(Updated as far as #213) SimonM - 11.05.2009

Solutions for the earlier STEPs are not available on the internet, so we're making our own.

Please submit a solution, if you can, to any problem which is currently unsolved (red above).

If you see any mistakes please point them out.

PM me for the papers.

The 1994 thread is still missing a couple of solutions.

N.B. The usual paper site has the papers mislabelled (in the wrong order), check before you waste your time

Also, the number of questions has changed, and there are different proportions of pure:mechanics:stats.

STEP I (Mathematics)

1: Solved by DFranklin

2: Scattered throughout the thread, summary by DFranklin

3: Solution by ukgea

4: Solution by Speleo

5: Solution by Speleo

6: Solution by Speleo

7: Solution by DFranklin

8: Solved by generalebriety

9: Solution by brianeverit

10: Solution by *bobo*

11: Solution by generalebriety

12: Solution by generalebriety

13: Solution by *bobo* (1), (2)

14: Solution by Glutamic Acid

15: Solution by nota bene

16: Solution by ben-smith

STEP II (Further Mathematics A)

1: Solution by nota bene

2: Solution by Rabite

3: Solution by insparato

4: Solution by generalebriety

5: Solution by Rabite

6: Solution by nota bene and insparato

7: Solution by Speleo

8: Solution by datr

9: Solution by khaixiang

10: Solution by generalebriety

11: Solution by *bobo* (1), (2), (3)

12: Solution by *bobo* (1), (2)

13: Solution by *bobo*

14: Solution by ben-smith

15: Solution by toasted-lion

16: Solution by deltinu

STEP III (Further Mathematics B)

1: Solution by generalebriety

2: Solution by justinsh

3: Solved by insparato

4: Solution by generalebriety

5: Solution by SimonM

6: Solution by insparato with comments by DFranklin

7: Solution by Rabite and Speleo

8: Solution by DFranklin

9: Solution by Speleo

10: Solution by DFranklin

11: Solution by DFranklin

12: Solution by *bobo* (1), (2)

13: Solution by brianeverit

14: Solution by Glutamic Acid

15: Solution by DFranklin

16: 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

Solutions for the earlier STEPs are not available on the internet, so we're making our own.

Please submit a solution, if you can, to any problem which is currently unsolved (red above).

If you see any mistakes please point them out.

PM me for the papers.

The 1994 thread is still missing a couple of solutions.

N.B. The usual paper site has the papers mislabelled (in the wrong order), check before you waste your time

Also, the number of questions has changed, and there are different proportions of pure:mechanics:stats.

STEP I (Mathematics)

1: Solved by DFranklin

2: Scattered throughout the thread, summary by DFranklin

3: Solution by ukgea

4: Solution by Speleo

5: Solution by Speleo

6: Solution by Speleo

7: Solution by DFranklin

8: Solved by generalebriety

9: Solution by brianeverit

10: Solution by *bobo*

11: Solution by generalebriety

12: Solution by generalebriety

13: Solution by *bobo* (1), (2)

14: Solution by Glutamic Acid

15: Solution by nota bene

16: Solution by ben-smith

STEP II (Further Mathematics A)

1: Solution by nota bene

2: Solution by Rabite

3: Solution by insparato

4: Solution by generalebriety

5: Solution by Rabite

6: Solution by nota bene and insparato

7: Solution by Speleo

8: Solution by datr

9: Solution by khaixiang

10: Solution by generalebriety

11: Solution by *bobo* (1), (2), (3)

12: Solution by *bobo* (1), (2)

13: Solution by *bobo*

14: Solution by ben-smith

15: Solution by toasted-lion

16: Solution by deltinu

STEP III (Further Mathematics B)

1: Solution by generalebriety

2: Solution by justinsh

3: Solved by insparato

4: Solution by generalebriety

5: Solution by SimonM

6: Solution by insparato with comments by DFranklin

7: Solution by Rabite and Speleo

8: Solution by DFranklin

9: Solution by Speleo

10: Solution by DFranklin

11: Solution by DFranklin

12: Solution by *bobo* (1), (2)

13: Solution by brianeverit

14: Solution by Glutamic Acid

15: Solution by DFranklin

16: 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

(edited 13 years ago)

Scroll to see replies

I'm on to Q1 STEP II (Further Pure A)

Is it just me or is it pure brute force?

Okay, if B knows A's numbers and they are of the form $a_1+a_2+a_3=9$ B can always chose $b_1+b_2+b_3=9$ so that he wins; the possible numbers are:

1+1+7

1+2+6

1+3+5

1+4+4

2+2+5

2+3+4

3+3+3 (and of course in other order as well...)

So here it is quite convincing B can always beat A if B knows A's numbers and the order of them.

edit: To make this clearer I'll list the possibilities explicitly like David told (so the person typing out these markschemes won't have to fill in half the answers themselves!)

A ------B

(117)(351)

(126)(351)

(135)(216)

(234)(441)

(333)(441)

(522)(144)

(144)(315)

Furthermore, if A choses 1+1+7 (or some other order of those numbers) it is impossible for A to win no matter what B choses (they can tie).

If now A gets to chose two triples and B are to find one triple that beats both it is quite clear from looking at the possible combinations that A shall not chose any of the combinations with few rearrangements (like 333 and 117 can easily be counteracted by 441). Many combinations leads to tied situations. From trial and error it is possible to eliminate all triples but some of the 531s. Testing these we can see 135/153 (by 171),135/531 (by 441), 135/513 (by 144), 135/351 (by 441), 135/315 (by 441), 153/531 (by 261), 153/513 (by 126), 153/351 (by 261), 153/315 (by 414), 531/513 (by 144), 531/351 (by 441), 531/315 (by NONE), 351/315 (by 414)

Conclusion: A shall chose 5+3+1 and 3+1+5 as his triples to always be sure of winning.

(okay and I am masochistic, here are all the stupid possible combinations:

117 (beats neither 531 nor 315)

711 (beats neither 531 nor 315)

171 (beats neither 531 nor 315)

144 (beats 531, not 315)

141 (beats neither 531 nor 315)

441 (beats 315, not 531)

135 (ties with both 315 and 531)

153 (beats 531, not 315)

531 (beats 315, ties with 531)

513 (ties with both 531 and 315)

351 (ties with both 531 and 315)

315 (beats neither 315 nor 531)

225 (beats neither 531 nor 315)

252 (beats 531, not 315)

522 (beats 315, ties with 531)

333 (ties with both 531 and 315)

234 (beats neither 531 nor 315)

243 (beats neither 531 nor 315)

432 (beats 315, ties with 531)

423 (beats 315, not 531)

342 (beats 531, ties with 315)

324 (beats neither 315 nor 531)

126 (beats 315, not 531)

162 (beats 531, not 315)

216 (beats neither 531 nor 315)

261 (beats neither 531 nor 315)

612 (beats 531, ties with 315)

621 (beats 315, ties with 531)

Is it just me or is it pure brute force?

Okay, if B knows A's numbers and they are of the form $a_1+a_2+a_3=9$ B can always chose $b_1+b_2+b_3=9$ so that he wins; the possible numbers are:

1+1+7

1+2+6

1+3+5

1+4+4

2+2+5

2+3+4

3+3+3 (and of course in other order as well...)

So here it is quite convincing B can always beat A if B knows A's numbers and the order of them.

edit: To make this clearer I'll list the possibilities explicitly like David told (so the person typing out these markschemes won't have to fill in half the answers themselves!)

A ------B

(117)(351)

(126)(351)

(135)(216)

(234)(441)

(333)(441)

(522)(144)

(144)(315)

Furthermore, if A choses 1+1+7 (or some other order of those numbers) it is impossible for A to win no matter what B choses (they can tie).

If now A gets to chose two triples and B are to find one triple that beats both it is quite clear from looking at the possible combinations that A shall not chose any of the combinations with few rearrangements (like 333 and 117 can easily be counteracted by 441). Many combinations leads to tied situations. From trial and error it is possible to eliminate all triples but some of the 531s. Testing these we can see 135/153 (by 171),135/531 (by 441), 135/513 (by 144), 135/351 (by 441), 135/315 (by 441), 153/531 (by 261), 153/513 (by 126), 153/351 (by 261), 153/315 (by 414), 531/513 (by 144), 531/351 (by 441), 531/315 (by NONE), 351/315 (by 414)

Conclusion: A shall chose 5+3+1 and 3+1+5 as his triples to always be sure of winning.

(okay and I am masochistic, here are all the stupid possible combinations:

117 (beats neither 531 nor 315)

711 (beats neither 531 nor 315)

171 (beats neither 531 nor 315)

144 (beats 531, not 315)

141 (beats neither 531 nor 315)

441 (beats 315, not 531)

135 (ties with both 315 and 531)

153 (beats 531, not 315)

531 (beats 315, ties with 531)

513 (ties with both 531 and 315)

351 (ties with both 531 and 315)

315 (beats neither 315 nor 531)

225 (beats neither 531 nor 315)

252 (beats 531, not 315)

522 (beats 315, ties with 531)

333 (ties with both 531 and 315)

234 (beats neither 531 nor 315)

243 (beats neither 531 nor 315)

432 (beats 315, ties with 531)

423 (beats 315, not 531)

342 (beats 531, ties with 315)

324 (beats neither 315 nor 531)

126 (beats 315, not 531)

162 (beats 531, not 315)

216 (beats neither 531 nor 315)

261 (beats neither 531 nor 315)

612 (beats 531, ties with 315)

621 (beats 315, ties with 531)

nota bene

I'm on to Q1 STEP II (Further Pure A)

Is it just me or is it pure brute force?

Is it just me or is it pure brute force?

nota bene

If now A gets to chose two triples and B are to find one triple that beats both it is quite clear from looking at the possible combinations that A shall not chose any of the combinations with few rearrangements (like 333 and 117 can easily be counteracted by 441).

Spoiler

I'll have a go at (what is labelled as) STEP III question 8. (Paper reference 9465.)

Ok, here we go. (Some bits refused to be LaTeX-ed, so I've taken the tags out because I'm lazy instead of fixing the code.)

\overrightarrow{DE} \cdot \overrightarrow{EF} = DE \cdot EF \cos \angle DEF \\

1 = 5 \cos \angle DEF \\

\angle DEF = \cos^{-1}(1/5).

\text{From here it is simply a case of using a calculator to calculate } DE \text{ and applying a property of the scalar product.}

Ok, here we go. (Some bits refused to be LaTeX-ed, so I've taken the tags out because I'm lazy instead of fixing the code.)

Unparseable latex formula:

[br]\displaystyle[br]\text{Sphere } S \text{ intersects plane }\Pi. \\[br]S: x^2 + y^2 + (z-a)^2 = b^2\\[br]\Pi : z = 0 \\[br]\text{Solve simultaneously:} \\[br]x^2 + y^2 = b^2 - a^2. \\[br]\text{If } b^2 > a^2\text{, then } x^2 + y^2 = c^2 \longleftarrow \text{circle} \\[br]\text{If } b^2 = a^2\text{, then } x^2 + y^2 = 0 \longleftarrow \text{point } (0, 0, 0) \\[br]\text{If } b^2 < a^2\text{, then } x^2 + y^2 = -c^2 \longleftarrow \text{empty set.} \\ \\[br][br]\text{The vectors i, j and k are such that } \overrightarrow{AC} = 2\mathbf{i},\; \overrightarrow{AD} = 2\mathbf{k},\; \\[br]\text{and i, j and k form a right handed system of orthonormal base vectors.} \\[br]\overrightarrow{AC} = 2\mathbf{i}, \overrightarrow{AD} = 2\mathbf{k}, \overrightarrow{BE} = 3\mathbf{k}, \overrightarrow{CF} = 4\mathbf{k} \\[br]\overrightarrow{AB} = 2(\cos \frac{\pi}{3} \mathbf{i} + \sin \frac{\pi}{3} \mathbf{j}) = \mathbf{i} + \sqrt{3} \mathbf{j} \\[br]\overrightarrow{DE} = \overrightarrow{DA} + \overrightarrow{AB} + \overrightarrow{BE} = \begin{pmatrix} 1 \\ \sqrt{3} \\ 1 \end{pmatrix} \\[br]\text{Similarly, } \overrightarrow{EF} = \begin{pmatrix} 1 \\ -\sqrt{3} \\ 1 \end{pmatrix} \\[br]\overrightarrow{DE}\cdot \overrightarrow{EF} = \begin{pmatrix} 1 \\ \sqrt{3} \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -\sqrt{3} \\ 1 \end{pmatrix} = -1 \\

\overrightarrow{DE} \cdot \overrightarrow{EF} = DE \cdot EF \cos \angle DEF \\

1 = 5 \cos \angle DEF \\

\angle DEF = \cos^{-1}(1/5).

Unparseable latex formula:

[br]\displaystyle[br]\text{The pole } AD \text{ is shifted.} \\[br]\overrightarrow{AD} = \overrightarrow{AB} \cos \frac{\pi}{4} + 2\mathbf{k} \cos \frac{\pi}{4} \\[br]= \begin{pmatrix} 1/\sqrt{2} \\ \sqrt{3/2} \\ 1/\sqrt{2} \end{pmatrix}.[br]\overrightarrow{DE} = -\begin{pmatrix} 1/\sqrt{2} \\ \sqrt{3/2} \\ 1/\sqrt{2} \end{pmatrix} + \begin{pmatrix} 1 \\ \sqrt{3} \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ 0 \\ 3 \end{pmatrix} \\[br] = \frac{1}{\sqrt{2}} \begin{pmatrix} \sqrt{2} - 1 \\ \sqrt{3}(\sqrt{2} - 1) \\ 3\sqrt{2} - 1 \end{pmatrix}. \\ \\[br][br]\overrightarrow{DE} \cdot \overrightarrow{EF} = \frac{1}{\sqrt{2}} (\sqrt{2} - 1 - 3(\sqrt{2} - 1) + 3\sqrt{2} - 1) = 1 + \frac{1}{\sqrt{2}}.

\text{From here it is simply a case of using a calculator to calculate } DE \text{ and applying a property of the scalar product.}

DFranklin

Possible spoiler:

Spoiler

Whatever xD (I proved they were the only ones as well lol...) Actually I'm very surprised if I just had to verify that I did the question in <10 minutes .

EDIT: STEP 1 Question 1

There are 7 ways of throwing 6:

1+5

2

2

3

3

3

3

Okay, I'll write up the steps slowly so I can follow my own train of thought...

2. we know there is no 1+1, so there is a 1 only on one die (may be multiple 1s though)

3.There are two ways of getting 3 this means there is a duplication of either 2s or 1s

4. Only one combination and as we know there is at least one 1 there must be at least one 3(not on the same die as the one/s) OR there must be a two on one die and a two on the other, however the latter leads up the wrong creek pretty soon after testing it.

5. Again only one combination and we therefore know there must be at least one 4 (not on the same die as the one/s)

Summary: Die I has a 1 and Die II has one 3 and one 4 and no 1s. (also we know there is a duplication of either a 1 or a 2 (on different dice), which implies there is no 2 on the die with the one/s) As there is only one way of getting a sum of 4 there must only be one 1, therefore duplicate 2s. So we now have Die I: 1 and Die II: 2 2 3 4 (no 1s)

6. For there to be four ways of getting six it is either 4 on die I and 55 on die II or 5555 on die II but the latter obviously fails if we attempt to continue on that track.

7. There are three ways of getting the sum 7 and we have one hitherto (and die II we have all positions filled) which means there is one 5 on die I.

8-12. We fill up with 6s and check that the conditions holds, which they seem to do; therefore the dice had the numbers 145666 respectively 223455.

STEP II Question 3

a)

$\frac{dy}{dx} - y - 3y^2 = -2$

$y = -\frac{1}{3u}\frac{du}{dx}$

$\frac{dy}{dx} = -\frac{1}{3u} \frac{d^2u}{dx^2} + \frac{1}{3}\left(\frac{du}{dx}\right)^2$

So

$-\frac{1}{3u} \frac{d^2u}{dx^2} + \frac{1}{3}\left(\frac{du}{dx}\right)^2 + \frac{1}{3u}\frac{du}{dx} - \frac{1}{3u^2}\left(\frac{du}{dx}\right)^2 = -2$

$-\frac{1}{3u} \frac{d^2u}{dx^2} + \frac{1}{3u}\frac{du}{dx} + 2 = 0$

Multiply by -3u

$\frac{d^2u}{dx^2} - \frac{du}{dx} - 6u$

$m^2 - m - 6 = 0$

$(m - 3)(m + 2) = 0$

$u = Ae^{3x} + Be^{-2x}$

$y = -\frac{1}{3u} \frac{du}{dx}$

$\frac{du}{dx} = 3Ae^{3x} - 2Be^{-2x}$

$y = \frac{-(3Ae^{3x} - 2Be^{-2x})}{3u}$

$3u = \frac{-3Ae^{-3x} + 2Be^{-2x}}{y}$

$u = \frac{-3Ae^{3x} + 2Be^{-2x}}{3y}$

$\frac{-3Ae^{3x} + 2Be^{-2x}}{3y} = Ae^{3x} + 2Be^{-2x}$

$3y = \frac{-3Ae^{3x} + 2Be^{-2x}}{Ae^{3x} + 2Be^{-2x}}$

$y = \frac{-3Ae^{3x} + 2Be^{-2x}}{3Ae^{3x} + 2Be^{-2x}}$

$y = \frac{-3Ae^{5x} + 2B}{3Ae^{5x} + 2B}$

b)

$x^2 \frac{dy}{dx} + xy + x^2y^2 = 1$

$y = \frac{1}{x} + \frac{1}{v}$

$\frac{dy}{dx} = -\frac{1}{x^2} - \frac{1}{v^2}\frac{dv}{dx}$

Substitute back in

$x^2\left(-\frac{1}{x^2} - \frac{1}{v^2}\frac{dv}{dx}\right) + x(\frac{1}{x} + \frac{1}{v} + x^2\left(\frac{1}{x} + \frac{1}{v}\right)^2 = 1$

$-1 - \frac{x^2}{v^2} \frac{dv}{dx} + 1 + \frac{x}{v} + x^2\left(\frac{1}{x^2} + \frac{2}{vx} + \frac{1}{v^2}\right) = 1$

$-\frac{x^2}{v^2}\frac{dv}{dx} + \frac{3x}{v} + \frac{x^2}{v^2} + 1 = 1$

$-x^2 \frac{dv}{dx} + 3xv + x^2 = 0$

$\frac{dv}{dx} - \frac{3v}{x} = 1$

Integrating factor

$x^{-3} \frac{dv}{dx} - \frac{3v}{x^4} = x^{-3}$

$\frac{d}{dx}(\frac{v}{x^3}) = \frac{1}{x^3}$

$\frac{v}{x^3} = -\frac{1}{2x^2} + C$

$v = -\frac{x}{2} + Cx^3$

$y = \frac{1}{x} + \frac{1}{v}$

$y - \frac{1}{x} = \frac{1}{v}$

$\frac{1}{v} = \frac{xy-1}{x}$

$v = \frac{x}{xy - 1}$

$\frac{x}{xy-1} = -\frac{x}{2} + Cx^3$

$xy - 1 = \frac{x}{-\frac{x}{2} + Cx^3}$

$xy = 1 + \frac{2x}{2Cx^3-x}$

$xy = \frac{2cx^3-x + 2x}{2Cx^3-x}$

$y =\frac{2cx^3+x}{2Cx^4-x^2}$

$y = \frac{x(2cx^2+1)}{x^2(2cx^2-1)}$

$y = \frac{2cx^2+1}{x(2cx^2-1)}$

a)

$\frac{dy}{dx} - y - 3y^2 = -2$

$y = -\frac{1}{3u}\frac{du}{dx}$

$\frac{dy}{dx} = -\frac{1}{3u} \frac{d^2u}{dx^2} + \frac{1}{3}\left(\frac{du}{dx}\right)^2$

So

$-\frac{1}{3u} \frac{d^2u}{dx^2} + \frac{1}{3}\left(\frac{du}{dx}\right)^2 + \frac{1}{3u}\frac{du}{dx} - \frac{1}{3u^2}\left(\frac{du}{dx}\right)^2 = -2$

$-\frac{1}{3u} \frac{d^2u}{dx^2} + \frac{1}{3u}\frac{du}{dx} + 2 = 0$

Multiply by -3u

$\frac{d^2u}{dx^2} - \frac{du}{dx} - 6u$

$m^2 - m - 6 = 0$

$(m - 3)(m + 2) = 0$

$u = Ae^{3x} + Be^{-2x}$

$y = -\frac{1}{3u} \frac{du}{dx}$

$\frac{du}{dx} = 3Ae^{3x} - 2Be^{-2x}$

$y = \frac{-(3Ae^{3x} - 2Be^{-2x})}{3u}$

$3u = \frac{-3Ae^{-3x} + 2Be^{-2x}}{y}$

$u = \frac{-3Ae^{3x} + 2Be^{-2x}}{3y}$

$\frac{-3Ae^{3x} + 2Be^{-2x}}{3y} = Ae^{3x} + 2Be^{-2x}$

$3y = \frac{-3Ae^{3x} + 2Be^{-2x}}{Ae^{3x} + 2Be^{-2x}}$

$y = \frac{-3Ae^{3x} + 2Be^{-2x}}{3Ae^{3x} + 2Be^{-2x}}$

$y = \frac{-3Ae^{5x} + 2B}{3Ae^{5x} + 2B}$

b)

$x^2 \frac{dy}{dx} + xy + x^2y^2 = 1$

$y = \frac{1}{x} + \frac{1}{v}$

$\frac{dy}{dx} = -\frac{1}{x^2} - \frac{1}{v^2}\frac{dv}{dx}$

Substitute back in

$x^2\left(-\frac{1}{x^2} - \frac{1}{v^2}\frac{dv}{dx}\right) + x(\frac{1}{x} + \frac{1}{v} + x^2\left(\frac{1}{x} + \frac{1}{v}\right)^2 = 1$

$-1 - \frac{x^2}{v^2} \frac{dv}{dx} + 1 + \frac{x}{v} + x^2\left(\frac{1}{x^2} + \frac{2}{vx} + \frac{1}{v^2}\right) = 1$

$-\frac{x^2}{v^2}\frac{dv}{dx} + \frac{3x}{v} + \frac{x^2}{v^2} + 1 = 1$

$-x^2 \frac{dv}{dx} + 3xv + x^2 = 0$

$\frac{dv}{dx} - \frac{3v}{x} = 1$

Integrating factor

Unparseable latex formula:

e^{\int -\frac{3}{x} \hspace5 dx} = e^{-3lnx} = x^{-3}

$x^{-3} \frac{dv}{dx} - \frac{3v}{x^4} = x^{-3}$

$\frac{d}{dx}(\frac{v}{x^3}) = \frac{1}{x^3}$

Unparseable latex formula:

\frac{v}{x^3} = \int \frac{1}{x^3} \hspace5 dx

$\frac{v}{x^3} = -\frac{1}{2x^2} + C$

$v = -\frac{x}{2} + Cx^3$

$y = \frac{1}{x} + \frac{1}{v}$

$y - \frac{1}{x} = \frac{1}{v}$

$\frac{1}{v} = \frac{xy-1}{x}$

$v = \frac{x}{xy - 1}$

$\frac{x}{xy-1} = -\frac{x}{2} + Cx^3$

$xy - 1 = \frac{x}{-\frac{x}{2} + Cx^3}$

$xy = 1 + \frac{2x}{2Cx^3-x}$

$xy = \frac{2cx^3-x + 2x}{2Cx^3-x}$

$y =\frac{2cx^3+x}{2Cx^4-x^2}$

$y = \frac{x(2cx^2+1)}{x^2(2cx^2-1)}$

$y = \frac{2cx^2+1}{x(2cx^2-1)}$

I really hate LaTeX.

I've done STEP II (paper A), question 2, 3, 8 and STEP III (paper B), question 1, 2, 4, 9 awhile ago. So I won't do them again and will try other questions. Here's question 9 of STEP II:

Sorry, I know this is very ugly, if it's too much of an eyesore I will remove it for conformity's sake.

This question seems way too easy compared to others, so perhaps I've overlooked something.

Sorry, I know this is very ugly, if it's too much of an eyesore I will remove it for conformity's sake.

This question seems way too easy compared to others, so perhaps I've overlooked something.

Ok, done one question.

II/7

I've decided that writing questions out in text is less legible and more time consuming than just scanning them in, and latex takes far too long for me.

Tell me if the writing's too faint, I'm going to use pen next time anyway.

I've decided that writing questions out in text is less legible and more time consuming than just scanning them in, and latex takes far too long for me.

Tell me if the writing's too faint, I'm going to use pen next time anyway.

STEP I (paper 9465) Q3

I shall do this question in a lot of cases. Not knowing really how to indent properly on TSR, I'll use a "case 1.2.4" notation, where case 1. Hope you'll understand.

(i)

Assume integers p, q, r solve the equation.

Case 1: $p = 1$

Then

$\displaystyle \frac{1}{p} + \frac{1}{q} + \frac{1}{r} > 1$

contradiction.

Case 2: $p = 2$

Then

$\displaystyle \frac{1}{q} + \frac{1}{r} = \frac{1}{2}$

Also, we have $r \geq q \geq 2$

Case 2.1: $q = 2$

Then

$\displaystyle \frac{1}{q} + \frac{1}{r} > \frac{1}{2}$

contradiction.

Case 2.2: $q = 3$

From this follows r = 6, and so (p, q, r) = (2, 3, 6) is a solution.

Case 2.3: $q = 4$

From this follows r = 4, and so (p, q, r) = (2, 4, 4) is a solution.

Case 2.4: $q \geq 5$

Then

$r \geq 5$

and so

$\displaystyle \frac{1}{q} + \frac{1}{r} \leq \frac{2}{5}$

contradiction.

(Now we've covered all the cases that can arise when $p = 2$.)

Case 3: $p = 3$

Then

$\displaystyle \frac{1}{q} + \frac{1}{r} = \frac{2}{3}$

and

$r \geq q \geq 3$

Case 3.1 $q = 3$

Then r = 3, and we've found the solution (p, q, r) = (3, 3, 3)

Case 3.2 $q \geq 4$

Then

$r \geq 4$

and

$\displaystyle \frac{1}{q} + \frac{1}{r} \leq \frac{1}{2}$

contradiction.

(Now we've covered all the cases that can arise when $p = 3$.)

Case 4:$p \geq 4$

Then $q, r \geq 4$

and

$\displaystyle \frac{1}{p} + \frac{1}{q} + \frac{1}{r} \leq \frac{3}{4}$

contradiction.

Now we've covered all the cases, and the only solutions are thus (p, q, r) = (2, 3, 6), (2, 4, 4) and (3, 3, 3).

(ii)

Assume integers p, q, r satisfy the inequality.

Case 1: $p = 1$

This has already been given in the question. (Besides i think there's a typo in the paper, it should be $1\leq n \leq m$. But who cares...)

Case 2: $p = 2$

Then the inequality becomes

$\displaystyle \frac{1}{q} + \frac{1}{r} > \frac{1}{2}$

and we of course also have $r \geq q \geq 2$.

Case 2.1: $q = 2$

Then the inequality becomes

$\displaystyle \frac{1}{r} > 0$

which is satisfied by any integer $r \geq 2$.

Case 2.2: $q = 3$

Then the inequality becomes

$\displaystyle \frac{1}{r} > \frac{1}{6}$

which for positive r is equivalent to r < 6

and hence all solutions for this case are given by (p, q, r) = (2, 3, 3), (2, 3, 4) and (2, 3, 5).

Case 2.3: $q \geq 4$

Then $r \geq 4$ and

$\displaystyle \frac{1}{q} + \frac{1}{r} \leq \frac{1}{2}$

Contradiction.

(Now we have covered all the cases that can arise when q = 2.)

Case 3: $p = 3$

Then $q, r \geq 3$ and so

$\displaystyle \frac{1}{p} + \frac{1}{q} + \frac{1}{r} \leq \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1$

contradiction.

Now all the cases have been covered, and hence the only solutions other than (1, m, n) are

(2, 2, r) where r can be any integer larger than or equal to 2,

(2, 3, 3)

(2, 3, 4)

and

(2, 3, 5).

I shall do this question in a lot of cases. Not knowing really how to indent properly on TSR, I'll use a "case 1.2.4" notation, where case 1. Hope you'll understand.

(i)

Assume integers p, q, r solve the equation.

Case 1: $p = 1$

Then

$\displaystyle \frac{1}{p} + \frac{1}{q} + \frac{1}{r} > 1$

contradiction.

Case 2: $p = 2$

Then

$\displaystyle \frac{1}{q} + \frac{1}{r} = \frac{1}{2}$

Also, we have $r \geq q \geq 2$

Case 2.1: $q = 2$

Then

$\displaystyle \frac{1}{q} + \frac{1}{r} > \frac{1}{2}$

contradiction.

Case 2.2: $q = 3$

From this follows r = 6, and so (p, q, r) = (2, 3, 6) is a solution.

Case 2.3: $q = 4$

From this follows r = 4, and so (p, q, r) = (2, 4, 4) is a solution.

Case 2.4: $q \geq 5$

Then

$r \geq 5$

and so

$\displaystyle \frac{1}{q} + \frac{1}{r} \leq \frac{2}{5}$

contradiction.

(Now we've covered all the cases that can arise when $p = 2$.)

Case 3: $p = 3$

Then

$\displaystyle \frac{1}{q} + \frac{1}{r} = \frac{2}{3}$

and

$r \geq q \geq 3$

Case 3.1 $q = 3$

Then r = 3, and we've found the solution (p, q, r) = (3, 3, 3)

Case 3.2 $q \geq 4$

Then

$r \geq 4$

and

$\displaystyle \frac{1}{q} + \frac{1}{r} \leq \frac{1}{2}$

contradiction.

(Now we've covered all the cases that can arise when $p = 3$.)

Case 4:$p \geq 4$

Then $q, r \geq 4$

and

$\displaystyle \frac{1}{p} + \frac{1}{q} + \frac{1}{r} \leq \frac{3}{4}$

contradiction.

Now we've covered all the cases, and the only solutions are thus (p, q, r) = (2, 3, 6), (2, 4, 4) and (3, 3, 3).

(ii)

Assume integers p, q, r satisfy the inequality.

Case 1: $p = 1$

This has already been given in the question. (Besides i think there's a typo in the paper, it should be $1\leq n \leq m$. But who cares...)

Case 2: $p = 2$

Then the inequality becomes

$\displaystyle \frac{1}{q} + \frac{1}{r} > \frac{1}{2}$

and we of course also have $r \geq q \geq 2$.

Case 2.1: $q = 2$

Then the inequality becomes

$\displaystyle \frac{1}{r} > 0$

which is satisfied by any integer $r \geq 2$.

Case 2.2: $q = 3$

Then the inequality becomes

$\displaystyle \frac{1}{r} > \frac{1}{6}$

which for positive r is equivalent to r < 6

and hence all solutions for this case are given by (p, q, r) = (2, 3, 3), (2, 3, 4) and (2, 3, 5).

Case 2.3: $q \geq 4$

Then $r \geq 4$ and

$\displaystyle \frac{1}{q} + \frac{1}{r} \leq \frac{1}{2}$

Contradiction.

(Now we have covered all the cases that can arise when q = 2.)

Case 3: $p = 3$

Then $q, r \geq 3$ and so

$\displaystyle \frac{1}{p} + \frac{1}{q} + \frac{1}{r} \leq \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1$

contradiction.

Now all the cases have been covered, and hence the only solutions other than (1, m, n) are

(2, 2, r) where r can be any integer larger than or equal to 2,

(2, 3, 3)

(2, 3, 4)

and

(2, 3, 5).

I/4

I like pencil vv

N.B. these are rough solutions, I'd write much more explanation of what was happening in the real exam.

I like pencil vv

N.B. these are rough solutions, I'd write much more explanation of what was happening in the real exam.

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