AIMO Past Papers

[brightsky]

Does anyone have any AIMO past papers (preferably with solutions) that they are willing to post up (excluding the practice paper on the AMT website)? Thanks.

Also, good luck to anyone doing the AIMO this Thursday. 

[brightsky]

is a perfect square where is an integer. Find the largest possible value of . Any help appreciated.

[brightsky]

Are you able to give me a scanned copy of just 2007-2008? I'm working on them now.  

[brightsky]

is a perfect square where is an integer. Find the largest possible value of . Any help appreciated.

AIMO 2007 Q7.

Suppose x^2-19x+94 = m^2 where m is a non-neg integer. Then x^2-19x+94-m^2 = 0

Go from there.

Ah, solved. Thanks Azure!

Next:

5. Find where and are non-zerp solutions of the system of equations:




9. Find a prime, p, with the property that for some larger prime number, q, both 2q - p and 2q + p are prime numbers. Prove that there is only one such prime p.

[kamil9876]

Probably not the best way, but the equations given are ugly anyway :

Subtract from the first equation the second times 5. You get:








So either x=0. y=5x, or y=17, lolz. Now you can plug these in to the original equation to see what u get in each case.

[brightsky]

Probably not the best way, but the equations given are ugly anyway :

Subtract from the first equation the second times 5. You get:








So either x=0. y=5x, or y=17, lolz. Now you can plug these in to the original equation to see what u get in each case.

Ooh, thanks kamil!

I got a related question:

Real numbers are linked by the two equations:





Determine the largest value for .

[/0]

The Arithmetic-Quadratic mean inequality states that:



Using this,

So



Hmm well if this happens to be right so far then I guess you could solve for e.

EDIT: Oh wait, AM-QM only works for positive numbers

[kamil9876]

9. Find a prime, p, with the property that for some larger prime number, q, both 2q - p and 2q + p are prime numbers. Prove that there is only one such prime p.

p=3, q=5 works.

Now suppose that there exists some other prime with this property, call it p. Thus (it is easy to see p cannot be 2)

Thus



Case 1:

Then one of p or -p is -1 mod 3. Thus one of 2q-p or 2q+p is a multiple of 3. Note also that 3<q<2q-p<2q+p  Since q>q.
It follows that one of these numbers must be a multiple of 3 greater than 3, hence not prime.

The other case is basically the same.

[Ahmad]

Real numbers are linked by the two equations:





Determine the largest value for .

First I tried to reason by analogy, if instead of a, b, c, d, e we only had c, d, e then we can think about it geometrically in R^3. We can have e as the vertical axis, then we have a plane (first equation) and a sphere (second equation) intersecting in some sort of rotated circle in R^3 and we're trying to find the highest point on the circle, which we visually see occurs when c=d. By analogy we may expect that equality occurs when a=b=c=d, and generally when dealing with symmetric inequalities this is a good idea.

If a=b=c=d then the first equation gives us that a = b = c = d = 10 - e/2. We try to analyse the effect of deviation of the values of a, b, c, d away from 10 - e/2 (our suspected best point). So we let a = 10 - e/2 + a', and similarly for b,c,d, and we think of a' as representing a small deviation. Now we know that a' + b' + c' + d' = 0 from the first equation, this is very convenient. Plugging these into the second equation results in massive cancellation, and we're left with (assuming I did this right), which tells us our maximum occurs when 5e - 80 = 0, or e = 16.

It's interesting to think about why symmetry causes the massive cancellation, and why performing a change of coordinates (a,b,c,d) -> (a',b',c',d') to localize our variables at our suspected maximum is a good idea.

[shrek27]

https://drive.google.com/open?id=1E3l2RGQvOg4C0V-LtzacIIKc--psVw2z

These has all the past AIMO papers with SOLUTIONS from 1998 to 2017.

Enjoy.