STEP Maths I, II, III 1997 Solutions

I
Q1

Possibilities to obtain 10p by using coins of 1, 2, 5 and 10p

10
5+5
5+2+2+1
5+2+1+1+1
5+1+1+1+1+1
2+2+2+2+2
2+2+2+2+1+1
2+2+2+1+1+1+1
2+2+1+1+1+1+1+1
2+1+1+1+1+1+1+1+1
1+1+1+1+1+1+1+1+1+1

Possibilities to obtain 20p by using 1, 2, 5 and 10p coins:
20=10+10 Hence we can see that there are 11 possibilities to obtain it if added with a 10.

Now we will have;
5+5+5+5
5+5+5+2+2+1 etc. 2 times more
5+5+2+2+2+2+2, 5+5+2+2+2+2+1+1 etc. 4 times more
5+2+2+2+2+2+2+2+1 etc. 7 times more
2+2+2+2+2+2+2+2+2+2 etc. 10 times more

i.e. total answer =41
And, as DFranklin pointed out later, I missed the obvious 20p :p:

Edited: corrected stupid thinking mistake, should not just work on a STEP question for 5min :p:

Edit2: Now it shall be OK, and when checking in Siklos I find he writes something that is wrong (only finds 6ways to get 5+....) but the final answer in there is the same.

Reply 2

Ive got STEP II Question 3 to write up. Ill do it abit later on :smile:

.

I didnt see you posting this oops! Ill get a mod to delete the one i started.

Reply 3

dvs

10

Where are the papers?

You guys are going to eat up all my free time. :p:

Reply 4

Papers coming, upload is sloooow... :p:

insparato: Actually yours was first, so I should be the one to say sorry. But I just thought that you might want some relief from having to update the thread, I mean I just happen to got too much free time right now so I might just as well do it, you concentrate on doing the questions instead. :p:

Reply 5

Ukgea can attach them or you could pm me and ill give you the link.

I dont think im allowed to provide you it on TSR threads.

As Nicholas has tried to do above :smile:

.

Reply 6

ukgea

Papers coming, upload is sloooow... :p:

insparato: Actually yours was first, so I should be the one to say sorry. But I just thought that you might want some relief from having to update the thread, I mean I just happen to got too much free time right now so I might just as well do it, you concentrate on doing the questions instead. :p:

Hehe no prob :p:

sure if you've got the free time. I managed to get Question 3 on STEP II out today so gonna type that out now.

Reply 7

Meh!

"STEP I 1997.pdf:
Your file of 1.32 MB bytes exceeds the forum's limit of 609.7 KB for this filetype.
STEP II 1997.pdf:
Your file of 1.17 MB bytes exceeds the forum's limit of 609.7 KB for this filetype.
STEP III 1997.pdf:
Your file of 1.33 MB bytes exceeds the forum's limit of 609.7 KB for this filetype."

How were your files so small, insparato?

Reply 8

There were all under 600kb just lucky i guess.

Reply 9

shall we upload them to rapidshare or something instead? (Although that would mean we had to download the files to the harddrive...)

Reply 10

Kolya

17

STEP II Question 1

Note that the first term + last term, 2000+5777 = 7777. The second term + last-1'th term, 2002+5775 = 7777, and so on.

We only have to consider the number of terms in the 2000's and 5000's. As we are adding the terms up, we need to find the number of terms in the 2000's.

Looking at the progression:
2000
2002
2005
2007

2020
2022
2025
2027

2050
2052
2055
2057

2070
2072
2075
2077
etc...

There are 16 terms with the first two digits, and this occurs 4 times with each different hundreds number, so there are 64 "sets" of 7777.

Therefore:

64\cdot 7777 = 2^6 \cdot 7 \cdot 1111 = 2^6 \cdot 7 \cdot 11 \cdot 101

Done.

Reply 11

Perhaps. If you want to do it, then fine. Otherwise we could just have people PMing around the links, besides I think most people who are actually doing STEP revision know where to find them anyway.

Reply 12

ukgea

Perhaps. If you want to do it, then fine. Otherwise we could just have people PMing around the links, besides I think most people who are actually doing STEP revision know where to find them anyway.

Yes, I think so too, but it seemed like dvs wanted them, I'll see if I upload them to rapidshare soon, otherwise someone could PM him, and others wanting it (if that isn't done...)

Reply 13

dvs

10

nota bene

Yes, I think so too, but it seemed like dvs wanted them, I'll see if I upload them to rapidshare soon, otherwise someone could PM him, and others wanting it (if that isn't done...)

I was sent the link. :smile:

Will look over them after I have some lunch.

Reply 14

DFranklin

18

Nota Bene - I think you're just missing the "obvious" solution of just using the 20p! As I happen to have an approach I think can be useful in similar problems, I hope no-one minds if I give it here:

Step I, Q1, 1997.

Let P(n,k) be the no. of ways of making n using coins <=k. We make the following observations:

P(n,1) = 1 for all n.

P(1,2) = 1, P(2,2) = 2, P(n,2)=floor(n/2)+1 (using 0,1,2,...,floor(n/2) 2's).

P(10,5) = P(10,2)+P(5,2)+1 = 6 + 3 + 1 = 10 (first term is using no 5's, second term is using one 5, last term is using 2 fives. We will use a similar construction repeatedly later).
P(10,10) = P(10,5)+1 = 11. (first term using no 10s, 2nd term using 1 10).

For the 2nd part:

P(20,5)=P(20,2)+P(15,2)+P(10,2)+P(5,2)+1 = 11 + 8 + 6 + 3 + 1 = 29
P(20,10) = P(20,5)+P(10,5)+1 = 29 + 10 + 1 = 40
P(20,20) = P(20,10)+1 = 41.

Reply 15

DFranklin

18

I'll do Paper I Q3,6,7.

Reply 16

DFranklin

Nota Bene - I think you're just missing the "obvious" solution of just using the 20p!

Yeah, LOL

Reply 17

STEP II Question 3

\frac{ax+b}{x^2+2x+2} + \frac{cx+d}{x^2-2x+2} = \frac{1}{x^4+4}

Therefore

(ax+b)(x^2-2x+2) + (cx+d)(x^2+2x+2) = 1

Expand out

ax^3 -2ax^2 + 2ax + bx^2 -2bx + 2b + cx^3 +2cx^2 + 2cx + dx^2 + 2dx + 2d = 1

(a+c)x^3 (-2a+b+2c+d)x^2 + (2a-2b+2c+2d)x + 2b + 2d = 1

Comparing coefficients

a + c = 0

- 1

-2a+b+2c+d = 0

- 2

2a-2b+2c+2d = 0

- 3

2b+2d = 1

- 4

a = -c

- Substituting this into 2 and 3

b + 4c + d = 0

b = d

- Substituting this into 4

4b = 1

b = \frac{1}{4}

4d = 1

d = \frac{1}{4}

Substituting values of b and d into 2

\frac{1}{2} + 4c = 0

c = -\frac{1}{8}

- Substituting back into 1

a = \frac{1}{8}

So

a = \frac{1}{8}

b = \frac{1}{4}

c = -\frac{1}{8}

d = \frac{1}{4}

Therefore

\frac{1}{x^4+4} = \frac{\frac{1}{8}x+\frac{1}{4}}{x^2+2x+2} + \frac{\frac{1}{4}-\frac{1}{8}x}{x^2-2x+2}

= \frac{x+2}{8x^2+16x+16} + \frac{2-x}{8x^2-16x+16}

\displaystyle \int \frac{1}{x^4+4} = \int \frac{x+2}{8x^2+16x+16} + \frac{2-x}{8x^2-16x+16}

\displaystyle = \frac{1}{8}\int \frac{x+2}{x^2+2x+2} + \frac{2-x}{x^2-2x+2}

Consider

\displaystyle \int \frac{x+2}{x^2+2x+2}

\displaystyle \int \frac{x+2}{x^2+2x+2} = \int \frac{2x+2}{x^2+2x+2} - \frac{x}{x^2+2x+2}

= ln(x^2+2x+2) - \int \frac{x}{x^2+2x+2}

Consider

\displaystyle \int \frac{x}{x^2+2x+2}

\displaystyle \int \frac{x}{x^2+2x+2} = \int \frac{2x+2}{x^2+2x+2} - \frac{2}{x^2+2x+2} - \frac{x}{x^2+2x+2}

\displaystyle 2\int \frac{x}{x^2+2x+2} = ln (x^2+2x+2) - \int \frac{2}{x^2+2x+2}

\displaystyle = ln (x^2+2x+2) - \int \frac{2}{(x+1)^2+1}

let u = x+1

\frac{du}{dx} = 1

du = dx

= ln (x^2+2x+2) - 2\int\frac{1}{u^2+1}

= ln(x^2+2x+2) - 2arctanu

= ln(x^2+2x+2) - 2arctan(x+1)

\displaystyle 2\int \frac{x}{x^2+2x+2} = ln(x^2+2x+2) - 2arctan(x+1)

\displaystyle \int \frac{x}{x^2+2x+2} = \frac{1}{2}ln(x^2+2x+2) - arctan(x+1)

Therefore

\displaystyle \int \frac{x+2}{x^2+2x+2} = ln(x^2+2x+2) - [\frac{1}{2}ln(x^2+2x+2) - arctan(x+1)]

= \frac{1}{2}ln(x^2+2x+2) + arctan(x+1)

Consider

\displaystyle \int \frac{2-x}{x^2-2x+2}

\displaystyle \int \frac{2-x}{x^2-2x+2} = \int \frac{2x-2}{x^2-2x+2} +\int \frac{4}{x^2-2x+2} - \int \frac{3x}{x^2-2x+2}

\displaystyle = ln (x^2-2x+2) + 4\int \frac{1}{(x-1)^2+1} - 3 \int \frac{x}{x^2-2x+2}

Consider

\displaystyle \int \frac{1}{(x-1)^2+1}

u = x-1

du = dx

\displaystyle \int \frac{1}{(x-1)^2+1} = \int \frac{1}{u^2+1}

= arctan(x-1)

\displaystyle \int \frac{2-x}{x^2-2x+2} = ln(x^2-2x+2) +4arctan(x-1) - \int \frac{3x}{x^2-2x+2}

Consider

\displaystyle \int \frac{3x}{x^2-2x+2}

\displaystyle \int \frac{3x}{x^2-2x+2} = \int \frac{2x-2}{x^2-2x+2} + \int \frac{x}{x^2-2x+2} + \int \frac{2}{x^2-2x+1}

2\displaystyle \int \frac{x}{x^2-2x+2} = ln (x^2-2x+2) + 2arctan(x-1)

\displaystyle \int \frac{x}{x^2-2x+2} = \frac{1}{2}ln (x^2-2x+2) + arctan(x-1)

Therefore

\displaystyle \int \frac{2-x}{x^2-2x+2} = ln(x^2-2x+2) +4arctan(x-1) -3[\frac{1}{2}ln (x^2-2x+2) + arctan(x-1)]

Finally

\displaystyle \int \frac{1}{x^4+4} = \frac{1}{8} \int \frac{x+2}{x^2+2x+2} + \frac{2-x}{x^2-2x+2}

\\ = [\frac{1}{8}ln(x^2+2x+2) -\frac{1}{8}(\frac{1}{2}ln(x^2+2x+2)-arctan(x+1)) + \frac{1}{8}ln(x^2-2x+2) + \frac{1}{2}arctan(x-1) - \frac{3}{8}(\frac{1}{2}(ln(x^2-2x+2) + arctan(x-1)]_0^1

\\ = [\frac{1}{8}ln5 - \frac{1}{8}(\frac{1}{2}ln5 - arctan2)+ \frac{1}{8}ln1 + \frac{1}{2}ln0 - \frac{3}{8}(\frac{1}{2}ln1+arctan0)] - [\frac{1}{8}ln2 - \frac{1}{8}(\frac{1}{2}ln2 - arctan(1))+\frac{1}{8}ln2 + \frac{1}{2}arctan(-1) -\frac{3}{8}(\frac{1}{2}ln2+arctan(-1))

\\= [\frac{1}{16}ln5 +\frac{1}{8}arctan(2)] - [\frac{1}{16}ln2 + \frac{\pi}{32} + \frac{1}{8}ln2 -\frac{\pi}{8} - \frac{3}{16}ln2 +\frac{3\pi}{32}]

= \frac{1}{16}ln5 + \frac{1}{8}arctan2

This has been a nightmare typing this up as i dont exactly write in a logical order when its a big question on paper. Ill read through it just to check it. Although it does get the answer.

Reply 18