jb4 Posted May 17, 2006 Share Posted May 17, 2006 hey rust8y this is from the maths challenge 2006 intermediate booklet and it does say clearly no help from other people unless they r doing the same booklet as u... I'm doing the same booklet mate. Link to comment Share on other sites More sharing options...
ashennell Posted May 17, 2006 Share Posted May 17, 2006 Yeah' date=' the answer to question 2 is 190- its factors are 1, 2, 5, 38, 19, 10Fenders- 1, 2, 5, 8, 9, 0[/quote'] Don't you need to show that all numbers with these fenders will have four more? Instead of just an example? Link to comment Share on other sites More sharing options...
girl Posted May 18, 2006 Share Posted May 18, 2006 a. 1-fender which is composite is 121 b. 90 c. 270 Link to comment Share on other sites More sharing options...
jb4 Posted May 18, 2006 Share Posted May 18, 2006 Don't you need to show that all numbers with these fenders will have four more? Instead of just an example? Fair point mate. Link to comment Share on other sites More sharing options...
ashennell Posted May 18, 2006 Share Posted May 18, 2006 Fair point mate. Althought, I think the fenders you got are the right ones. i.e. 1,2,5,8,9,0 ...it took me a while to work that out. It isnt too hard to show that there are always these fenders. I'm not doing this problem sheet thing so I shouldnt provide answers. Link to comment Share on other sites More sharing options...
123456789 Posted May 18, 2006 Share Posted May 18, 2006 is there any numbers with ONLY 0,1,2,3,4,5,6,7 and 9 as fenders??? plz reply Link to comment Share on other sites More sharing options...
ashennell Posted May 18, 2006 Share Posted May 18, 2006 is there any numbers with ONLY 0,1,2,3,4,5,6,7 and 9 as fenders??? plz reply No, because as part of Q2 we need to show that any number that has 0,9 as fenders will also ahve four more. I one which is 8. So any number with 0 an 9 as fender must have 8 and hence the combination you require is impossible. Edit : sorry this is wrong, I will leave it here anyway. Edit : no it is correct after all. I need more coffee .... Link to comment Share on other sites More sharing options...
ashennell Posted May 18, 2006 Share Posted May 18, 2006 The proof (bearing in mind im not a mathematician at all) is something like: Any factors ending in 9, times 2, is a factor ending in 8, e.g. 19*2=38. 2 and 5 are both factors of 10 (and any number ending in 0). So any number with 9 and 0 as fenders, denoted a and b, can be rexpressed as a*2*(b/2) Link to comment Share on other sites More sharing options...
ashennell Posted May 18, 2006 Share Posted May 18, 2006 While im on this thread, I want to apoplogise to Matt for the comment I made earlier. It was very rude of me. I was just having one of those days where everything seemed to annoy me. Now that the fender thing has 'clicked' it is pretty obvious from the first post. I had never come across the 'ender' terminology and more importantly wouldnt of thought that grouping numbers by their final digit was significant or even useful. Perhaps this is why I couldnt get it initially. Link to comment Share on other sites More sharing options...
jeska Posted May 18, 2006 Share Posted May 18, 2006 well, who isn't doing the math challenge. alas, me too. perhaps i can help with question 4. i got as a 9 fender under 1000 540 (0,1,2,3,4,5,6,8,9) but you should check though.... Link to comment Share on other sites More sharing options...
blorg glorg Posted May 20, 2006 Share Posted May 20, 2006 u said that its fender is 8 cuz its divisble by 38? well i randomly tried 159x140 and divided it by 38 but it didnt work cuz it went into heaps of decimal places but i no 8 is rite Link to comment Share on other sites More sharing options...
blorg glorg Posted May 20, 2006 Share Posted May 20, 2006 u said that its fender is 8 cuz its divisble by 38? well i randomly tried 159x140 and divided it by 38 but it didnt work cuz it went into heaps of decimal places but i no 8 is rite oh this last message was to jb4 Link to comment Share on other sites More sharing options...
blorg glorg Posted May 20, 2006 Share Posted May 20, 2006 oh and can any1 help on q4 four vehicles i cant get bii) the question is: a car a van a truck and a bike r all travelling in the same direction on the same road, each at its own constant speed. at 10am, the car overtakes the van; at noon, it overtakes the truck;at 2pm it overtakes the bike. at 4pm the truck overtakes the bike and, at 6pm the van overtakes the truck. a) the speed of the car is 120km/h and the speed of the trck is 80km/h. i) find the speed of the van and the bike. ii)find the time at which the van overtakes the bike. b)let c and T represent the speeds, in km.h, of the car and the truck respectively. i) find the speed of the van and bike in terms of c and T. ii) show that the time when the van overtakes the bike is the same, regardless of the speed of the car and the truck.(this is the question im stuck on) Link to comment Share on other sites More sharing options...
novusnet3 Posted May 24, 2006 Share Posted May 24, 2006 Hey! People should've asked "Rust8y" why he needed the answers to these problems. Why? He is parts of the Australian Maths Challenge and wishes to win a cash prize by getting the answers from the internet. Rust8y, I reckon the Australian Mathematics Trust will be pretty interested in this site. Link to comment Share on other sites More sharing options...
novusnet3 Posted May 24, 2006 Share Posted May 24, 2006 jeska, ashnell, u guys r unaustralian. thats not the way we do things here. go back to where-ever. Link to comment Share on other sites More sharing options...
novusnet3 Posted May 24, 2006 Share Posted May 24, 2006 evr1's an immigrant Link to comment Share on other sites More sharing options...
blorg glorg Posted May 24, 2006 Share Posted May 24, 2006 no1s really interested in this thread anymore so whocares Link to comment Share on other sites More sharing options...
Phi for All Posted May 24, 2006 Share Posted May 24, 2006 Hey! People should've asked "Rust8y" why he needed the answers to these problems. Why? He is parts of the Australian Maths Challenge and wishes to win a cash prize by getting the answers from the internet. Rust8y, I reckon the Australian Mathematics Trust will be pretty interested in this site.Yes, because we have oodles of time to follow up on these kinds of things. Let's close this one. Link to comment Share on other sites More sharing options...
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