-2) + ... + 3 + 2 + 1

We stated n was odd so we can ignore all powers of 2 from this identity and it follows that if we want n to divide (n-1)! then n must have prime divisors that are present in the product;

\prod^{r}_{k=1} T_{2k-1}

T_1 = 1

so really we are only concerened with the product;

\prod^{m}_{k=2} T_{2k-1}

However the nth triangle umber is the sum for the first n numbers;

\sum^{n}_{i=1} i = \frac{n(n+1)}{2}

We want n to have prime factors that are present in the product

T_1 \times T_3 \times T_5 ... \times T_{2r-1}

So given n, we need its prime divisors to be present in the product;

1 \times 3 \times 5 \times ... \times \frac{(n-2)(n-1)}{2}

ofcourse you might expect the product of the first (n-2) odds to be a bitch - it certainly it is.

This dosen't even mention even n.

This post is just some thoughts and any information that anybodyd else attempting might find useful when tackling.

Reply 186

DFranklin

DeanK22: I suggest you start off by looking at the first few n and seeing if n | (n-1)! (you shouldn't need to actually evaluate (n-1)! if you think about it).

That should be enough to give a hypothesis (and probably also a plan of how to prove it).

Reply 187

DFranklin

DeanK22: I suggest you start off by looking at the first few n and seeing if n | (n-1)! (you shouldn't need to actually evaluate (n-1)! if you think about it).

That should be enough to give a hypothesis (and probably also a plan of how to prove it).

Wrote a few but not really getting the idea.

If A is the set of n such that n | (n-1)! then A' = { n | 4 or prime}

not really progressed from this though? any hints?

Reply 188

OP

If n is composite, n = ab, a<(n-1),b<(n-1)

Reply 189

SimonM

If n is composite, n = ab, a<(n-1),b<(n-1)

i tried this but couldn't show that ab could be represented completely by (n-1)!

unless i am being mega retarded and it is dead simple

(for example consider n and (n-1)! , the prime factors of n are

p_1^n \times p_2^m \times ...

and then I couldn't show that there wouldn't be a case when the exponent for a prime factor of n would exceed the exponent of the same prime factor present in (n-1)! (except when n is prime which is a trivial case)

Reply 190

OP

DeanK22

i tried this but couldn't show that ab could be represented completely by (n-1)!

Why not?

Reply 191

SimonM

Why not?

still having diffiulties - do you have a solution?

Spoiler

Reply 192

DFranklin

Dean: I would split into two cases (for composite n>4). n can be written as ab with

a \neq b

. Or n can only be written as b^2 (in which case b must be prime, although that doesn't matter).
The first case should be obvious. The second case is almost as obvious.

big spoiler

Reply 193

OP

STEP III, Question 1

Spoiler

Reply 194

DFranklin

Dean: I would split into two cases (for composite n>4). n can be written as ab with

a \neq b

. Or n can only be written as b^2 (in which case b must be prime, so it doesn't matter).
The first case should be obvious. The second case is almost as obvious.

big spoiler

oh yea that was kind of a simple question i was making a hash of. cheers DF

Reply 195

OP

STEP III, Question 2

Spoiler

Reply 196

OP