UK Maths Challenge problems 23 and 25

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
Sign up Log in
8 replies · 2K views
Trail_Builder
Messages
148
Reaction score
0
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

Homework Statement



numbers 23 and 25

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

Homework Equations





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
 

Discussion

Physics news on Phys.org
For 23, note that they are similar triangles.
 
yeh i got that, but i still can't go from there :(
 
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.
 
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.
 
Jesus Australians in Yr 12 don't know anything about mods, Englands smart >.<
 
thnx, i should be able to do them now, thnx
 
wooo i did 23, thnx for you help buddy, ill get better as spotting stuff like that with practise :D

as for 25, we havnt done mods yet, and if that's 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?
 
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.