[/quote]
,
of natural numbers such that for every pair of natural numbers 
and 
,![\[
\frac{1}{2} < \frac{\gcd(a_r, a_s)}{\gcd(r, s)} < 2
\]](http://latex.artofproblemsolving.com/1/6/7/167679c1707b957d87311298ea5b72347a9bdc45.png)
such that : 
and 
be positive real numbers such that 
Prove that:
table filled with natural numbers, we say it is a divisor table if:
-
,
-
,
for every 
.
is given. Determine the smallest natural number 
,
satisfying
Prove
has a right angle at 
.
be the point on line 
such that 
and 
lies between 
.
is chosen so that 
and 
is the bisector of 
.
is chosen so that 
and 
is the bisector of 
.
be the midpoint of 
be the point such that 
is a parallelogram. Prove that 
and 
are concurrent.
have altitudes 
and 
.
is 
.
meet 
at 
and 
at 
.
at 
.
.
such that 
.![$$2 \sqrt[4]{\frac{a}{b+c-a}}\ge 2 +\frac{2a^2-b^2-c^2}{(a+b)(a+c)}.$$](http://latex.artofproblemsolving.com/f/0/7/f0731fbe5c83aedc8474975f930f7194c08cc070.png)
,
meets the circumcircle of 
again at 
,
meets it again at 
.
,
and 
are concurrent.
.
!
candy bars. Let N & M be positive integers, what is N+M?
However, I quickly saw that 
and 
intersect at the point 
.
?
,
.
are chosen from 0 to 1. What is the probability that their positive difference is more than 
?
?
families, there are 
children respectively. If a random child from any of the families is chosen, what is the probability that the child has 
siblings? Express your answer as a common fraction.[/quote]
,
people, exactly 
such that 
.
,
.
.
has 
,
,
.
.
Y by RainbowJessa, Sharkship, RollingPanda4616, huajun78, PikaPika999, HacheB2031, GodGodGodGodGoose, Soupboy0, jkim0656, booking, dragon888
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!
Problems usually get harder throughout the week, so Sunday is the easiest and Saturday is the hardest!
Please hide solutions and answers, hints are fine though!
Problems usually get harder throughout the week, so Sunday is the easiest and Saturday is the hardest!
This post has been edited 7 times. Last edited by Leeoz, Apr 14, 2025, 4:51 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
/ 
is divided by 2025?
.
.
so we just get 1 and 1 mod 2025 = 
is for division im pretty sure
.
