# STEP I, II, III 2002 Solutions

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**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

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**STEP III, Question 1**

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#5

__STEP II 2002 Question 3__
Spoiler:

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.

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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.

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#6

__Question 1, STEP I, 2002__
Spoiler:

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We 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 x-axis.

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:

- 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 x-axis.

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:

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#7

Some attached

Spoiler:

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(the end of the question 2 is the reason why I should not do maths late at night :/

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#12

**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) .

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#15

I'm not checking if anyone posted the solutions, cause I completed these not long ago.

as required.

and

, as required.

Using , , we obtain:

**2002 II Question 1**as required.

and

, as required.

Using , , we obtain:

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#16

**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

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#17

**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.

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#18

**2002 II Question 13**

,

since since it's a cumulative distribution function.

We know

We know

Using the condition found before, we deduce that

.

We know , so .

We also know that

So

, as required.

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