AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
non-zero digits such that deleting consecutively the digit of the left, in each step, we obtain a divisor of the previous number.
there exists an -digit number divisible by 
all of whose digits are odd.
be an acute triangle with 
. Let 
be the centroid of triangle 
and let 
be the foot of the perpendicular from 
to side 
. The median 
intersects the circumcircle 
of triangle at a second point 
. Let 
be the point where intersects . The line 
intersects the circle at a point 
, such that lies between and . Prove that the points 
and lie on a circle.
atoms called usamons. Each usamon either has one electron or zero electrons, and the physicist can't tell the difference. The physicist's only tool is a diode. The physicist may connect the diode from any usamon 
to any other usamon 
. (This connection is directed.) When she does so, if usamon has an electron and usamon does not, then the electron jumps from to . In any other case, nothing happens. In addition, the physicist cannot tell whether an electron jumps during any given step. The physicist's goal is to isolate two usamons that she is sure are currently in the same state. Is there any series of diode usage that makes this possible?
All I can do now is hope and pray for HCSSiM. So far I'm 0 for 3 for summer programs lol
All I can do now is hope and pray for HCSSiM. So far I'm 0 for 3 for summer programs lol
All I can do now is hope and pray for HCSSiM. So far I'm 0 for 3 for summer programs lol
All I can do now is hope and pray for HCSSiM. So far I'm 0 for 3 for summer programs lol
All I can do now is hope and pray for HCSSiM. So far I'm 0 for 3 for summer programs lol
,
.
.
,
.
.
.
Let 
.![$$ \frac{1}{2}\geq \frac{a}{a^2+3 }+ \frac{b}{b^2+3} \geq \frac{1}{4}(\frac{1}{\sqrt[3]{3}}+\sqrt[3]{3}-1)$$](http://latex.artofproblemsolving.com/f/9/1/f9119e071c12778037616c624c7b9ca75356e8e3.png)
![$$ \frac{1}{2}\geq \frac{a}{a^3+3 }+ \frac{b}{b^3+3} \geq \frac{1}{2\sqrt[3]{9}}$$](http://latex.artofproblemsolving.com/2/b/2/2b2d8a011229fe88a81cb75f9b018ad3a6245faa.png)
![$$ \frac{1}{2}\geq \frac{a}{a^2+ab+2}+ \frac{b}{b^2+ ab+2} \geq \frac{4\sqrt[3]{3}+3\sqrt[3]{9}-6}{17}$$](http://latex.artofproblemsolving.com/5/f/6/5f673cd494b1909ca237db63e96f949c24b6f2ad.png)
![$$ \frac{1}{2}\geq \frac{a}{a^3+ab+2}+ \frac{b}{b^3+ ab+2} \geq \frac{\sqrt[3]{3}}{5}$$](http://latex.artofproblemsolving.com/b/c/3/bc343a07c88dbd98ebffdfb78f311b7b82683d65.png)
