STEP I:
1: Solution by Unbounded
2: Solution by Unbounded
3: Solution by nota bene
4: Solution by SimonM
5: Solution by nota bene
6: Solution by Unbounded
7: Solution by sonofdot
8: Solution by Unbounded
9: Solution by cliverlong
10: Solution by Dadeyemi
11:Solution by Unbounded
12: Solution by Robbie10538
13: Solution by Unbounded
14: Solution by brianeverit
STEP II:
1: Solution by sonofdot
2: Solution by dadeyemi
3: Solution by sonofdot
4: Solution by dadeyemi
5: Solution by dadeyemi
6: Solution by welshenglish
7: Solution by Glutamic Acid
8: Solution by Glutamic Acid
9: Solution by tommm
10: Solution by tommm
11: Solution by tommm
12: Solution by brianeverit
13: Solution by brianeverit
14: Solution by brianeverit
STEP III:
1: Solution by SimonM
2: Solution by Dadeyemi
3: Solution by Daniel Freedman
4: Solution by Dadeyemi
5: Solution by Dadeyemi
6: Solution by Elongar
7: Solution by Dadeyemi
8: Solution by Dadeyemi
9: Solution by brianeverit
10: Solution by brianeverit
11: Solution by brianeverit
12: Solution by brianeverit
13: Solution by brianeverit
14: Solution by brianeverit
Solutions written by TSR members:
1987  1988  1989  1990  1991  1992  1993  1994  1995  1996  1997  1998  1999  2000  2001  2002  2003  2004  2005  2006  2007
STEP I, II, III 2002 Solutions
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STEP II 2002 Question 3
Spoiler:Show
If k=1, then and , therefore it is true for k=1.
If k=2, then and , therefore it is true for k=2.
If k=3, then and , therefore it is true for k=3.
Assume, for some k
Therefore by induction, for all positive integers k.
Now consider and with r<s. Assume that and have a common factor p
Since all Fermat numbers are odd, and cannot share a common factor greater than one, hence by contradiction, no two Fermat numbers have a common factor greater than 1.
All Fermat numbers are either prime or made up of prime factors. Since no Fermat numbers share a common factor greater than one, no Fermat numbers can have the same prime factor. Since there are infinitely many Fermat numbers, there must therefore be infinitely many prime numbers. 
Question 1, STEP I, 2002
Spoiler:ShowWe need to find the points of intersection between the two ellipses:
 ellipse 1
 ellipse 2
making y^2 the subject in the equation for ellipse 1:
and substitution into the equation for ellipse 2:
Subbing back into the equation for y^2:
Therefore the coordinates of intersection are (2,1) and (2,1)
From a sketch, we can see that this means that the centre of a circle passing through those two points must lie on the xaxis.
Letting the coordinates of the centre of the circle be (a,0):
where r is the radius of the circle.
Substitute the point (2,1) in:
So the equation of the circle now becomes:

Some attached
Spoiler:Show(the end of the question 2 is the reason why I should not do maths late at night :/ 
Some more;
Did these quite a while ago I'm afraid some may be partial solutions. 
Question 6, STEP I, 2002

STEP II 2002 Question 1

II/8:
For x < 0, dy/dx = y ==> 1/y dy/dx = 1 ==> ln y = x +c ==> y = Ae^(x); as y = a when x = 1, a = Ae so A = a/e; therefore for x < 0.
For > 0, dy/dx = y ==> 1/y dy/dx = 1 ==> ln y = x + c ==> y = Be^x; as y = b when x = 1, b = Be so B = b/e therefore for x > 0.
If a = b, there is no jump.
Note that e^x  1 is positive if x > 0, and negative if x < 0.
for x > 0.
Given that y = e^e when x = 1, e^e = Ae^(e  1) so A = e;
for x < 0.
At x = 0, y = using the solution for x > 0. So for continuity, this must the same result when substituted into x < 0, so
(i)
(ii) . 
II/7: (Scary scary vectors.)

Question 2, STEP I, 2002

2002 II Question 11
Total weight (to check our calculus)
Total moment
.
We know that total moment , so , as required.
Draw out a diagram: A touching the ground, with R_A going up, F_A going right. B touching the wall, with F_B going up, R_B going left. Somewhere ebtween A and B, we have the centre of mass. We denote to be the distance of c.o.m. from A. Angle going from the horizontal to the line AB.
Equating the horizontal forces: . Noting that , we get . Noting that , we get
Equating the vertical forces: .
Equating the moments:
as required.
Not sure if I'm right, but I'll give it a go:
, and 
2002 II Question 12
probability such that more than l coins of the L coins will produce less than m heads
Estimating q: , where z is the probability of a coin producing less than m heads.
We can say that , which gives:
for small z.
We then substitute to get as required.
z is the probability that less than m heads are made after M throws.
.
as required.
K = 7, k = 2, L = 500, l = 4, M = 100, m = 48, p = 0.6. , which is a large z, so we are not allowed to approximate as needed. 
Question 10, STEP I, 2002

Question 11, STEP I, 2002
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