[/quote]
,
,
.
lies on 
such that 
.
passing trough 
at 
.
.
such that
,
.
and 
the midpoints of 
and 
.
and 
meet 
and 
at 
and 
,
,
bisects both angles 
,
,
.
be the circumcircle of triangle 
,
be the tangent line of 
passing 
.
be the points each on side 
such that 
.
meets 
.
meets 
passing 
.
and tangent to line 
.
![\[
\text{Find all prime numbers } (p, q) \text{ such that } p^2 + 3pq + q^2 \text{ is a fifth power of an integer.}
\]](http://latex.artofproblemsolving.com/c/f/8/cf855cb087102024a50d41f49c67e7fd1ef22767.png)
be an integer, and let 
be positive integers. Show that there exist positive integers 
satisfying the following three conditions:
for 
the remainders of 
denotes the integer part of real number 
,
be the set of positive integers. A function 
satisfies the equation ![\[\underbrace{f(f(\ldots f}_{f(n)\text{ times}}(n)\ldots))=\frac{n^2}{f(f(n))}\]](http://latex.artofproblemsolving.com/2/e/7/2e73ff2f75b103e9d6a4004a42e4ed7f4ee64a67.png)
for all positive integers 
.
Y by Adventure10
A positive integer divisor of 
is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as 
, where 
and 
are relatively prime positive integers. What is 
?
is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as 
, where 
and 
are relatively prime positive integers. What is 
?
This post has been edited 1 time. Last edited by LauraZed, Jan 31, 2020, 6:14 PM
Reason: fixed phrasing to match test, added answer choices
Reason: fixed phrasing to match test, added answer choices
divisors of 
,
factors. However, if 
,
,
factors. Thus the answer is 
giving 
.
by prime factorizing. @OP please include the answer choices.
case.
divisors. Use prime factorization to take a square root but round up exponents that don't divide by 
evenly. It's 
for 
.
.
total divisors. All perfect square divisors of 
choices for the power of 
or 
.
choices for the power of 
,
for the powers of 
and 
.
.
.
![\[12!=2^{6+3+1}\cdot3^{4+1}\cdot5^2\cdot7\cdot11=2^{10}\cdot3^5\cdot5^2\cdot7\cdot11,\]](http://latex.artofproblemsolving.com/7/2/1/7219fb8204eb06284bc409a4e57814a5c06662e6.png)
which has 
divisors.
.
can be any positive integer divisor of 
,
possible values of ![\[\frac{36}{11\cdot6\cdot3\cdot4}=\frac{1}{22}\implies\boxed{\textbf{(E)}\ 23}\]](http://latex.artofproblemsolving.com/5/d/8/5d8e0c83098aa8fd03106e04006911acfaa56d16.png)
.
So, it has 
we take the floor function when each exponent is divided by 
so 
Now finding all the divisors, 
So, 
ive found the prime factorization but im confused how to go on...
(this is a formula, look up "formula for amount of divisors" if you don't know it). Our divisor is in the form 
,
.
for 
and 
and one way for 
,
desirable cases. Answer 
.
