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no if you view the thread a number of times it counts towards the total as different views.

ok there's no full solution cos it's just a guess: my guess is X(3) = 11 where X(N) = max X for N weighings. And the recurrence is X(n+1) = 3X(n) - 1 or something like that someone can solve the recurrence for me :D

well?

rules are: Ox#(H) = +1 Ox#(O) = -2 And sum of Ox# = charge on ion. So: a) H2S2O7 2Ox#(H) + 2Ox#(S) + 7Ox#(O) = 0 2Ox#(S) = 0 - 2*(+1) - 7*(-2) = 12 Ox#(S) = 6 b) SO3(2-) Ox#(S) + 3Ox#(O) = -2 Ox#(S) = -2 - 3*(-2) = 4

Yes and yes However you don't need to learn the 2unit course (unlike 4u english), even tho it may be considered pre-requisite knowledge.

seems like 56 now...

sorry, u're right with 15, 18, 51, 81. Silly mistake number 17365 for me. As for case 3, there are only 3 cases: either 1, 2, or 3 digits out of the three, that are not divisible by 3, 1 digit definitely won't give us what we want cos if we only had one digit not divisible by 3, the whole...

Well firstly I did work the problem out properly, so X = 2^N is not a guess. I just haven't posted the whole solution so others can have a go. And besides the reason why you can't do 9 coins when N=3 is the same reason why you can't do 5 coins with N=2. Show me how to do 9 coins with 3...

1) For divisibility test for 3 is that the sum of its digits must be divisible by 3. It doesn't matter which order the digits are in. Also, 1 gives remainder 1, and 5, 8 both give remainder 2 when divided by 3. Case1: single figures 3, 9 # = 2 Case2: double figures 39, 93 # = 2...

how?

X = 2^N ?

You mean you want P(AB, BC, AC, ABC) ? Hmm...this is 1 - P(A,B,C separate) P(A,B,C separate) = 4*3*2/4*4*4 = 6/16 = 3/8 So P(AB, BC, AC, ABC) = 1-3/8 = 5/8 I don't know how you got 20 arrangements, there should be 64 (4*4*4), 4 different options to place each of A,B,C.

What things? Depends which subject you mean. For the sciences, understanding and remembering may not be enough. You'll prolly want to write model answers to common questions, which involves going thru past trial papers and doing them. For 4umaths, once you've solved a hard problem, go...

i knew this guy, who's a good friend of mine, who seriously believed that work was the only thing that he needed to do to get good marks. So he did (like to the extreme of 6+ hours), but there were subjects that he still screwed up, because he wasn't studying properly (he didn't do a single...

hell no, i mean i know a lot of ppl that didn't need to work as hard... work varies with people, and i know a chronic bludger who i reckon was smart enough to get 99+ if he bothered but he didn't, but other than that extreme I also know a lot of people who i reckon didn't need to work as...

I SO disagree with this. Well, depends what you mean by "hard" work, but we can't generalise that everyone who got 99 has worked "hard" by their standards. Most people need to work hard, but some people seriously don't.

isn't Alan Fels "structurally unemployed"?

Re: here's another............ ridiculously ambiguous. you can interpret this question in many different ways. Assuming the people are called A, B, C "any one of 3 people", does that mean any of A, B, C? or does it mean any particular person (e.g. just A)? "be in the same group" as...

dunno, haven't tried, dun wanna even think about doing so...arrggh :chainsaw:

did you double the answer cos you need to consider the area of the whole loop, not just the half above the x axis?