Search found 136 matches
- Thu Mar 06, 2014 12:05 pm
- Forum: Junior Level
- Topic: RMO-2010/3
- Replies: 9
- Views: 10445
Re: RMO-2010/3
Sorry my solution was not complete. I didn't notice the word "non-zero digits". Without that condition it's $1125$ and a python program ensures that. Here's another one: we have $18$ multiples of $4$ in range $[1,100]$ with non-zero digits. When the last number formed by the last two digits is of th...
- Tue Mar 04, 2014 1:14 pm
- Forum: Algebra
- Topic: Generalization of an IMO Problem!
- Replies: 1
- Views: 2272
Generalization of an IMO Problem!
1. Prove that: \[\cos \dfrac\pi7 - \cos\dfrac {2\pi}{7}+\cos\dfrac{3\pi}{7}=\dfrac{1}{2}\]
2. Generalization (and a useful hint for the above): Prove that, $\forall ~k\geq 3$ such that $k$ is an odd integer: \[\sum_{n=1}^{(k-1)/2}\cos\dfrac{(2n-1)\pi}{k}=\dfrac{1}{2}\]
2. Generalization (and a useful hint for the above): Prove that, $\forall ~k\geq 3$ such that $k$ is an odd integer: \[\sum_{n=1}^{(k-1)/2}\cos\dfrac{(2n-1)\pi}{k}=\dfrac{1}{2}\]
- Mon Mar 03, 2014 12:35 pm
- Forum: Junior Level
- Topic: RMO-2010/3
- Replies: 9
- Views: 10445
Re: RMO-2010/3
Since, $[4,8]=8$, by PIE, the answer is: \[\left.\left(\left\lfloor \dfrac{9999}{4} \right\rfloor-\left\lfloor \dfrac{999}{4} \right\rfloor\right)\right/2~=~\dfrac{2250}2~ = ~\boxed{1125}\]
- Sat Feb 22, 2014 9:05 pm
- Forum: Number Theory
- Topic: BdMO 2014: Secondary - Q5
- Replies: 1
- Views: 2185
BdMO 2014: Secondary - Q5
For a positive integer \(n\), find the greatest \(p\) such that \(n\) can be rewritten as sum of \(p\) consecutive positive integers.
- Fri Feb 21, 2014 8:25 pm
- Forum: Number Theory
- Topic: infinite primes
- Replies: 4
- Views: 3348
Re: infinite primes
By FLT, any odd prime $p\mid 2^{p-1}-1$. So just consider $p=q$ and infinitude of primes completes the rest.
- Thu Feb 20, 2014 1:04 pm
- Forum: Algebra
- Topic: IMO - 1987 - 4
- Replies: 9
- Views: 6034
Re: IMO - 1987 - 4
f(f(2014)) = 2014 + 2014 = 4028