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.

Yeah' date=' the answer to question 2 is 190- its factors are 1, 2, 5, 38, 19, 10

Fenders- 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?

a. 1-fender which is composite is 121

b. 90

c. 270

Don't you need to show that all numbers with these fenders will have four more? Instead of just an example?

 

Fair point mate.

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.

is there any numbers with ONLY 0,1,2,3,4,5,6,7 and 9 as fenders??? plz reply

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 ....

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)

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.

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....

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

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

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)

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.

jeska, ashnell, u guys r unaustralian. thats not the way we do things here. go back to where-ever.

evr1's an immigrant

no1s really interested in this thread anymore so whocares

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.