AMC 12A 2015
by djmathman, Jan 31, 2015, 2:25 AM
is this Tuesday. My aim is a 120+ so that I don't need to pull an unrealistic performance on the AIME to qualify for USAMO.
I'm actually legitimately scared, oops
In the meantime, here's an OMO problem from the last contest that I finally got around to writing a complete solution to last week.
I'm actually legitimately scared, oops
In the meantime, here's an OMO problem from the last contest that I finally got around to writing a complete solution to last week.
OMO Fall 2014.18 wrote:
We select a real number 
uniformly and at random from the interval 
. Define
![\[ S = \frac{1}{\alpha} \sum_{m=1}^{1000} \sum_{n=m}^{1000} \left\lfloor \frac{m+\alpha}{n} \right\rfloor. \]](//latex.artofproblemsolving.com/0/9/9/099efafef67403350c582ffa561a65e586668764.png)
Let
denote the probability that 
. Compute 
.
uniformly and at random from the interval 
. Define![\[ S = \frac{1}{\alpha} \sum_{m=1}^{1000} \sum_{n=m}^{1000} \left\lfloor \frac{m+\alpha}{n} \right\rfloor. \]](http://latex.artofproblemsolving.com/0/9/9/099efafef67403350c582ffa561a65e586668764.png)
Let
denote the probability that 
. Compute 
.![\[S=\dfrac1\alpha\sum_{n=1}^{1000}\sum_{m=1}^n\left\lfloor\dfrac{m+\alpha}n\right\rfloor=\dfrac1\alpha\sum_{n=1}^{1000}\left[\sum_{m=1}^n\dfrac{m+\alpha}n-\sum_{m=1}^n\left\{\dfrac{m+\alpha}n\right\}\right].\]](http://latex.artofproblemsolving.com/7/0/9/709f35be0f002962f9b84b7488677f6301fca5bd.png)
and 
be integers such that 
and 
for every 
.![\[\sum_{m=1}^n\left\{\dfrac{m+\alpha}n\right\}=\sum_{m=1}^n\left\{\dfrac{nq_m+(r_m+\{\alpha\})}n\right\}=\sum_{m=1}^n\left\{\dfrac{r_m+\{\alpha\}}n\right\}.\]](http://latex.artofproblemsolving.com/a/b/f/abf288af5d7623848eac0c7e541443ea1589a149.png)
consists of 
consecutive integers, the sequence 
must consist of the integers 
(due to the fact that the remainders will "wrap around" at some point). Therefore the leftover sum can be easily rewritten as![\[\sum_{m=1}^n\left\{\dfrac{r_m+\{\alpha\}}{n}\right\}=\sum_{m=0}^{n-1}\dfrac{m+\{\alpha\}}n.\]](http://latex.artofproblemsolving.com/8/1/5/815c5af41da9d2ec14c4e72e5a7399ac912e9646.png)
![\begin{align*}\dfrac1\alpha\sum_{n=1}^{1000}\sum_{m=1}^n\left\lfloor\dfrac{m+\alpha}n\right\rfloor&=\dfrac1\alpha\sum_{n=1}^{1000}\left[\sum_{m=1}^n\dfrac{m+\alpha}n-\sum_{m=0}^{n-1}\dfrac{m+\{\alpha\}}n\right]\\&=\dfrac1\alpha\sum_{n=1}^{1000}\dfrac{n+n(\alpha-\{\alpha\})}n\\&=\dfrac{1000(1+\lfloor\alpha\rfloor)}\alpha.\end{align*}](http://latex.artofproblemsolving.com/3/a/7/3a7aad97fa8b121d9f4f1c72e2577b65890e4d5e.png)
,
.
.
,
.
gives 
.
,
.![\[p=\dfrac1{500}\sum_{k=0}^5\dfrac16(5-k)=\dfrac1{500}\left(\dfrac{1+2+3+4+5}6\right)=\dfrac1{200}\]](http://latex.artofproblemsolving.com/0/4/9/049acf6f4fb1f49b98610a37118786fbccf48db9.png)
.This post has been edited 3 times. Last edited by djmathman, Mar 5, 2015, 3:42 AM
I don't feel like I have improved a bit at math since last year. I guess this week a big test for me.
April 2025
November 2024