The Ultimate Maths Competition Thread

Original post by physicsmaths

I will write something up about them later when I am free :smile:

.

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Thnx xD

Reply 141

Original post by gasfxekl

give an example? you must always plug in easy values like x,0, x,-x etc. then another principle is to make things cancel- for example if u have xf(y)+y somewhere try subbing x=a/f(y) so that u get a+y, etc.
if you have a symmetric expression switch around x and y.
Note that f(f(x))=x or 6x or whatever implies injectivity! try to get injectivity as it is often very useful.

could you help with bmo1 1992 q5 pls

Reply 142

Original post by 11234

could you help with bmo1 1992 q5 pls

So another thing you should keep in mind is finding important values (like

f(1)

).

In this case observe that

f(n+1)>f(n)

implies that

f(x+n)\geq f(x)+n

, and since

f(1) \geq 1

we have

f(n) \geq n

. In fact

f(n) > n

since

f(n)=n \Leftrightarrow f(f(n))=f(n)=3n=n

, contradiction.
Therefore

3n=f(f(n)) > f(n)

.
Furthermore

f(3n)=f(f(f(n))=3f(n)

.
Now

1<f(1)<3

, so

f(1)=2

f(2)=f(f(1))=3, f(3)=3f(1)=6, f(6)=3f(2)=9

, so

f(6)=9>f(5)>f(4)>f(3)=6

gives

f(4)=7, f(5)=8

We now spot something interesting, and prove it by induction:

3^k\leq n \leq 2\cdot 3^k \Rightarrow f(n)=3^k+n

.
For multiples of 3 we have

3^k \leq 3l \leq 2\cdot 3^k \Leftrightarrow 3^{k-1}\leq l \leq 2\cdot 3^{k-1}

, so

f(3l)=3f(l)=3(l+3^{k-1})

. Now we have shown it for the multiples of 3. Now

f(3l)=3l+3^k <f(3l+1)<f(3l+2)<f(3l+3)=3l+3+3^k

so we have it for all values.

Finally, if

2\cdot 3^k<n<3^{k+1}

, we obtain

f(n)=3(n-3^k)

by using the previous part and

f(f(n))=3n

.

So

f(1992)=3(1992-3^6)=3789

Reply 143

Original post by gasfxekl

So another thing you should keep in mind is finding important values (like

f(1)

).

In this case observe that

f(n+1)>f(n)

implies that

f(x+n)\geq f(x)+n

, and since

f(1) \geq 1

we have

f(n) \geq n

. In fact

f(n) > n

since

f(n)=n \Leftrightarrow f(f(n))=f(n)=3n=n

, contradiction.
Therefore

3n=f(f(n)) > f(n)

.
Furthermore

f(3n)=f(f(f(n))=3f(n)

.
Now

1<f(1)<3

, so

f(1)=2

f(2)=f(f(1))=3, f(3)=3f(1)=6, f(6)=3f(2)=9

, so

f(6)=9>f(5)>f(4)>f(3)=6

gives

f(4)=7, f(5)=8

We now spot something interesting, and prove it by induction:

3^k\leq n \leq 2\cdot 3^k \Rightarrow f(n)=3^k+n

.
For multiples of 3 we have

3^k \leq 3l \leq 2\cdot 3^k \Leftrightarrow 3^{k-1}\leq l \leq 2\cdot 3^{k-1}

, so

f(3l)=3f(l)=3(l+3^{k-1})

. Now we have shown it for the multiples of 3. Now

f(3l)=3l+3^k <f(3l+1)<f(3l+2)<f(3l+3)=3l+3+3^k

so we have it for all values.

Finally, if

2\cdot 3^k<n<3^{k+1}

, we obtain

f(n)=3(n-3^k)

by using the previous part and

f(f(n))=3n

.

So

f(1992)=3(1992-3^6)=3789

Thanks for the reply but Im still quite confused as to how it implies nf(n) dop you have any more questions i could try

Reply 144

Original post by 11234

Thanks for the reply but Im still quite confused as to how it implies nf(n) dop you have any more questions i could try

what do you mean how it implies nf(n)?
yes you can try: find all functions

f: \mathbb{Q} \to \mathbb{R}: f(x+y)=f(x)f(y)

or
find all functions

f: \mathbb{R} \to \mathbb{R}: f(xy)=xf(y)-yf(x)

or
find all functions

f: \mathbb{Q} \to \mathbb{Q}: f(x+y)=f(x)f(y)-f(x)-f(y)+1, f(1)=2

Reply 145

Original post by gasfxekl

what do you mean how it implies nf(n)?
yes you can try: find all functions

f: \mathbb{Q} \to \mathbb{R}: f(x+y)=f(x)f(y)

or
find all functions

f: \mathbb{R} \to \mathbb{R}: f(xy)=xf(y)-yf(x)

or
find all functions

f: \mathbb{Q} \to \mathbb{Q}: f(x+y)=f(x)f(y)-f(x)-f(y)+1, f(1)=2

How does f(3n) imply 3f(n)

Reply 146

Original post by 11234

How does f(3n) imply 3f(n)

well

f(f(a))=3a

, if

a=f(n)

this implies

f(f(f(n)))=3f(n)

, on the other hand

f(f(f(n)))=f(3n)

since

f(f(n))=3n

Reply 147

Original post by gasfxekl

well

f(f(a))=3a

, if

a=f(n)

this implies

f(f(f(n)))=3f(n)

, on the other hand

f(f(f(n)))=f(3n)

since

f(f(n))=3n

Could you help with f(xy)=f(x)+f(y) Ive got to f(x^n)=nf(x) but how do I solve for x. I cant see a substitution...

Reply 148

Original post by 11234

Could you help with f(xy)=f(x)+f(y) Ive got to f(x^n)=nf(x) but how do I solve for x. I cant see a substitution...

its the other way around! its f(x+y)=f(x)f(y)

Reply 149

Original post by gasfxekl

its the other way around! its f(x+y)=f(x)f(y)

How would I solve the other one apparently the answer is f(x)=f(1)log(x)

Reply 150

Anyone else planning on spending the summer doing BMO problems?

Reply 151

Original post by Maths465Man

Anyone else planning on spending the summer doing BMO problems?

Yep definitely need to improve do you wanna discuss any problems

Reply 152

Original post by 11234

Yep definitely need to improve do you wanna discuss any problems

Definitely. I going to make my summer plan tomorrow which should work out at about three hours a day on BMO problems. I'm thinking of starting at some of the older papers as there are less BMO1 questions on each older paper (only 5) and I'm guessing the latest papers will be harder. For BMO2, I'll start as soon as I'm getting 50+ on each BMO1 paper. I've managed it on a few BMO1 papers, but I've been really inconsistent lately with my scores making me think I need to focus more time on BMO1 again.

Reply 153

Original post by Maths465Man

Silly question but how can you get 50+ if there are only questions each carrying 10 marks

Reply 154

Original post by 11234

Silly question but how can you get 50+ if there are only questions each carrying 10 marks

Well there are 6 questions each worth 10 marks and I'm hoping to get at least 50 or more from the possible 60 marks.

Reply 155

Original post by Maths465Man

Well there are 6 questions each worth 10 marks and I'm hoping to get at least 50 or more from the possible 60 marks.

Oh I'm sure some papers had 5...

Reply 156

username1162972

Original post by 11234

Oh I'm sure some papers had 5...

No those are rather old.

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Reply 157

Aldrstream

@Maths465Man Nice little thread you've started here, you seem to be pretty active in olympiad maths, out of curiosity did you get into any of the summer schools?

Reply 158

Original post by physicsmaths

No those are rather old.

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I know this is unrelated but what jobs are you looking at after a maths degree because I really wanna do a maths degree but Im not sure on employment opportunities

Reply 159