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I want to see negative rep introduced

Re: HSC 2014 4U Marathon - Advanced Level hmmmm yeah i'm assuming there is some other condition to it? edit: thanks sy :)

Because all divisors of a number are essentially combinations of their prime factors

for each of the prime factors take the index plus 1 and times then together so 4x3x5x2=120

rawrliongirl i think the problem might be dealt with a bit if you don't keep checking your answers when you are doing hw You need to back yourself, commit to an answer, move on, do as many as you can then check them all later. Be honest with yourself where you are going wrong when you do then...

Re: HSC 2014 4U Marathon for part iii it's similar to part ii let x be any positive integer x= a1 + a2.10 + a3.10^2.........an.(10^(n-1)) = a1 + 10(a2 + a3.10 + a4.10^2 +.......... + an(10^n-2)) 10 is divisible by 5 therefore 10(a2 + a3.10 + a4.10^2 +.......... + an(10^n-2)) is divisible...

Re: HSC 2014 4U Marathon for the first part take any number x x= a1 + a2.10 + a3.10^2.........an.(10^(n-1)) = a1 + 9.a2 + a2 + 99a3 + a3...........+ 999999(n-2 9s)an + an = (a1 +a2 +a3 +.......+an) + 3(3a1 + 33a2 +...........33333333(n-2 3s)an) 3(3a1 + 33a2 +...........33333333(n-2 3s)an)...

they said they are in year 11, meaning they have already passed beyond the limit of AIMO which is, as you said, for students up to year 10 level

Re: HSC 2014 4U Marathon - Advanced Level where did you get it from btw? are you sure there is a solution in the scope of the 4u course haha?

Re: HSC 2014 4U Marathon (also im having a brain fart on that problem you posted in the Advanced thread, i can't seem to think of how to approach it)

Re: HSC 2014 4U Marathon haha it's ok :) i do think the mod arithmetic approach is better in terms of efficiency but obviously your solution is more within the scope of the course and so better for this thread

Re: HSC 2014 4U Marathon How so? Doesn't every odd number fit into those congruencies listed?

Re: HSC 2014 4U Marathon alternatively u can take the approach that every odd number is congruent to 1 mod 8 3 mod 8 5 mod 8 7 mod 8 Squaring it becomes 1 mod 8 9 mod 8 or 1 mod 8 25 mod 8 or 1 mod 8 49 mod 8 or 1 mod 8 Subtracting 1 from each gives 0 mod 8 i.e. all are divisible by 8

what makes you say that?

you know blowing people's avatars up isn't very nice

are all the jokes in that thread as bad as yours? :P

Anyone else aiming for a 99+ ATAR btw?

Probably tennis but i enjoy swimming too and most other sports

Oh well that seems like pretty legitimate criticism to be honest, i find it kind of hard not to get carried away with description at times

what's so dangerouss (i see what you did there) about it lol?