2 problems

1. Jan 26, 2007

Trail_Builder

I had to do this past paper from the uk maths challenge and could do all of them bar 23 and 25. Any help or starting points would be great :D

1. The problem statement, all variables and given/known data

numbers 23 and 25

http://www.wpr3.co.uk/UKMT/imc.html

2. Relevant equations

3. The attempt at a solution

23. I did some pythagorus, and found things in terms of PR, and ended up going in circles a few times, but fear i have missed a key concept. can someone tell me where to begin?

25. not sure where to begin. I 'could' in theory do in the long and very slow way of trial and error, but in a exam an hour long there must be a way to do it... just advice on where to begin would be nice :D

thnx

2. Jan 26, 2007

cristo

Staff Emeritus
For 23, note that they are similar triangles.

3. Jan 26, 2007

Trail_Builder

yeh i got that, but i still cant go from there :(

4. Jan 26, 2007

cristo

Staff Emeritus
Well, given that QPR and RPS are equal angles, then you can form two equations for the cosine of the angle, each expression containing the side PR, which is what you want to calculate.

5. Jan 26, 2007

Dick

For 25), I don't want to completely spoil your fun, so I'll just drop some hints.

i) Clearly j,k<=4 and l,m<=3.

ii) Since the powers are adding to an even number, k=0. Put that in and reduce the equation mod 5. This tells you what l is.

iii) Put that in and reduce mod 11. This gives you j. Now you know m. Bingo.

6. Jan 27, 2007

Gib Z

Jesus Australians in Yr 12 don't know anything about mods, Englands smart >.<

7. Jan 27, 2007

Trail_Builder

thnx, i should be able to do them now, thnx

8. Jan 27, 2007

Trail_Builder

wooo i did 23, thnx for ya help buddy, ill get better as spotting stuff like that with practise :D

as for 25, we havnt done mods yet, and if thats the only way ya'd do it in the real thing, ill just give it a pass.

unless mods are easily explainable and just a case of knowing it?

9. Jan 27, 2007

Dick

Modulo arithmetic is elementary. Let's do the first step of 25. Since j=0 we have 5^j+7^l+11^m=2005. Reducing this modulo 5 is just considering only remainders after division by 5. Assuming j<>0 this becomes 2^l+1^m=0 (since 7mod5=2 and 11mod5=1. This gives 2^l=(-1) mod 5. Why, 2^2=4, which has a remainder of -1 when divided by 5. So I'll try l=2. No other small number will work. Etc. This can be REALLY handy for these smart ass type questions.