![\[ \log_{10}(\tan 1^{\circ})+ \log_{10}(\tan 2^{\circ})+ \log_{10}(\tan 3^{\circ})+ \cdots + \log_{10}(\tan 88^{\circ})+\log_{10}(\tan 89^{\circ}). \]](http://latex.artofproblemsolving.com/a/9/1/a91984f21cb9567f93101bda9f55dccac5cfe935.png)
cottages. At time 
an angry peasant arises from each cottage, and every 
minutes 
seconds to either burn a peasant or to burn a cottage, but Trodgor cannot begin burning cottages until all the peasants around him have been burned. How many seconds does it take Trodgor to burn down the entire village?
?
algebra books (all the same thickness) and 
geometry books (all the same thickness, which is greater than that of an algebra book) to completely fill a certain shelf. Also, 
of the geometry books would fill the same shelf. Finally, 
of the algebra books alone would fill this shelf. Given that 
are distinct positive integers, it follows that 
be reals such that 
Prove that
Let 
Prove that
Prove that
.
Let 
.
Let 
.
Let 
and 
Prove that
Prove that
Prove that
are positive real numbers such that 
.
.
Y by samrocksnature, FaThEr-SqUiRrEl, icematrix2, GoodMorning, HWenslawski, megarnie
A finite set 
of positive integers has the property that, for each 
and each positive integer divisor 
of 
, there exists a unique element 
satisfying 
. (The elements and 
could be equal.)
Given this information, find all possible values for the number of elements of.
of positive integers has the property that, for each 
and each positive integer divisor 
of 
, there exists a unique element 
satisfying 
. (The elements and 
could be equal.)Given this information, find all possible values for the number of elements of
.This post has been edited 1 time. Last edited by P_Groudon, Apr 15, 2021, 5:19 PM
works
.
,
.
with 
primes each, all 
of them being distinct. Label the primes 
and 
.![\[c_1\cdot c_2\cdot \cdots \cdot c_n\]](http://latex.artofproblemsolving.com/0/d/5/0d53f51aec9169784e2e468269ab14e9082835a8.png)
where 
or 
satisfy the problem conditions (clearly 
since there are two choices for each 
)
(since we already explicitly showed 
also be in 
,
which contradicts the unique condition.
in 
where 
is the number of divisors of 
since each divisor of 
,
such that 
,
pairs 
,
dividing 
.
of the elements in 
,
,
.
,
works.
.
distinct primes 
,
.
digits. Let 
,
if ![$B[i] = 0$](http://latex.artofproblemsolving.com/b/f/a/bfa853873313263c18174aab7415b0160d1467c7.png)
,
if ![$B[i] = 1$](http://latex.artofproblemsolving.com/9/8/5/98505848efa52555b17ba6e6f693c8637a81ee41.png)
.
where 
possible bit strings is a valid 
where 
such that ![$B'[i]=0$](http://latex.artofproblemsolving.com/9/2/8/928dc79286b4e962ca408567ef499c5c05043a47.png)
if 
,
(where 
denotes XOR).
where 
has a prime factor 
,
.
denote the 
of positive integer divisors of 
we have some 
such that 
,
is unique. Follows that 
.
.
such that 
.
be prime factorization of 
,
,
.
such that 
.
,
is some 
such that 
.
.
,
.
since 
,
It would be different if the problem had said "If 
.
,
,
,
.
exceeds ![\[
n = (\alpha_1+1)\cdots(\alpha_k+1).
\]](http://latex.artofproblemsolving.com/6/0/9/6090b88176d04fc87c752fe071e74b111f0d51c2.png)
Now, let 
be such that 
,
.
.![\[
p_1^{\alpha_1-1}p_2^{\alpha_2}\cdots p_k^{\alpha_k} \mid t.
\]](http://latex.artofproblemsolving.com/0/1/7/0179fa94c4fabdbd52e9b6986e20dd0ce1adb62f.png)
Moreover, we also know that there is an element of 
be this element. Since 
,
with 
,![\[
qp_1^{\alpha_1-1}p_2^{\alpha_2}\cdots p_k^{\alpha_k} \mid t.
\]](http://latex.artofproblemsolving.com/9/2/c/92c8c7851c7da784e09cc0ea0424f523d7515abc.png)
However, notice now that the number of divisors of 
which contradicts the Lemma.
,
be a set of ![$[k]=\{1,\cdots,k\}$](http://latex.artofproblemsolving.com/6/6/f/66fd074a55580cb666e66e74d9eaecb9c4ad6c56.png)
.![\[
S = \left\{\Bigg(\prod_{i \in E}p_i\Bigg)\Bigg(\prod_{j \in [k]\setminus E}q_j\Bigg) : E \subseteq [k]\right\}.
\]](http://latex.artofproblemsolving.com/d/c/1/dc16a0d069d382f56131f8d979cc904e7a432083.png)
To help clarify the above definition of 
:![\[
\begin{tabular}{ccc|ccc||c} $p_1$ & $p_2$ & $p_3$ & $q_1$ & $q_2$ & $q_3$ & Element \\ \hline
1 & 1 & 1 & 0 & 0 & 0 & $p_1p_2p_3$ \\
1 & 1 & 0 & 0 & 0 & 1 & $p_1p_2q_3$ \\
1 & 0 & 1 & 0 & 1 & 0 & $p_1p_3q_2$ \\
0 & 1 & 1 & 1 & 0 & 0 & $p_2p_3q_1$ \\
1 & 0 & 0 & 0 & 1 & 1 & $p_1q_2q_3$ \\
0 & 1 & 0 & 1 & 0 & 1 & $p_2q_1q_3$ \\
0 & 0 & 1 & 1 & 1 & 0 & $p_3q_1q_2$ \\
0 & 0 & 0 & 1 & 1 & 1 & $q_1q_2q_3$ \\
\end{tabular}
\]](http://latex.artofproblemsolving.com/8/a/a/8aaf6aaea41bc89950a2c8b1bdc7fb1ff281b276.png)
![$U \subseteq [k]$](http://latex.artofproblemsolving.com/5/4/5/545fd2163f729f5472034cd66cd0230f1279898a.png)
and ![$U' = [k]\setminus U$](http://latex.artofproblemsolving.com/6/4/5/64556808cf8cea0ce777b184310b56ddcf0a03b5.png)
as described in the construction.
and 
for which 
and 
,
means that 
and 
.
.![$V \subseteq [k]$](http://latex.artofproblemsolving.com/3/7/1/371479544150a8c9467aa2e9b9ab6b669d107072.png)
and ![$V' = [k] \setminus V$](http://latex.artofproblemsolving.com/a/1/7/a170941e9158da766669e51ceed7704ea5029458.png)
be the sets that correspond to 
(resp. 
)
(resp. 
)
if and only if 
,
where 
is a non-negative integer.
is the set with 
elements satisfying the conditions. Then, we will construct a set 
that works as well:
be two distinct primes that do not divide any of the elements in 
,
and 
to 
.
such that 
(which means 
and 
are in 
(since 
.
so we are done.
is a divisor of 
.
.
.
Therefore, some 
.
.
Now, let 
and 
be such that
Now, note that if 
,
which is a contradiction.
,
and 
.
This contradicts the uniqueness guaranteed by the problem statement, so we are done.
.
.
.
where 
,
of the elements of 
,
iff 
.
,
we have some 
.
is constant over all 
, but mostly I'm too lazy to writeup.
,
.
for 
for 
and integer 
.
,
for 
.
to have a number in 
,
(With Z relatively prime to everything else). However, now 
a contradiction.
would be nonconstant). This gives 
where we either any number 
is either divisibly by the 
or 
,
,
;
over all possibilities suffices.
for any 
.
.
of the elements in 
;
of the elements in 
,
for some 
for some 
.
(and 
and 
of primes, with
where 
Now, notice that, if you take some 
and 
which denote the sets of prime factors of some number 
and a divisor 
which has prime factors 
and 
we can simply take the set 
and its corresponding number 
This is because
and we can see that 
and 
So, if we consider 
and 
to be the universal sets, then we have that 
Now, considering the primes not in 
we see that 
and 
must be empty. But, by Demorgan’s Law, that implies that the sets 
and 
are universal opposites. Thus, the set 
Every element of 
)
for some number 
is defined to be 
the number which satisfies 
then 
is unique, 
.
Next, assume for the sake of contradiction that 
for some number 
Notice that this number 
and 
we have that
by the fact that 
is an integer. But, since 
this cannot be an integer. Thus, we are done.
for nonnegative 
clearly works.
for any prime 
for some element 
so that 
,
and 
which is a contradiction as 
is letting each element be 
where we have two groups of distinct primes 
and 
.
,
"
then 
works, otherwise consider two disjoint sets of primes 
and 
,
simply let either 
or 
be a prime factor of the element, it is easy to see there are 
be the number of divisors function, we claim that for all 
.
.
by Pigeonhole Principle there exists two elements 
where 
are both relatively prime to 
.
.
.
,
is absurd. Now consider the element 
,
multiples of 
for all 
that 
,
.
with 
and 
.
with 
,
,
.
.
.
,
and 
(note 
for some 
with 
)
.
,
would represent the product 
.
to be another 
must be distinct if 
.
such that there is a prime 
.
be the prime factorization of 
,
.
such that 
and there must exist an element 
such that 
.
for 
.
because if not, 
which is a contradiction, as the inequality holds due to 
.
cannot have any primes besides 
that divide it that are not in 
and 
cannot have any primes besides 
which gives the desired contradiction. 
.
to be distinct primes and we let 
.
,
.
,
and similar for 
.
,
,
,
,
,
,
,
,
.
,
with 
such that 
,
and 
,
,
,
.
such that 
,
and 
,
.
by 
and that 
c
,
,
.
,
is constant. Hence, for each 
and prime 
,
;![\[\tfrac{v_p(s)+1}{v_p(s)}=\tfrac{\tau(s)}{\tau(s/p)}=
\tfrac{\tau(t)}{\tau(s/p)}=\tfrac{\tau(s/p\cdot pt/s)}{\tau(s/p)}=
\tfrac{\tau(s/p)\tau(pt/s)}{\tau(s/p)}=\tau(\tfrac{pt}{s})\in\mathbb{Z},\]](http://latex.artofproblemsolving.com/9/3/c/93c7b7177442529c2ffd3db240c1d865a231d9db.png)
meaning 
.
,
.
. Then we can toggle all 
'
Then each element in 
with this product. Therefore there must be :
total elements in 
The products of the exponents + 1of all the primes in the number that shares this 
By a similar argument all 
.
For every element ![\[\tau(s) = |S|.\]](http://latex.artofproblemsolving.com/3/1/3/313fbf1611ebacc2609e02235bbb97a4ab0b48db.png)
be distinct prime numbers. Then, put 
Due to the symmetry of this construction, we only need to check the condition is satisfied for 
Now, all positive divisors of 
where 
The only element 
is ![\[ t = \left( \prod\limits_{i \in T} p_i\right)\left(\prod\limits_{i \in (\{1, 2, 3, \ldots k \} \backslash T)} p_{i+k}\right).\]](http://latex.artofproblemsolving.com/2/7/a/27aba67fbd81c3f6f370f9c13039834f541e4033.png)
This is because, for all 
if 
must divide 
Indeed this value of 
where all the 
Because 
we must have 
for some 
for that 
Therefore 
the 
which we know is true by the given condition. We also know that there is 
such that 
So we have 
If there exists some 
and 
then we have ![\[ \tau(c) \ge e_1(e_k + 1)\prod\limits_{i=2}^m (e_i + 1) > \frac{e_k + 1}{2}\prod\limits_{i=2}^m(e_i + 1) \ge \tau(a) = |S|.\]](http://latex.artofproblemsolving.com/7/6/3/7630ffa82090d90161130b8d2f8521ec904bdcbd.png)
This contradicts the fact that 
for all 
So we must have ![\[\gcd(b, a) = \gcd(b, c) = 1\]](http://latex.artofproblemsolving.com/3/b/1/3b18c3b86f77ca8c2b2fcbc51aabbb343699aaac.png)
which contradicts the uniqueness in the given condition, and we're done.
such that every number is the product of 
.
and the other half vice versa. Then just do the 
construction on each of these halves with the other 
primes completing the induction.
.
.
.
then we know there exists a distinct number 
.
In particular, the LHS is an integer which implies 
.
for every 
denotes the number of divisors of 
for all 
.
.
such that 
and 
.
,
as desired.
