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HSC 2014 MX2 Marathon (archive) (3 Viewers)

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integral95

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Re: HSC 2014 4U Marathon

Gonna skip lots of lines, especially in the earlier parts, because this is extremely long.


hahaha got the equation too long message and had to split up part v



Lol for part iv) i think you meant the points (1,1) and (e,0)
 

Stygian

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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) is divisible by 3

(a1 +a2 +a3 +.......+an) is the sum of the digits of x

therefore if the sum of the digits of x is divisible by 3 then = (a1 +a2 +a3 +.......+an) + 3(3a1 + 33a2 +...........33333333(n-2 3s)an) is divisible by 3 and thus x is divisible by 3

for part ii

let x be any positive integer

x= a1 + a2.10 + a3.10^2.........an.(10^(n-1))

= a1 +a2.10 + 10^2(a3 + a4.10 + a5.10^2 +.......... + an(10^n-3))

10^2 is divisible by 4 (since 4 times 25 is 100)

so 10^2(a3 + a4.10 + a5.10^2 +.......... + an(10^n-3)) is divisible by 4

the last two digits of the number are given by a1 +a2.10

therefore if the last two digits of the number are divisible by 4, a1 +a2.10 is also divisible by 4 and therefore a1 +a2.10 + 10^2(a3 + a4.10 + a5.10^2 +.......... + an(10^n-3)) is divisible by 4

therefore xi s divisible by 4
 

Stygian

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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) is divisible by 3

(a1 +a2 +a3 +.......+an) is the sum of the digits of x

therefore if the sum of the digits of x is divisible by 3 then = (a1 +a2 +a3 +.......+an) + 3(3a1 + 33a2 +...........33333333(n-2 3s)an) is divisible by 3 and thus x is divisible by 3

for part ii

let x be any positive integer

x= a1 + a2.10 + a3.10^2.........an.(10^(n-1))

= a1 +a2.10 + 10^2(a3 + a4.10 + a5.10^2 +.......... + an(10^n-3))

10^2 is divisible by 4 (since 4 times 25 is 100)

so 10^2(a3 + a4.10 + a5.10^2 +.......... + an(10^n-3)) is divisible by 4

the last two digits of the number are given by a1 +a2.10

therefore if the last two digits of the number are divisible by 4, a1 +a2.10 is also divisible by 4 and therefore a1 +a2.10 + 10^2(a3 + a4.10 + a5.10^2 +.......... + an(10^n-3)) is divisible by 4

therefore xi s divisible by 4
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 by 5

if a1 is 5 then it is divisible by 5 and a1 + 10(a2 + a3.10 + a4.10^2 +.......... + an(10^n-2)) is divisible by 5

if a1 is 0 then a1 + 10(a2 + a3.10 + a4.10^2 +.......... + an(10^n-2)) is divisible by 5

therefore x is divisible by 5
 

aDimitri

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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) is divisible by 3

(a1 +a2 +a3 +.......+an) is the sum of the digits of x

therefore if the sum of the digits of x is divisible by 3 then = (a1 +a2 +a3 +.......+an) + 3(3a1 + 33a2 +...........33333333(n-2 3s)an) is divisible by 3 and thus x is divisible by 3

for part ii

let x be any positive integer

x= a1 + a2.10 + a3.10^2.........an.(10^(n-1))

= a1 +a2.10 + 10^2(a3 + a4.10 + a5.10^2 +.......... + an(10^n-3))

10^2 is divisible by 4 (since 4 times 25 is 100)

so 10^2(a3 + a4.10 + a5.10^2 +.......... + an(10^n-3)) is divisible by 4

the last two digits of the number are given by a1 +a2.10

therefore if the last two digits of the number are divisible by 4, a1 +a2.10 is also divisible by 4 and therefore a1 +a2.10 + 10^2(a3 + a4.10 + a5.10^2 +.......... + an(10^n-3)) is divisible by 4

therefore xi s divisible by 4
part 4 is essentially the same as part 1.
separate a1, a3, a5 etc. then assume that a1+ a3+ a5 etc = a2 + a4 + a6 + etc.
you can factorise out an 11 from there.

for the difference bit, you just add a2+a4+a6+... to the series and factor out an 11, in the process you have to subtract a2+a4+a6+...
this means you're left with an 11 factored out of one bit, and as a remainder you have a1 - a2 + a3 - a4 + a5 - a6...
therefore if the difference is divis by 11, the whole thing is divis by 11
 

go_cadel

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Re: HSC 2014 4U Marathon



... is this at all close? All I can think of doing is:

The first part was on the right track.
Rtp: (x+y)^2/xy + 2 >=6
Rtp: (x+y)^2>=4xy
But (x-y)^2>=0
Hence x^2 + y^2 >= 2xy
And (x+y)^2 >= 4xy as required
 

Kurosaki

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Re: HSC 2014 4U Marathon

Hey Sy, is the answer for this 0?

Drawing up a circle centre O and radius r and choosing any 3 random points for A, B, C, connect chords AB, BC and AC. Let angle ACB be .
Applying the sine rule and the fact that the angle at the centre of the circle is twice the angle at the circumference subtended by the same arc:

, which simplifies to give:

.

A simple application of the same procedure for the other chords yields the same inequality as above. but then that means a triangle will not exist as their sum will be less than pi.
 
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Kurosaki

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Re: HSC 2014 4U Marathon

Wait wait wait, I think I screwed up, it's also possible that you can have which guarantees they are all less than the radius...hmmm... about then? Don't know how to calculate it exactly though..
One would have to divide by 3 though as it could be any angle in the triangle, which would make it about ?

Idk I'm probably completely wrong.
 
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glittergal96

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Re: HSC 2014 4U Marathon

Simple trig shows that two points on the circumference have distance at most r if and only if the angle subtended by their respective radii is at most pi/3.

Because of rotational symmetry, the situation is the same as picking two numbers randomly (the signed angles BOA and COA), and asking what the probability is that all three angles AOB, BOC, COA are smaller than pi/3 in size.

For a given , must lie in an interval of length .

So visualising the full probability space as the rectangle in the plane, we calculate the probability by finding the area of the subregion that obeys the angle condition and dividing by the rectangles area. This calls for integration:



(I could probably explain this better but hopefully it's clear what I'm doing.)
 

Sy123

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Re: HSC 2014 4U Marathon

I got glittergal96's answer

Originally when I posted this question I simply had a glance and thought it was simple common sense, but the problem is more complex than that and probably should of been in the advanced thread

What I do is fix one point (symmetry), then I let one point be a variable point, let its angle between it and the original fixed point be . In order for a length to be less than the radius, the differences between angles must be within . Which yields an integral that is the same as glittergal96

However what I did was:



So when doing integration, I did:





So lets take A as the fixed point, the probability that C is within range of A and B for some B in the range of A is 1/4 (i.e. 1/4 is equal to the bracket product)

But we still have to compute the probability that B is IN range of A, and that is simply 1/3 (the Pr outside of brackets)

Multiply them and we get 1/12
 
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Re: HSC 2014 4U Marathon

Need assistance with an inequality question (have not done this with my teacher yet, only looking through past papers and understanding the solutions) and this might be easy to you guys.

You are given - a + 1/a greater than or equal to 2.

Show that if a>0, b>0, c>0 are real numbers, then

i) b+c/a + c+a/b + a+b/c is greater than or equal to 6.
 

go_cadel

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Re: HSC 2014 4U Marathon

Need assistance with an inequality question (have not done this with my teacher yet, only looking through past papers and understanding the solutions) and this might be easy to you guys.

You are given - a + 1/a greater than or equal to 2.

Show that if a>0, b>0, c>0 are real numbers, then

i) b+c/a + c+a/b + a+b/c is greater than or equal to 6.
b/a + a/b>=2 (using the given a+1/a >=2)
now similarly c/a + a/c >=2
and b/c+c/b >=2
Summing the above gives the required inequality.
 
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Re: HSC 2014 4U Marathon

b/a + a/b>=2 (using the given a+1/a >=2)
now similarly c/a + a/c >=2
and b/c+c/b >=2
Summing the above gives the required inequality.
Ohhhhhhhhhhhh how stupid of me, can't believe i didn't see that. b/a + a/b = b^2 + a^2/ab which is greater than two from (a-b)^2>0
And then similarly for the rest
 
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