,
,
,
,
or even expressions of those terms combined as if that would make them defined. I have made this post to answer these questions once and for all, and I politely ask everyone to link this post to threads that are talking about this issue.
,
isn't even rigorously defined in this expression. What exactly do we mean by 
.
,
does not exist in the first place, although you will see why in a calculus course. So the point is that 
?
indeterminate? In analysis (the theory behind calculus and beyond), limits involving an algebraic combination of functions in an independent variable may often be evaluated by replacing these functions by their limits. However, if the expression obtained after this substitution does not provide sufficient information to determine the original limit, then the expression is called an indeterminate form. For example, we could say that 
,
or ![\[0/0, \infty/\infty, \infty-\infty, \infty \times 0\]](http://latex.artofproblemsolving.com/2/6/0/260f97a1cf1ea8722ff97a29b83bac7bfe83e577.png)
etc etc. So what makes something undefined? In the broader scope, something being undefined refers to an expression which is not assigned an interpretation or a value. A function is said to be undefined for points outside its domain. For example, the function 
given by the mapping 
is undefined for 
.
is undefined because dividing by ![\[\overline{\mathbb{R}}=\mathbb{R}\cup \{-\infty,+\infty\}.\]](http://latex.artofproblemsolving.com/9/9/3/993c78b62ec2c84b5d77daa8e9a8cd6f70fcf904.png)
So instead of treating infinity as an idea, we define infinity (positively and negatively, mind you) as actual numbers in the reals. The advantage of doing this is for two reasons. The first is because we can turn this thing into a totally ordered set. Specifically, we can let 
for each 
which means that via this order topology each subset has an infimum and supremum and 
is therefore compact. While this is nice from an analytic standpoint, extending the reals in this way can allow for interesting arithmetic! In ![\begin{align*}
a + \infty = \infty + a & = \infty, & a & \neq -\infty \\
a - \infty = -\infty + a & = -\infty, & a & \neq \infty \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \pm\infty, & a & \in (0, +\infty] \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \mp\infty, & a & \in [-\infty, 0) \\
\frac{a}{\pm\infty} & = 0, & a & \in \mathbb{R} \\
\frac{\pm\infty}{a} & = \pm\infty, & a & \in (0, +\infty) \\
\frac{\pm\infty}{a} & = \mp\infty, & a & \in (-\infty, 0).
\end{align*}](http://latex.artofproblemsolving.com/6/d/8/6d8fc1eca5c791d430a7168cd3b37a02104fa318.png)
So addition, multiplication, and division are all defined nicely. However, notice that we have some indeterminate forms here which are also undefined,![\[\infty-\infty,\frac{\pm\infty}{\pm\infty},\frac{\pm\infty}{0},0\cdot \pm\infty.\]](http://latex.artofproblemsolving.com/a/0/b/a0b67076efcc014c78b2023be1151c84b57c1503.png)
So while we define certain things, we also left others undefined/indeterminate in the process! However, in the context of measure theory it is common to define 
as greenturtle3141 noted below. I encourage to reread what he wrote, it's great stuff! As you may notice, though, dividing by ![\[\mathbb{C}^*=\mathbb{C}\cup\{\tilde{\infty}\}\]](http://latex.artofproblemsolving.com/8/4/e/84ed3d6ab237df970333b87beaef5311caa46185.png)
which we call the Riemann Sphere (it actually forms a sphere, pretty cool right?). As a note, 
means complex infinity, since we are in the complex plane now. Here's the catch: division by ![\[\frac{z}{0}=\tilde{\infty},\frac{z}{\tilde{\infty}}=0.\]](http://latex.artofproblemsolving.com/1/6/0/160cb939707708ff4e0930724df2928af11d7fd1.png)
where 
and 
Furthermore, we actually have some nice properties with multiplication that we didn't have before. In 
it holds that![\[\tilde{\infty}\times \tilde{\infty}=\tilde{\infty}\]](http://latex.artofproblemsolving.com/8/e/0/8e035b4841d20c5ad671c46355cc667ca01e533b.png)
but 
and 
are left as undefined (unless there is an explicit need to change that somehow). One could define the projectively extended reals as we did with ![\[{\widehat {\mathbb {R} }}=\mathbb {R} \cup \{\infty \}.\]](http://latex.artofproblemsolving.com/c/0/4/c044ff37849acad3b8b280b9753539225a1276ca.png)
They behave in a similar way to the Riemann Sphere, with division by 
with no parallel sides is inscribed in a circle 
.
are inscribed in triangles 
,
and 
,
,
,
.
be a quadrilateral inscribed in a circle 
such that 
.
and 
be the midpoints of 
and 
respectively. The line 
meets 
.
are concurrent.
be a continuous function. A chord is defined as a segment of integer length, parallel to the x-axis, whose endpoints lie on the graph of 
.
chords, one of which has length 2025. Find the minimum possible value of 
has been partitioned into disjoint pairs 
)
,
equals 
.![\[ |a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}| \]](http://latex.artofproblemsolving.com/0/4/4/04422a7d2266b92a6b3f215b698f078cda85091f.png)
ends in the digit 
.
![$\sqrt[2]{2}$](http://latex.artofproblemsolving.com/1/8/6/1869ac018fb72a26a12ab14499cbf1b1c13b4a10.png)
,![$\sqrt[3]{3}$](http://latex.artofproblemsolving.com/1/7/a/17a6bb5dc11696b8650b6d3a11e770f861f4c326.png)
,![$\sqrt[5]{5}$](http://latex.artofproblemsolving.com/a/b/9/ab90f2d42dc24065e2a64988fe76f22eb9359799.png)
.
such that![\[
\sqrt{x - 2011} + \sqrt{2011 - x} + 10
\]](http://latex.artofproblemsolving.com/4/8/b/48b82d13b733947420a0d763c6af74636770835a.png)
is an integer.
Y by ktong, WindTheorist, nguyendangkhoa17112003, Mr.Chagol, karitoshi, Guardiola, Tom_Hulla, Davrbek, Vietjung, Mathman12334, Carpemath, MathGenius_, Kagebaka, dangerousliri, Davi-8191, fastandfuriuos, mathleticguyyy, Hedy, amar_04, vsamc, fidgetboss_4000, OlympusHero, itslumi, pog, samrocksnature, math31415926535, centslordm, JacksonJunior, megarnie, microsoft_office_word, ZHEKSHEN, HWenslawski, jasperE3, rayfish, MathLuis, Math4Life2020, Lamboreghini, A_Crabby_Crab, TheHawk, aidan0626, Adventure10, Sedro, deplasmanyollari, NicoN9, ItsBesi, cubres, Soupboy0
Find all pairs 
of positive integers such that ![\[ k!=(2^n-1)(2^n-2)(2^n-4)\cdots(2^n-2^{n-1}). \]](//latex.artofproblemsolving.com/3/5/3/353bbb15bac205053c26208af8cac7c32a296f55.png)
Proposed by Gabriel Chicas Reyes, El Salvador
of positive integers such that ![\[ k!=(2^n-1)(2^n-2)(2^n-4)\cdots(2^n-2^{n-1}). \]](http://latex.artofproblemsolving.com/3/5/3/353bbb15bac205053c26208af8cac7c32a296f55.png)
Proposed by Gabriel Chicas Reyes, El SalvadorThis post has been edited 4 times. Last edited by v_Enhance, Jul 18, 2019, 6:26 PM
Reason: add $2^n-4$ term
Reason: add $2^n-4$ term
is more important than a missed comma. This can be confused to 
'
,
.
.
'
by the LTE lemma, so 
.
,
.
)
and 
for left and right side respectively.
.
.
,
.
,
,
,
.
are the only solutions.
When 
i.e. 
as 
.
to get 
and 
,
.
,
,
which is trivially false for 
as there are 
terms on both sides, and the other case is very similar, though a little more complicated-the answers will be 
for 
respectively).
.![\[ \nu_2 (k!) = \nu_2 \left( \prod_{i=0}^{n-1} (2^n - 2^i) \right) = \frac{(n(n-1)}{2} . \]](http://latex.artofproblemsolving.com/7/2/3/7234d348576e0e3c8ff02ea030007fe79248badd.png)
Thus 
.![\[ \prod_{i=0}^{n-1} (2^n - 2^i) > 2^{n^2}. \]](http://latex.artofproblemsolving.com/b/8/d/b8d2a4745f458fcc08dcbda034a5c8109da4205f.png)
It is easy to see that these two facts are impossible for sufficiently large 
.
will suffice since
Then we induct up:![\[ \left(\frac{n(n+1)}{2}\right)! > 2^{n^2} \cdot 6^{n} > 2^{(n+1)^2} \]](http://latex.artofproblemsolving.com/3/4/d/34d88799270ef7ae5bc81222228fcedf36de1c5e.png)
for all 
.
.
are solutions, but when 
,
has no solution, note that the right handside is divisible by 31, so then 
,
But we also have:
Then we must have 
So we must have :
This holds only for n<7. We can check 
by hand.
,
= n(n-1)/2 thus 
, now RHS<
thus k!<
>
> 
if 
Now, from LTE, we have that 
.
By checking manually we see that 
and 
,
and 
to see they do not have an integer solution for 
.
.
,
contributes at least one factor of 
.
.
is 
and 
,
,
.
for all 
,
of both sides we have,
Thus, 
.
if and only if 
.
.
Thus, as a result of our previous bound we have,
which implies,![\[\frac{n}{6}-1<\lfloor \frac{n}{6} \rfloor\le \nu_3((\lfloor \frac{n}{2} \rfloor)!) \le \frac{\lfloor \frac{n}{2} \rfloor}{n-2}\]](http://latex.artofproblemsolving.com/b/b/a/bba35b874f0f61eece62e9a7f540737caed7d8a1.png)
from which it is not hard to see that we require 
.
we see that 
and 
work yielding 
and 
respectively and others don't 
,
.
as well. So, 29 is a fraction of some expression of the form 
which is clearly impossible for any 
which completes the check.
be the sum of digits in the base 
expansion of 
.![\[ \nu_2(Q) = \frac{n(n-1)}{2}\]](http://latex.artofproblemsolving.com/a/6/7/a6720ea5962b7d0ea0948953ee2d25a289d4caed.png)
That shows 
.![\[ \nu_p(k!) \leq \lfloor \frac{2^n}{p-1}\rfloor \leq \nu_p(Q)\]](http://latex.artofproblemsolving.com/1/3/3/133030946003860c1aaa3ae201d8d9b209cd8c00.png)
Clearly for all primes the 
makes no diffrence. now the equality should hold for that 
we see that only 
work
and 
.
to balance out the other prime factors. Specifically, in addition to examining 
.![\[\nu_2(k!) = \sum_{i = 0}^{n - 1} \nu_2(2^n - 2^i) = 1 + 2 + \dots + (n - 1) = \frac{n(n - 1)}{2}\]](http://latex.artofproblemsolving.com/7/2/2/722ef53213b209dda2ef9fd441696edfffeede2b.png)
and ![\[\nu_3(k!) = \nu_3\left(\prod_{i = 0}^{n - 1} (2^n - 2^i)\right) = \sum_{i = 1}^n \nu_3(2^i - 1) = \sum_{i = 1}^{\lfloor n/2 \rfloor} \nu_3(4^i - 1) = \left\lfloor \frac n2 \right\rfloor + \sum_{i = 1}^{\lfloor n/2 \rfloor} \nu_3(i)\]](http://latex.artofproblemsolving.com/1/5/b/15b3c78038390215e3ca158812562630e0a22d66.png)
![\[ = \left\lfloor \frac n2 \right\rfloor + \nu_3\left(\left\lfloor \frac n2 \right\rfloor!\right)\]](http://latex.artofproblemsolving.com/c/e/a/cea539d00d3c035e331b12a5765f3a90ae6bf928.png)
by LTE and the fact that 
is 
,![\[\nu_3\left(\left(\left\lfloor \frac n2 \right\rfloor + 1\right) \cdot \left(\left\lfloor \frac n2 \right\rfloor + 2\right) \cdots k\right) = \left\lfloor \frac n2 \right\rfloor.\]](http://latex.artofproblemsolving.com/e/9/c/e9cac8a489be49fa9801c86ff47de09d73000ebd.png)
At least 
terms in the product on the left side are divisible by 
,![\[\left\lfloor \frac n2 \right\rfloor \geq \left\lfloor \frac{k - \lfloor n/2 \rfloor}{3} \right\rfloor \geq \frac{k - \lfloor n/2 \rfloor}{3} - 1 \implies k \leq 4 \left\lfloor \frac n2 \right\rfloor + 3 \leq 2n + 3.\]](http://latex.artofproblemsolving.com/8/3/c/83ca8e2cf43596a8c68d6d62d174d57d2cb4d613.png)
But from before we have ![\[\frac{n(n - 1)}{2} = \nu_2(k!) = k - s_2(k) < k,\]](http://latex.artofproblemsolving.com/6/3/7/63736929ca5258a66bede6fb9edde5a5519be0ba.png)
so 
.
,
.
and 
work as desired. 
then no such 
.
if 
and we can check that 
if 
.
which is equivalent to the lemma. ////
.
Now we state another lemma.
for 
.
iff 
.
Since 
it follows that 
.
So from 
and 
it follows that 
as claimed.
we can manually check and find that only 
only, it is trivial to verify that both of these work.
,
which is just 
,
,
.
.
we have 
,
,
is true for most 
the inequality 
is true, which is just combining the three highest terms with the three lowest. This is just obviously true by computation and the fact that the function is increasing.
![\[\frac{n(n-1)}2=\nu_2\left(\prod_{i=0}^{n-1}(2^n-2^i)\right)=\nu_2(k!)\le k\]](http://latex.artofproblemsolving.com/0/3/7/037b148c6003a97b31b0b09ec7a7670951ea423c.png)
where the last step follows from evaluating Legendre's formula as a geometric series.![\[\frac{k}3-1<\nu_3(k!)=\nu_3\left(\prod_{i=0}^{n-1}(2^n-2^i)\right)=\nu_3\left(\prod_{i=1}^{n}(2^i-1)\right).\]](http://latex.artofproblemsolving.com/6/0/9/6097e98727e3b3950cdac8dc73e482f91e453a47.png)
Now 
if 
is odd and 
if ![\[\frac{k}3-1<\nu_3\left(\prod_{i=1}^{n}(2^i-1)\right)\le\frac{n}2+\sum_{i=1}^{\left\lfloor\frac{n}2\right\rfloor}\nu_3(2i)\le\frac{n}2+\nu_3(n!)<n\]](http://latex.artofproblemsolving.com/a/2/5/a25e3d650555c80829594f60c3797e3a18ee916f.png)
where the last step again follows from evaluating Legendre's formula as a geometric series.
,
.
.
and 
.
and 
.
,![\[v_5(k!) \geq v_7(k!)\]](http://latex.artofproblemsolving.com/f/2/b/f2beb3f8c83a8e34c2b44ba6d3ba6f0cbb3d534f.png)
This is trivial by Legendre's. Now, let![\[P(n) = \prod_{i = 0}^{n - 1} (2^n- 2^i)\]](http://latex.artofproblemsolving.com/4/8/8/488abc99a72779408a434c9d239d6b8414fa8115.png)
Thus, if there were a solution pair 
we would need![\[v_5(P(n)) \geq v_7(P(n))\]](http://latex.artofproblemsolving.com/d/0/7/d0798640f22222ed0a3f244e9766cdb3ef809216.png)
Now, note that the order of 
is 
respectively. So, by LTE![\[v_5(2^{4k} - 1) = v_5(k) + 1\]](http://latex.artofproblemsolving.com/3/a/6/3a6c3ed57b749485437b8ca1efc8689460834ea8.png)
![\[v_7(2^{3k} - 1) = v_7(k) + 1\]](http://latex.artofproblemsolving.com/e/1/e/e1ee9b93c860552121438fe02782428da1eebc17.png)
Thus, we in fact have![\[v_5(P(n)) = \left\lfloor \frac{n}{4} \right\rfloor + \left\lfloor \frac{n}{4 \cdot 5} \right \rfloor + \cdots\]](http://latex.artofproblemsolving.com/5/7/b/57b285a50087b6d9edc4b9280654950eece4f1e5.png)
![\[v_7(P(n)) = \left\lfloor \frac{n}{3} \right\rfloor + \left\lfloor \frac{n}{3 \cdot 7} \right \rfloor + \cdots\]](http://latex.artofproblemsolving.com/9/3/b/93b9dad4af0ee0c3a133b7710c9169be095ed9c5.png)
Now, notice that![\[v_5(P(n)) = \left\lfloor \frac{n}{4} \right\rfloor + \left\lfloor \frac{n}{4 \cdot 5} \right\rfloor + \cdots < \frac{n}{4} + \frac{n}{4 \cdot 5} + \cdots = \frac{5n}{16}\]](http://latex.artofproblemsolving.com/6/e/e/6ee765ae81d6a8045a7dec3ea7a079830554b92b.png)
Furthermore, if 
,![\[v_7(P(n)) > \frac{n}{3} r + \cdots + \frac{n}{3 \cdot 7^m} - (m + 1) \geq \frac{n}{18}(\frac{7^{m + 1} - 1}{7^m}) - (m + 1)\]](http://latex.artofproblemsolving.com/7/2/b/72b0c6b9676153b313c9b182bee91dafd4083535.png)
Assume 
,![\[v_7(P(n)) > \frac{n}{3} r + \cdots + \frac{n}{3 \cdot 7^m} - (m + 1) \geq \frac{n}{18}(\frac{7^{m + 1} - 1}{7^m}) - (m + 1) > \frac{8n}{21} - m - 1\]](http://latex.artofproblemsolving.com/9/e/b/9ebce375a5d242af0392a6ef0cc8dbd7f9651e7d.png)
Thus, we have![\[\frac{5n}{16} > v_5(P(n)) \geq v_7(P(n)) > \frac{8n}{21} - \log_7(n) - 1\]](http://latex.artofproblemsolving.com/9/8/c/98c14c3ecb4923096e3762ddc189be32951406fc.png)
![\[\log_7(n) + 1 > \frac{23n}{16 \cdot 21}\]](http://latex.artofproblemsolving.com/4/4/5/445c3cf3bacf1f3481d755ac3632918b4b222c27.png)
![\[7n > 7^{\frac{23n}{16 \cdot 21}}\]](http://latex.artofproblemsolving.com/a/7/4/a745c9dfc41a112cff0b333d80749ac4713b0fa8.png)
Since the RHS grows faster than the LHS, 
doesn't satisfy the above, so we must have 
.
and 
can be manually calculated, giving 
as the possible candidates, of which only 
work, so we are done.
,
.
.![\[k = \sum_{i=1}^{\infty} \frac{k}{2^i} \geq \sum_{i=1}^{\infty} \left\lfloor\frac{k}{2^i}\right\rfloor = v_2(k!) = \frac{n(n-1)}{2}\]](http://latex.artofproblemsolving.com/d/b/f/dbf1dd3f23db32cb7478c44379ce5510e252032d.png)
Hence we obtain![\[k! \geq \frac{n(n-1)}{2}! \Longrightarrow (2^n-1)(2^n-2)(2^n-4)\cdots(2^n-2^{n-1}) \geq \frac{n(n-1)}{2}!\]](http://latex.artofproblemsolving.com/6/c/a/6cab87f1f6870738beed60c17595c98db1fc87e8.png)
However it is obvious that![\[2^{n^2-1} \geq (2^n-1)(2^n-2)(2^n-4) \cdots (2^n-2^{n-1})\]](http://latex.artofproblemsolving.com/9/f/d/9fde65b1103d659ec97e84c1e80d7e77c7203ae7.png)
the 
appears since 
,![\[2^{n-1} \cdot 4^{\frac{n(n-1)}{2}} = 2^{n^2-1} \geq \frac{n(n-1)}{2}!\]](http://latex.artofproblemsolving.com/7/b/0/7b0a9bbf5b8a9e24802c3519f57f60c9f3e00622.png)
we must have 
,
.
it suffices to prove![\[2^{17} = 2^5 \cdot 4^{15-9} < 10(11)(12)(13)(14)(15)\]](http://latex.artofproblemsolving.com/b/d/6/bd65193dccdda2619af03f8b7c9f1cfe8750917d.png)
However this is obvious since 
and 
.
to obtain the values above (note 
is easy since 
implying 
which is absurd).
:
- this is a corollary of the fact that 
:
.
A cool thing we can do from here is take factorials of both sides (since we already have an expression for 
)
From here, we can rather quickly verify that 
for all 
(induction suffices), therefore all that remains is to check 
.
Meanwhile by LTE and Legendre's 
Thus, 
Testing yields the only solutions 
