[/quote]
,
be a point in the plane, distinct from the vertices of 
.
the reflections of 
and 
,
are collinear if and only if 
coincide, prove that 
given that there exist positive integers 
satisfying
and 
,
as follows:
where 
are all the positive divisors of 
.
Find all positive integers 
.
Find all positive integers 
.
be an odd prime number. How many 
of 
are there, the sum of whose elements is divisible by 
be real numbers satisfying
be a fixed square and 
a variable point on segment 
.
is constructed such that 
is on segment 
and 
is on segment 
.
be the intersection point of lines 
and 
.
board, two opposite sides have been joined, forming a cylinder. Determine whether it is possible to place 
.
.
.
denote the number of positive integers less than 
has at least 
solutions in 
that can be written in the form
where 
are pairwise coprime positive integers such that 
,
,
.
and 
intersect at the point 
.
?
,
.
are chosen from 0 to 1. What is the probability that their positive difference is more than 
?
?
Y by RainbowJessa, Sharkship
Every day, I will try to post a new problem for you all to solve! If you want to post a daily problem, you can! 
Please hide solutions and answers, hints are fine though!
The first problem is:
Please hide solutions and answers, hints are fine though!
The first problem is:
March 21st Problem wrote:
Alice flips a fair coin until she gets 2 heads in a row, or a tail and then a head. What is the probability that she stopped after 2 heads in a row? Express your answer as a common fraction.
This post has been edited 4 times. Last edited by Leeoz, Mar 30, 2025, 12:08 AM
and 
.
So, 
.
are 6 cases, which add up t 
.
contribute a total of 
.
contribute 
.
contribute 
.
and 
are the same number, as are 
and 
,
and 
,
.
in the product (prime factorization) although it's still kind of a bash
'
Since each is relatively prime, I can do the simple calculation 
it must have the die rolls 
and 
and 
and 
This gives 
and 
This gives 
gives 
The other person made a mistake
![[asy]
unitsize(1inch);
fill(unitsquare,red);
fill((0.5,0)--(1,0)--(1,0.5)--cycle,green);
fill((0,0.5)--(0,1)--(0.5,1)--cycle,green);
draw((0,0)--(2,0),EndArrow(5));draw((0,0)--(-1,0),EndArrow(5));
draw((0,0)--(0,2),EndArrow(5));draw((0,0)--(0,-1),EndArrow(5));
draw((1,1)--(1,0));
draw((1,1)--(0,1));
draw((0,0.5)--(0.5,1));
draw((0.5,0)--(1,0.5));
label("$a$",(2,0),E);
label("$b$",(0,2),N);
label("$1$",(1,0),S);
label("$\frac12$",(0.5,0),S);
label("$1$",(0,1),W);
label("$\frac12$",(0,0.5),W);
label("$0$",(0,0),SW);
[/asy]](http://latex.artofproblemsolving.com/3/7/1/3711bd1d3d55fee2884ebe8cded2a82f5f2a7928.png)
and 
so its area is ![\[\frac12\cdot\frac12\cdot\frac12=\frac18.\]](http://latex.artofproblemsolving.com/d/2/3/d230c392b5d78ce504403b48f1f961a6c6053bce.png)
The area at the bottom is also a right triangle with side lengths ![\[\frac18+\frac18=\boxed{\frac14}.\]](http://latex.artofproblemsolving.com/a/3/7/a375a931c177ee5051947f057ab34ecc1c74bd0f.png)
,
.
,
a negative number. I tried some random negative numbers, and I found that when 
,
,
or 
.
,
and we want the least possible value of it. Therefore we can complete the square to get 
In the extreme case of which 
(because any square must be nonnegative so the minimum value it can be is 0), we have the value is 
.
,
and 
-
,
inclusive. now, we have 3 cases
and 
as 
and 
then there is a 729/1000 chance no 1s occur in a number, meaning 271 numbers from 0 to 999. also 0 doesn't count anyway so it doesn't affect this number.
is an entire function that is not constant, by Liouville's Theorem, it is not bounded. Therefore, the function never attains a minimum 
First, the number of numbers without a 
There are 
choices for the units, 
numbers without a 
)
to 
and 
So, there are 
numbers from 
without a 
The combination of the results gives 
and subtracting from 
so 
