(Updated as far as #86.) SimonM  23.03.2009
For explanation, scroll down:
STEP Mathematics I
1: Solution by ukgea
2: Solution by Speleo
3: Solution by Speleo
4: Solution by Datr
5: Solution by DFranklin
6: Solution by Speleo
7: Solution by Speleo
8: Solution by Speleo
9: Solution by dr_98_98
10: Solution by brianeverit
11: Solution by ad absurdum
12: Solution by brianeverit
13: Solution by Datr
14: Solution by brianeverit
STEP Mathematics II:
1: Solved in Siklos Booklet on p20
2: Solved in Siklos Booklet on p24//Solution by Speleo
3: Solution by DFranklin
4: Solution by Datr
5: Solution by Speleo
6: Solution by ukgea
7: Solution by DFranklin
8: Solution by Speleo (1), (2)
9: Solution by brianeverit
10: Solution by brianeverit
11: Solution by brianeverit
12: Solution by Datr
13: Solution by brianeverit
14: Solution by brianeverit
STEP Mathematics III
1: Solution by DFranklin
2: Solution by Insparato (Incomplete)
3: Solution by Rabite (1), (2)
4: Solution by ukgea
5: Solution by dvs
6: Solution by dvs
7: Solution by DFranklin
8: Solution by Dvs
9: Solution by DFranklin
10: Solution by DFranklin
11: Solution by DFranklin
12: Solution by brianeverit
13: Solution by DFranklin
14: Solution by DFranklin
Explanation:
In this thread, we will try to collectively provide solutions for the questions from the STEP Maths I, II and III papers from 1996, seeing as no solutions are available on the net.
If you want to contribute, simply find a question which is marked with "unsolved" above (you might want to check for new solutions after the time the list was last updated as well), solve it, and post your solution in the thread. I will try to update the list as often as possible. If you have a solution to a question that uses a different idea from one already posted, submit it as well, as it's often useful to see different approaches to the same problem.
If you find an error in one of the solutions, simply say so, and hope that whoever wrote the solution in the first place corrects it.
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 Maths I, II, III 1996 Solutions
Announcements  Posted on 



Im going to search for any answers in Siklos booklet.

Nevermind, apparently Siklos did that one too.


(Original post by datr)
I'm just typing up STEP II Q4 
Yeah it's a nice question, most of the integration/differentiation ones are  I normally just go straight for them first.

Trying to post it all in one go isn't working.
STEP I Question 3
i) f(n) = n^5  n^3 = n^3(n^2  1)
= n^3(n+1)(n1)
One of n1, n, n+1 must be divisible by three, therefore f(n) is.
If n is even, n^3 is divisible by 8, so f(n) is divisible by 24.
If n is odd, n+1, n1 are divisible by 2 and one of them is divisible by 4.
Therefore f(n) is divisible by 2*4*3 = 24.
ii) g(n) = (2^n + 1)(2^n  1)
2^n is not divisible by 3, but one of (2^n  1), 2^n and 2^n + 1 must be.
Therefore g(n) must be divisible by 3.
iii) h(n) = n^3  1 = (n1)(n^2+n+1). n1 = 0 mod 3. n = 1 mod 3.
n^2 + n + 1 = 1 + 1 + 1 = 0 mod 3.
h(n) has two factors each divisible by 3 and is therefore divisible by 9. 
This one looks like it may be wrong :/
STEP I Question 7
i) dy/dt = ky; k>=1
lny = C  kt
y = e^(Ckt) = Ae^(kt)
When t = 0, y = 1
1 = A
y = e^(kt)
y = [e^(k)]^t
Let b = e^(k)
y = b^t
Since k>=1, b<=1/e, b<1
ii) dy/dt = a  ky
dy/dt = a + ylnb
dy/dt  ylnb = a
Integrating factor = e^(tlnb)
e^(tlnb)dy/dt + e^(tlnb)y(lnb) = ae^(tlnb)
d/dt[ye^(tlnb)] = ae^(tlnb)
ye^(tlnb) = (a/lnb)e^(tlnb) + C
When t = 0. y = 1
1 = C  (a/lnb)
C = 1 + (a/lnb)
ye^(tlnb) = (a/lnb)e^(tlnb) + (a/lnb) + 1
y = (a/lnb) + (a/lnb)e^(tlnb) + e^(tlnb)
y = [1 + (a/lnb)]b^t  (a/lnb) 
STEP I Question 8
i) S = 0.38383838...
= 38/100 + 38/(100)^2 + 38/(100)^3 + ...
= [38/100]/[1(1/100)]
= (38/100)/(99/100)
= 38/99.
ii) x = 0.a1a2...aNb1b2b3...bkb1b2...bkb 1..
= 0.a1a2...aN + 0.000...000b1b2...bkb1...
= a1a2...aN/(10^N) + 0.b1b2...bkb1.../(10^N)
But 0.b1b2...bkb1...
= b1b2...bk/(10^k) + b1b2...bk(10^k)^2 + ...
= [b1b2...bk/10^k]/[1  1/(10^k)]
= b1b2...bk/(10^k  1)
So x = a1a2...aN/(10^N) + b1b2...bk/(10^N)(10^k  1)
Since a1a2...aN and b1b2...bk integers, x is rational. 
I haven't checked to see if the last part is consistent with the Siklos book.
STEP II Question 2
Let a = yz, b = zx, c = xy.
I: 2a + b  5c = 2
II: a  b + 2c = 1
III: a  2b + 6c = 3
I2II = 3b  9c = 0, b = 3c
I2III = 5b  17c = 4
2c = 4
c = 2
b = 6
a = 1  4 + 6 = 3
abc = 36
(xyz)^2 = 36
xyz = + 6
Factors of + 6 must be + 1, 2, 3 or 1, 1, 6.
Since yz = 3 and xy = 2; 1, 1, 6 is not possible.
Therefore a, b, c are + 1, 2, 3.
y = 1 clearly, so z = 3 and x = 2.
Since xy, yz, zx are all > 0,
(x,y,z) = (2,1,3) or (2,1,3). 
This is ridiculous, TSR won't post this in one go:
STEP II Question 8
g(0) = f(0) + f(0) = 2
g'(x) = f'(x)  f'(x)
g'(0) = 1 + 1 = 0
g''(x) = f''(x) + f''(x)
f''(x) = x + 3cos2x  f(x)
f''(x) = x + 3cos2x  f(x)
g''(x) = 6cos2x  f(x)  f(x)
g''(x) = 6cos2x  g(x)
g''(x) + g(x) = 6cos2x
Aux quadratic > m^2 + 1 = 0
m = +i
g(x) = Acosx + Bsinx + ...
g(x) = pcos2x
4p + p = 6
p = 2
g(x) = Acosx + Bsinx  2cos2x
g(0) = 2.
A  2 = 2
A = 4
g'(x) = 4sinx + Bcosx + 4sin2x
g'(0) = 0 = B
g(x) = 4cosx  2cos2x 
h(x) = f(x)  f(x)
h(0) = 0
h'(x) = f'(x) + f'(x)
h'(0) = 2
h''(x) = f''(x)  f''(x)
f''(x) = x + 3cos2x  f(x)
f''(x) = x + 3cos2x  f(x)
h''(x) = 2x  f(x) + f(x)
h''(x) = 2x + h(x)
h''(x)  h(x) = 2x.
Aux quadratic is m²1 = 0 ie m=±1.
If h(x) = kx, then h(x) = 2x.
So h(x) = Ae^x + Be^x 2x.
h(0) = 0=A+B 0
h`(0) = 2 =AB  2
So A+B = 0 and AB = 0, so A and B are both 0.
h(x) = 2x
2f(x) = g(x) + h(x)
f(x) = 2cosx  cos2x  x
Correction in italics thanks to Rabite 
After several attempts, I have a solution to Paper I, Q5. I think I must have missed something obvious though  this was very hard work. (Though largely because it is so easy to make a mistake; the written up answer probably won't look too bad).

Step QI, P5.
(i) . Suppose . Then
would give us a rational expression for , contradiction.
So .
As r is rational, . (i.e. our solutions are .
(ii) . So
. Again, we set up the quadratic and get:
. Use the quadratic formula to get:
Now .
So .
As in (i), we need to find the square root of this, so look for rational with .
We find and end up with a quadratic . Solve to find and since we want rational deduce and so the square root is .
Plugging this back in, we have
.
Since p is real, we take the latter root, dividing by 2 and using (i) we find . Use to get final solutions .
(iii) Again, use the quadratic formula:
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse..2z=2\pm\sqrt{4  4(1+i)(2\sqrt{3}2)} \\
\implies z = 1 \pm \sqrt{1  (1+i)(2\sqrt{3}2)} = 1 \pm \sqrt{32\sqrt{3}+2(1\sqrt{3})i)
Using (ii),
. 
On the first page of the tests:
"To be brought by the candidate: electronic calculators, [...]"
Wtf? Were they allowed to use calculators back then? On all the later papers it says "Candidates may not use electronic calculators" instead. But anyone know why were they allowed previously and not now? 
Step II, Q3:
F0 = 0, F1 =1, F2 = F0+F1 = 1, F3 = F1 + F2 = 2, F4 = F3 + F2 = 3, F5 = 5, F6 = 8, F7 = 13.
F0.F2F1^2 = 1, F1.F3F2^2 = 1, F2.F4F3^2 = 1.
Claim . Proof by induction on n.
By above, true for n<=3. Assume true for n = k. Then
So true for n=k+1 and so true for all n by induction.
Want to show . Assume true for 1 < k <=m. Then
.
So true for k=m+1.
Explictly when k=1 we have . So true for k=1, so true for all k. 
(Original post by ukgea)
On the first page of the tests:
"To be brought by the candidate: electronic calculators, [...]"
Wtf? Were they allowed to use calculators back then? On all the later papers it says "Candidates may not use electronic calculators" instead. But anyone know why were they allowed previously and not now?
I think Siklos also says there was a tendancy for the weaker candidates to just blindly throw everything into the calculator without thinking too much. 
Paper II question 6
We can easily see that the proper factors of 12 are 2, 3, 4, 6 and those of 16 are 2, 4, 8.
Let M be a positive factor of N (not necessarily proper), Then f can be written as
where for all .
For each of the i, there are ways of choosing the integers , for a total number of F possible factors of N:
Each of these must be unique since prime factorisation is unique. But now we've counted the two "improper" factors 1 and N as well, so we need to remove these, and the total number of proper factors is
.
(i) For the total number of proper factors to be 12, we must have
which means that, r must be 2 and , or vice versa, or In the latter case, the smallest integer that has only one prime factor repeated 13 times is . In the former case, clearly, to minimise N, we need to take as small prime factors as possible, and make sure that the smaller of the two factors is raised to power 6, i.e. the minimum is . This is less that , and so is our answer.
(ii) Okay. The smallest integer is . But it seems that every attempt to prove this either boils down to a huge awful grind, or a complete orgy of handwaving. The only slightly rigorous proof I've found is the one I'm going to make below, but it uses the inequality between the arithmetic and geometric means, which says (in the case of 3 numbers, which is the case I'm going to use):
For any positive reals a, b, c, we have
but then, I'm not sure I'm allowed to use this.
Anyway, basically, we first note that the number of factors depend only on the number of times the prime factors are occuring in N, but not on what the prime factors actually are. To minimise the number N, we thus need to make sure that we use the smallest prime factors available, and make sure that we have the highest exponents in our smallest prime factors, i.e. we want
and
Now, we note that satisfies the requirements (it has 14 proper factors). This shows that our minimum can't have prime factors higher than or equal to 7, because we have already established that the minimum satisfies , which means that the smallest possible N with a prime factor of 7 is . But this is larger than and hence is not the minimum. Thus we can assume that
.
Now, because we want 12 or more proper factors, we have to have
Together with the arithmeticgeometric means inequality, this means that
which means that
(calculators were allowed, remember? )
Now we basically list the numbers with in rising order, and note that none of the ones below can have 12 proper factors or more:
32 = 2^5 has 4 proper factors
48 = 2^4*3 has 8 proper factors
64 = 2^6 has 5 proper factors
96 = 2^5*3 has 10 proper factors
120 = 2^3 * 3 * 5 has 14 proper factors
and thus 120 is our desired minimum.
So, there ought to be a better way of showing this last part. Ideas, anyone? 
Paper II,1996, Q7:
Wlog, let the corners of the square be A(1,1), B(1,1), C(1,1), D(1,1).
Let the position of P be (x,y). Then p = y+1, q=x1, r=y1, s=x+1.
So either x=y, x=y or , and every line above is reversible so this proves points on these lines or circle satisfy pq=rs as required. 
(Original post by DFranklin)
Wlog, let the corners of the square be A(1,1), B(1,1), C(1,1), D(1,1).
Let the position of P be (x,y). Then p = y+1, q=x1, r=y1, s=x+1.
So either x=y, x=y or , and every line above is reversible so this proves points on these lines or circle satisfy pq=rs as required.
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