Solution $\boxed{15}$ Suppose $a_1,...,a_n$ are the number of gems of each type. If they are arranged serially like, all $a_1$ gems of type $1$ at first, all $a_2$ gems of type $2$ next, and so on till gems of type $n$, then clearly we'll need $n$ cuts to split evenly (each cut at the middle of eac...

- Thu Mar 02, 2017 9:20 pm
- Forum: Combinatorics
- Topic: Combi Marathon
- Replies:
**41** - Views:
**1579**

John Gabriel for the win. :3

Those who haven't heard of this great mathemagician, come to the path of light.

Those who haven't heard of this great mathemagician, come to the path of light.

- Sun Feb 26, 2017 11:57 pm
- Forum: Social Lounge
- Topic: Favorite mathematician?
- Replies:
**30** - Views:
**3021**

Solution $\boxed{52}$ Let $P(x,y)$ denote $g(f(x+y))=f(x)+(2x+y)g(y)$. If $g$ is constant then setting $g\equiv c$ we see that $f(x)=(1-2x-y)c$ for all $(x,y)\in\mathbb R^2$, implying $c=0$. So $f\equiv 0$ and $g\equiv 0$. This...

- Mon Feb 06, 2017 7:45 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies:
**184** - Views:
**20435**

Solution $\boxed{51}$ Screams probabilistic approach. Take a random set $B$. Clearly $|A+B|=N^2$, we let $X$ to be the number of distinct elements in $A+B$. For some fixed residue $r$ we have $N$ possible elements to complement with an element from $A$, precisely $r-A_i$ for $1\le i\le N$. So the p...

- Sat Feb 04, 2017 2:48 am
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies:
**184** - Views:
**20435**

Let $\triangle (a,b,c)$ denote that lengths $a,b,c$ form a valid triangle. Let $P(a,b)$ denote $\triangle\left(a,f(b),f\left(b+f(a)-1\right)\right)$. Easy to prove: $\triangle (1,a,b)\Rightarrow a=b$ and $\triangle (2,a,b)\Rightarrow \left|a-b\right|\le 1$. Now $P(1,b)\Rightarrow f\left(b+f(1)-1\rig...

- Wed Feb 01, 2017 9:04 pm
- Forum: Algebra
- Topic: 2009 IMO SL A3
- Replies:
**2** - Views:
**163**

Solution $\boxed{49}$ The key fact is that for any positive integer $d$ the set of edges divisible by $d$ forms a clique. Indeed, for any positive integer $d$ if there's only one edge divisible by $d$ then it's an induced $K_1$. Otherwise take two edges $\alpha d$ and $\beta d$. For any edg...

- Sun Sep 04, 2016 3:29 am
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies:
**184** - Views:
**20435**

The $\dbinom n 2$ consecutive natural numbers don't necessarily have to be the first $\dbinom n 2$ natural numbers.

- Sat Sep 03, 2016 3:56 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies:
**184** - Views:
**20435**

Solution $\boxed{48}$ Suppose $\displaystyle a_i = \prod_{j=1}^k p_j^{e_{i_j}}$ for each $1\le i\le n.$ Take $n$ distinct primes $d_1,d_2,...,d_n.$ Then we can take $b=\displaystyle\prod_{j=1}^k p_j^{\alpha_j}$ where $\alpha_j+e_{i_j}\equiv 0~(\bmod~d_i)~$ for each $1\le j \le k$ and $1\le ...

- Tue Aug 30, 2016 8:03 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies:
**184** - Views:
**20435**

Post a problem that's not lame.

- Thu Aug 11, 2016 10:04 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies:
**184** - Views:
**20435**

Let $f(1)=C$ and $P(n)~\Rightarrow ~f(1)+\cdots+f(n)=n^2 f(n)$. Then rearranging $P(n)-P(n-1)$ gives $\dfrac{f(n)}{f(n-1)}=\dfrac{n-1}{n+1}$. Multiplying similar expressions leads to \[\prod_{k=2}^n \dfrac{f(k)}{f(k...