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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Geometry
Lukariman   1
N 4 minutes ago by Primeniyazidayi
Given acute triangle ABC ,AB=b,AC=c . M is a variable point on side AB. The circle circumscribing triangle BCM intersects AC at N.

a)Let I be the center of the circle circumscribing triangle AMN. Prove that I always lies on a fixed line.

b)Let J be the center of the circle circumscribing triangle MBC. Prove that line segment IJ has a constant length.
1 reply
Lukariman
4 hours ago
Primeniyazidayi
4 minutes ago
Kingdom of Anisotropy
v_Enhance   24
N 12 minutes ago by deduck
Source: IMO Shortlist 2021 C4
The kingdom of Anisotropy consists of $n$ cities. For every two cities there exists exactly one direct one-way road between them. We say that a path from $X$ to $Y$ is a sequence of roads such that one can move from $X$ to $Y$ along this sequence without returning to an already visited city. A collection of paths is called diverse if no road belongs to two or more paths in the collection.

Let $A$ and $B$ be two distinct cities in Anisotropy. Let $N_{AB}$ denote the maximal number of paths in a diverse collection of paths from $A$ to $B$. Similarly, let $N_{BA}$ denote the maximal number of paths in a diverse collection of paths from $B$ to $A$. Prove that the equality $N_{AB} = N_{BA}$ holds if and only if the number of roads going out from $A$ is the same as the number of roads going out from $B$.

Proposed by Warut Suksompong, Thailand
24 replies
v_Enhance
Jul 12, 2022
deduck
12 minutes ago
Incentre-excentre geometry
oVlad   2
N 21 minutes ago by Double07
Source: Romania Junior TST 2025 Day 2 P2
Consider a scalene triangle $ABC$ with incentre $I$ and excentres $I_a,I_b,$ and $I_c$, opposite the vertices $A,B,$ and $C$ respectively. The incircle touches $BC,CA,$ and $AB$ at $E,F,$ and $G$ respectively. Prove that the circles $IEI_a,IFI_b,$ and $IGI_c$ have a common point other than $I$.
2 replies
oVlad
Yesterday at 12:54 PM
Double07
21 minutes ago
Great similarity
steven_zhang123   4
N 43 minutes ago by khina
Source: a friend
As shown in the figure, there are two points $D$ and $E$ outside triangle $ABC$ such that $\angle DAB = \angle CAE$ and $\angle ABD + \angle ACE = 180^{\circ}$. Connect $BE$ and $DC$, which intersect at point $O$. Let $AO$ intersect $BC$ at point $F$. Prove that $\angle ACE = \angle AFC$.
4 replies
steven_zhang123
5 hours ago
khina
43 minutes ago
Unexpected FE
Taco12   18
N an hour ago by lpieleanu
Source: 2023 Fall TJ Proof TST, Problem 3
Find all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that for all integers $x$ and $y$, \[ f(2x+f(y))+f(f(2x))=y. \]
Calvin Wang and Zani Xu
18 replies
1 viewing
Taco12
Oct 6, 2023
lpieleanu
an hour ago
Geometry
Lukariman   6
N 2 hours ago by Curious_Droid
Given circle (O) and point P outside (O). From P draw tangents PA and PB to (O) with contact points A, B. On the opposite ray of ray BP, take point M. The circle circumscribing triangle APM intersects (O) at the second point D. Let H be the projection of B on AM. Prove that $\angle HDM$ = 2∠AMP.
6 replies
Lukariman
Yesterday at 12:43 PM
Curious_Droid
2 hours ago
Powers of a Prime
numbertheorist17   33
N 2 hours ago by OronSH
Source: USA TSTST 2014, Problem 6
Suppose we have distinct positive integers $a, b, c, d$, and an odd prime $p$ not dividing any of them, and an integer $M$ such that if one considers the infinite sequence \begin{align*}
		ca &- db \\
		ca^2 &- db^2 \\
		ca^3 &- db^3 \\
		ca^4 &- db^4 \\
&\vdots
	\end{align*} and looks at the highest power of $p$ that divides each of them, these powers are not all zero, and are all at most $M$. Prove that there exists some $T$ (which may depend on $a,b,c,d,p,M$) such that whenever $p$ divides an element of this sequence, the maximum power of $p$ that divides that element is exactly $p^T$.
33 replies
numbertheorist17
Jul 16, 2014
OronSH
2 hours ago
Expected Intersections from Random Pairing on a Circle
tom-nowy   2
N 2 hours ago by lele0305
Let $n$ be a positive integer. Consider $2n$ points on the circumference of a circle.
These points are randomly divided into $n$ pairs, and $n$ line segments are drawn connecting the points in each pair.
Find the expected number of intersection points formed by these segments, assuming no three segments intersect at a single point.
2 replies
tom-nowy
3 hours ago
lele0305
2 hours ago
question4
sahadian   5
N 3 hours ago by Mamadi
Source: iran tst 2014 first exam
Find the maximum number of Permutation of set {$1,2,3,...,2014$} such that for every 2 different number $a$ and $b$ in this set at last in one of the permutation
$b$ comes exactly after $a$
5 replies
sahadian
Apr 14, 2014
Mamadi
3 hours ago
Find all functions $f$: \(\mathbb{R^+}\) \(\rightarrow\) \(\mathbb{R^+}\) such
guramuta   5
N 3 hours ago by jasperE3
Source: Balkan MO SL 2021
A5: Find all functions $f$: \(\mathbb{R^+}\) \(\rightarrow\) \(\mathbb{R^+}\) such that:
$$f(xf(x+y)) = xf(y) + 1 $$
5 replies
guramuta
4 hours ago
jasperE3
3 hours ago
number theory
frost23   3
N 3 hours ago by frost23
given any positive integer n show that there are two positive rational numbers a and b not equal to b which are such that a-b, a^2- b^2....................a^n-b^n are all integers
3 replies
frost23
4 hours ago
frost23
3 hours ago
partitioned square
moldovan   8
N 3 hours ago by cursed_tangent1434
Source: Ireland 1994
If a square is partitioned into $ n$ convex polygons, determine the maximum possible number of edges in the obtained figure.

(You may wish to use the following theorem of Euler: If a polygon is partitioned into $ n$ polygons with $ v$ vertices and $ e$ edges in the resulting figure, then $ v-e+n=1$.)
8 replies
moldovan
Jun 29, 2009
cursed_tangent1434
3 hours ago
Finding positive integers with good divisors
nAalniaOMliO   3
N 4 hours ago by nAalniaOMliO
Source: Belarusian National Olympiad 2025
For every positive integer $n$ write all its divisors in increasing order: $1=d_1<d_2<\ldots<d_k=n$.
Find all $n$ such that $2025 \cdot n=d_{20} \cdot d_{25}$.
3 replies
nAalniaOMliO
Mar 28, 2025
nAalniaOMliO
4 hours ago
Concurrent lines
MathChallenger101   4
N 4 hours ago by oVlad
Let $A B C D$ be an inscribed quadrilateral. Circles of diameters $A B$ and $C D$ intersect at points $X_1$ and $Y_1$, and circles of diameters $B C$ and $A D$ intersect at points $X_2$ and $Y_2$. The circles of diameters $A C$ and $B D$ intersect in two points $X_3$ and $Y_3$. Prove that the lines $X_1 Y_1, X_2 Y_2$ and $X_3 Y_3$ are concurrent.
4 replies
MathChallenger101
Feb 8, 2025
oVlad
4 hours ago
IMO ShortList 2008, Number Theory problem 4
April   20
N Apr 18, 2025 by megarnie
Source: IMO ShortList 2008, Number Theory problem 4
Let $ n$ be a positive integer. Show that the numbers
\[ \binom{2^n - 1}{0},\; \binom{2^n - 1}{1},\; \binom{2^n - 1}{2},\; \ldots,\; \binom{2^n - 1}{2^{n - 1} - 1}\]
are congruent modulo $ 2^n$ to $ 1$, $ 3$, $ 5$, $ \ldots$, $ 2^n - 1$ in some order.

Proposed by Duskan Dukic, Serbia
20 replies
April
Jul 9, 2009
megarnie
Apr 18, 2025
IMO ShortList 2008, Number Theory problem 4
G H J
Source: IMO ShortList 2008, Number Theory problem 4
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April
1270 posts
#1 • 4 Y
Y by narutomath96, Amir Hossein, Adventure10, Mango247
Let $ n$ be a positive integer. Show that the numbers
\[ \binom{2^n - 1}{0},\; \binom{2^n - 1}{1},\; \binom{2^n - 1}{2},\; \ldots,\; \binom{2^n - 1}{2^{n - 1} - 1}\]
are congruent modulo $ 2^n$ to $ 1$, $ 3$, $ 5$, $ \ldots$, $ 2^n - 1$ in some order.

Proposed by Duskan Dukic, Serbia
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daniel73
253 posts
#2 • 2 Y
Y by Adventure10, Mango247
You may just as well consider the remainders of $ \binom{2^n-1}{2k}$ for $ k=0,1,2,\dots,2^{n-1}-1$, and now,

Hint 1:
Click to reveal hidden text

Hint 2:
Click to reveal hidden text
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hxy09
602 posts
#3 • 1 Y
Y by Adventure10
induction on $ n$
and we prove the srtonger assertion:
$ \binom{2^{n}-1}{i}-\binom{2^{n}-1}{j}$ is not a multiple of $ 2^n$ for every $ (i,j)$ satisfies$ 1\le i<j \le 2^{n-1}-1$
and $ \binom{2^{n}-1}{i}+\binom{2^{n}-1}{j}$ is not a multiple of $ 2^{n+1}$ for every $ (i,j)$ satisfies$ 1\le i<j \le 2^{n-1}-1$
And I will post the full solution below(very sorry for being written in Chinese,I am glad to explain if anyone has questions) :)
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hxy09
602 posts
#4 • 3 Y
Y by narutomath96, Adventure10, Mango247
I am grateful if anyone can point out my mistakes or flaws :)
Attachments:
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Albanian Eagle
1693 posts
#5 • 2 Y
Y by Adventure10, Mango247
I solved it by proving

$ 1\le a\le b\le 2^{n-1}-1 \implies$
$ \prod_{x=a}^b \frac{2^n-x}{x}$ is not $ \equiv 1\pmod{2^n}$ and not $ \equiv -1\pmod{2^{n+1}}$
which is almost trivial by induction.
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peregrinefalcon88
299 posts
#6 • 2 Y
Y by Adventure10, Mango247
Let us note that $\binom{2^{n}-1}{k} = \prod_{j = 1}^{k}\frac{2^n-j}{j}$. Now, let us define the function F(n) = $\{ k \in \mathbb{Z} | 2^k | n, 2^{k+1} \nmid n\}$. We can see that F($2^n$-j) = F(j). From this we can see that $\binom{2^{n}-1}{k} \equiv 1$ mod $2$ $\forall$ $0 \le k \le 2^{n-1}-1$. We can also see that $F(j) \le 2^{n-2}$ since $j \le 2^{n-1}-1$. From this we can see that $\frac{2^n-j}{j} \equiv 3$ mod $4$. Now we notice that $\binom{2^n-2j}{2j} \equiv 1$ mod $4$ and $\binom{2^n-2j-1}{2j+1} \equiv 3$ mod $4$. Thus we now need only to show that $\binom{2^n-2j}{2j} \not\equiv \binom{2^n-2t}{2t}$ mod $2^n $ and $\binom{2^n-2j-1}{2j+1} \not\equiv \binom{2^n-2t-1}{2t+1}$ mod $2^n$ $\forall$ $j \neq t, 0 \le j, t \le 2^{n-2}-1$. To prove this we need only to show that $\prod_{j = a}^{b}\frac{2^n-j}{j} \not\equiv 1$ mod $2^n$ $\forall a \not\equiv b$ mod $2$. However, from Vieta's relatoins we can see that $\prod_{j = a}^{b}\frac{2^n-j}{j}$ can be rewritten as $\frac{2^n(2^n(x)+\sum_{j = a}^bj)}{\prod_{j = a}^bj}+1$ for some integer x. So we know want to show that $\frac{2^n(2^n(x)+\sum_{j = a}^bj)}{\prod_{j = a}^bj} \equiv 0$ mod $2^n$. However, we notice that $2^n(x)+\sum_{j = a}^bj$ is odd which means that there cannot be any factors of 2 in the denominator for $\frac{2^n(2^n(x)+\sum_{j = a}^bj)}{\prod_{j = a}^bj} $ to be divisible by $2^n$ and this is contradiction. Thus, the problem is proved.
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CatalystOfNostalgia
1479 posts
#7 • 5 Y
Y by DreamComeTrue, guptaamitu1, Adventure10, Mango247, and 1 other user
A different approach: as $\binom{2^{n}-1}{i}$ is clearly odd, we need to show that $\binom{2^{n}-1}{j}/\binom{2^{n}-1}{i}$ is not 1 modulo $2^{n}$, if $1\le i<j\le 2^{n-1}-1$. We can do this by writing this as the product of $\frac{2^{n}-k}{k}$ for $i+1\le k\le j$. Now, choose $k$ such that $v_{2}(k)=m$ is maximized: it is easy to check that $k$ is unique. From here, after canceling powers of 2, it is easy to see that each term except than the one indexed by $k$ is $-1$ modulo $2^{n-m+1}$, and it is also easy to check that the exceptional one is not $\pm1$. Thus $\binom{2^{n}-1}{j}/\binom{2^{n}-1}{i}$ is not 1 modulo $2^{n-m+1}$, so if $m>0$ certainly it is not 1 modulo $2^{n}$ either. If $m=0$, on the other hand, we only have one factor in the product, so the fraction is equal to $-1$ modulo $2^{n-m+1}$, so we are done in this case as well.
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shivangjindal
676 posts
#8 • 1 Y
Y by Adventure10
April wrote:
Let $ n$ be a positive integer. Show that the numbers
\[ \binom{2^n - 1}{0},\; \binom{2^n - 1}{1},\; \binom{2^n - 1}{2},\; \ldots,\; \binom{2^n - 1}{2^{n - 1} - 1}\]
are congruent modulo $ 2^n$ to $ 1$, $ 3$, $ 5$, $ \ldots$, $ 2^n - 1$ in some order.

Proposed by Duskan Dukic, Serbia

I finally solved it , after trying it since days! .
It is equivalent to show that $\binom{2^n-1}{1},\binom{2^n-1}{3}, \cdots , \binom{2^n-1}{2^{n}-1}$ leaves distinct remainders .
Suppose for the contrary that $\binom{2^n-1}{x} \equiv \binom{2^n-1}{y}\pmod{2^n}$. We can take $x>y$ . Using the binomial identities we get,
$\binom{2^n}{x}-\binom{2^n}{x-1}+\binom{2^n}{x-2}+\cdots-\binom{2^n}{y+1} \equiv 0 \pmod{2^n}$.
Now , since there are powers of $2$ , it is handy to use kummer's theorem . we will find $v_2{\binom{2^n}{k}}$ .
Suppose that $v_2(k)=q$ then using kummer';s theorem we get $v_2{\binom{2^n}{k}} = n-q$ . Note that , if k is odd then $q=0$. Thus , in this case $2^n$ will divide $\binom{2^n}{k}$ .
So we have to only check the sum $k$ which are even.
Also we note that if $k$ is even then $v_2{\binom{2^n}{k}} < n$ . Now , we choose a $k=a$ such that $v_2{\binom{2^n}{a}}=w$ is minimal. Then we will be able to take $2^w$ common , and the left term will be odd . so there will be a term where $k=b$ such that , $v_2{\binom{2^n}{a}} = v_2{\binom{2^n}{b}}$ . Now from earlier observation we see that , more the $v_2{k}$ less will be $v_2{\binom{2^n}{k}}$ . here we have , $v_2{\binom{2^n}{a}} = v_2{\binom{2^n}{b}} \implies v_2(a)=v_2(b)$. Suppose that $a<b$. Obviously $\exists c$ between $a$ and $b$ such that $v_2(a)<v_2(c)$ . Contradicting the minimality of $v_2{\binom{2^n}{x}}$ . So we are done! $\Box$
^^ Someone please check it , i am not sure it it is correct.
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maths.lover
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#9 • 2 Y
Y by Adventure10, Mango247
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WOuld you please explain what's $v_2{\binom{2^n}{k}}$ ?? I'm dying to understand this, especially kummer's theorem !??
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va2010
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#10 • 1 Y
Y by Adventure10
Hmm, I'd like to point out a solution without using valuations, for those who don't know it...

All such numbers are odd, which is easy to check.

Lemma 1: $\dbinom{2^n - 1}{2k} + \dbinom{2^n-1}{2k+1} \equiv 2^n \pmod{2^{n+1}}$

Proof: Observe that $\dbinom{2^n - 1}{2k}(1 + \frac{2^n - (2k+1)}{2k+1}) = \dbinom{2^n - 1}{2k}(\frac{2^n}{2k+1}) \equiv 0 \pmod{2^n}$, but it doesn't divide $2^{n+1}$, so it is equivalent to $2^n \pmod{2^{n+1}}$

Claim: $\dbinom{2^n - 1}{4k+3} \equiv \dbinom{2^{n-1}-1}{2k+1} \pmod{2^n}$ for defined values of $k$

Proof: We use induction. Check the result for $k = 0$. Now observe that for $k \rightarrow k+1$, it suffices to show that $\frac{\dbinom{2^n - 1}{4k+7}}{\dbinom{2^n - 1}{4k + 3}} \equiv \frac{\dbinom{2^{n-1} - 1}{2k+3}}{\dbinom{2^{n-1} - 1}{2k+1}} \pmod{2^n}$

Observe that the first product is equivalent to $(\frac{2^n - (4k+7)}{4k+7})(\frac{2^n - (4k+6)}{4k+6})(\frac{2^n - (4k+5)}{4k+5})(\frac{2^n - (4k+4)}{4k+4}) \equiv (-1)^2 (\frac{2^{n-1} - (2k+3)}{2k+3})(\frac{2^{n-1} - (2k+2)}{2k+2}) \equiv \frac{\dbinom{2^{n-1}-1}{2k+3}}{\dbinom{2^{n-1} - 1}{2k+1}} \pmod{2^n} $

Claim: $\dbinom{2^n - 1}{4k} \equiv \dbinom{2^{n-1} - 1}{2k} \pmod{2^n}$:
Proof: Similar to above.

Denote $P_n$ the assertion for $n$. We use induction. The result is true for $n = 1, 2, 3$, and $4$. We prove it for larger values.

So for induction, given $P_n$, observe that each term in $S_n$, the values of the binomial coefficients, are distinct mod $2^n$. When we use the above relations, we obtain a $P_{n+1}$, after using the lemma. How could there be a duplicate in this set? Well, if there is one, it would have to be a number from $S_n$, and then the lemma would produce another value. These two values would somehow be the same if the duplicate occurred. We show that this cannot happen. Observe that this implies that in $S_n$, there are two numbers such that, when summed, give $0 \pmod{2^{n+1}}$. First, the sum is obviously $0 \pmod{2^n}$, which each element in $S_n$ has only one pair for: one of its neighbors. But the sum, as shown, isn't $0 \pmod{2^{n+1}}$, so we are done.
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Zoom
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#11 • 1 Y
Y by Adventure10
April wrote:
Proposed by Duskan Dukic, Serbia

Actually his name is Dušan Djukić, or if you don’t have Serbian latin letters, Dusan Djukic is fine.
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rocketscience
466 posts
#12 • 1 Y
Y by Adventure10
Induct, with base cases trivial. Our inductive hypothesis implies that there exist precisely two solutions $x=m$ and $x=2^{n-1}-1-m$ satisfying $\binom{2^{n-1} - 1}{x} \equiv L \pmod{2^{n-1}}$ for any given odd number $L$. Note that
\[\binom{2^n-1}{2k} = \prod_{i=1}^{2k} \frac{2^n-i}{i} = \left( \prod_{i=1}^{k} \frac{2^{n-1}-i}{i} \right) \left(\prod_{\substack{j=1, \\ j \text{ odd}}}^{2k-1} \frac{2^n-j}{j}\right) = \binom{2^{n-1}-1}{k} \left(\prod_{\substack{j=1, \\ j \text{ odd}}}^{2k-1} \frac{2^n-j}{j}\right).\]We can safely take the left and right ends of this equality modulo $2^n$, whence $\binom{2^n-1}{2k} \equiv (-1)^k\binom{2^{n-1}-1}{k} \pmod{2^n}$. Now suppose that $\binom{2^n-1}{2i} \equiv \binom{2^n-1}{2j} \pmod {2^n}$ so that
\[(-1)^i\binom{2^{n-1}-1}{i} \equiv (-1)^j\binom{2^{n-1}-1}{j} \pmod{2^n}.\]Some casework:
  • If $i$ and $j$ are of the same parity, then the above binomial coefficients are also congruent modulo $2^{n-1}$, so by hypothesis either $i=j$ or $i=2^{n-1} -1- j$. Parity forces $i=j$.
  • If $i$ and $j$ are of different parity, we have
    \[\binom{2^{n-1}-1}{i} \equiv -\binom{2^{n-1}-1}{j} \pmod{2^n}.  \quad (\heartsuit)\]WLOG $i$ is even. This congruence $(\heartsuit)$ holds modulo $2^{n-1}$, but we can also check that
    \[\binom{2^{n-1}-1}{i} \equiv -\frac{2^{n-1}-(i+1)}{i+1} \binom{2^{n-1}-1}{i} = -\binom{2^{n-1}-1}{i+1}  \pmod{2^{n-1}}\]\[\implies \binom{2^{n-1}-1}{j} \equiv \binom{2^{n-1}-1}{i+1} \pmod{2^{n-1}}.\]By hypothesis, this forces either $j=i+1$ or $j=2^{n-1}-1-(i+1)$, but $i$ and $j$ have different parity so we must have $j=i+1$. Now examine $(\heartsuit)$. Rewrite the right-side binomial coefficient to get
    \[\binom{2^{n-1}-1}{i} \equiv -\frac{2^{n-1}-(i+1)}{i+1} \binom{2^{n-1}-1}{i} \pmod{2^n} \]\[\implies \frac{2^{n-1}-(i+1)}{i+1} \equiv -1 \implies 2^{n-1} \equiv 0 \pmod{2^n},\]which is absurd. So this case is impossible.
The above work shows that $\binom{2^n-1}{2k}$ for $k=0, 1, \dots, 2^{n-1}-1$ are distinct odd numbers modulo $2^n$, which easily converts to the desired claim.
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HKIS200543
380 posts
#13
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Our solution proceeds in two parts. First we show that $\binom{2^n-1}{a}$ is odd for any integer $a$. Then we show that
\[ \binom{2^n-1}{a} \not\equiv \binom{2^n-1}{b} \pmod{2^n} \]for any $0 \leq a < b \leq 2^{n-1}-1$. It is easy to see that these two claims put together give the result.

The first part is rather simple. By Legendre,
\[ \nu_2 \left( \binom{2^n-1}{a} \right) = s_2(a) + s_2(2^{n-1}-1-a) - s_2(2^n-1) \]Since no carries occur, this is just zero.

Most of the problem is in the second part. It suffices to show that
\[ \prod_{k=b+1}^a \frac{2^n - k}{k} \not\equiv 1 \pmod{2^n} .\]Let $M = \underset{b < k \leq a}{\max} \nu_2(k)$. Denote by $t$ the unique integer in that range where the maximum is taken.Clearly, $M \leq n-2$. Let $t = 2^M m$. Then
\begin{align*} \prod_{k=b+1}^a \frac{2^n - k}{k} &= \prod_{k=b+1}^a \frac{2^{n- \nu_2(k)} - k/\nu_2(k)}{k /\nu_2(k} \\
& \equiv (-1)^{a-b-1} \frac{2^{n-M} - m}{m}  \\
& \equiv \pm (2^{n-M} - 1) .  \pmod{2^{n-M+1}} . 
\end{align*}At the last step, we make use of the fact that $m^{-1}$ is odd. Since $2^{n-M} \geq 2$, it is impossible that
\[ 2^{n-M} - 1 \equiv \pm 1 \pmod{2^{n-M + 1}}. \]Since
\[ \binom{2^n-1}{a} = \binom{2^n-1}{b} \prod_{k=b+1}^a \frac{2^n - k}{k} \]we are done.
This post has been edited 2 times. Last edited by HKIS200543, Aug 9, 2020, 2:37 AM
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yayups
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#14
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Before starting with the proof, we establish that all the desired binomial coefficients are odd, which allows more ease in manipulating them modulo $2^n$.

Claim: We have that $\binom{2^n-1}{k}$ is odd for $0\le k\le 2^n-1$.

Proof: Let $s_2(m)$ denote the sum of the digits of $m$ written in binary. It is well known that \[v_2\left(\binom{a+b}{a}\right) = s_2(a+b)-s_2(a)-s_2(b).\]Using this, we see that \[v_2\left(\binom{2^n-1}{k}\right) = s_2(2^n-1) - s_2(k) - s_2((2^n-1)-k).\]Since $2^n-1$ is $n$ ones written in binary and $0\le k\le 2^n-1$, we see that the set of ones in the binary representation of $2^n-1-k$ is exactly the complement of those in the binary representation for $k$, so in fact \[s_2(2^n-1) - s_2(k) - s_2((2^n-1)-k)=0.\]This proves the claim. $\blacksquare$

We proceed by induction on $n$, with $n=1$ trivial. Now suppose the problem is true for $n$. Before proving $n+1$, we need to prove the following intermediate claim.

Claim: The numbers \[(-1)^a\binom{2^n-1}{a}\]generate all the odd residues modulo $2^{n+1}$ as $a$ varies from $0$ to $2^n-1$.

Proof: It suffices to show that \[(-1)^{a_1}\binom{2^n-1}{a_1} \equiv (-1)^{a_2}\binom{2^n-1}{a_2}\pmod{2^{n+1}}\]implies that $a_1=a_2$. The inductive hypothesis implies that in fact, $\binom{2^n-1}{2b}$ generates all the odd residues mod $2^n$ as $b$ varies from $0$ to $2^{n-1}-1$, and a similar statement holds if we replace $2b$ with $2b+1$. Thus, if $a_1$ and $a_2$ have the same parity, we are done trivially by simply looking modulo $2^n$.

So WLOG suppose that $a_1$ is even and $a_2$ is odd. Note that \[\binom{2^n-1}{a_2}=\frac{2^n-a_2}{a_2}\binom{2^n-1}{a_2-1},\]and that \[\frac{2^n-a_2}{a_2} = \frac{1}{a_2}2^n - 1 \equiv 2^n-1\pmod{2^{n+1}},\]so \[\binom{2^n-1}{a_2}\equiv (2^n-1) \binom{2^n-1}{a_2-1}\pmod{2^{n+1}}.\]Taking our original equation mod $2^n$ combined with the above implies that \[\binom{2^n-1}{a_1}\equiv \binom{2^n-1}{a_2-1}\pmod{2^n},\]and since both $a_1$ and $a_2-1$ are even, the inductive hypothesis implies that $a_1=a_2-1$. However, this them implies \[\binom{2^n-1}{a_2}\equiv (1-2^n)\binom{2^n-1}{a_2-1}\pmod{2^{n+1}},\]which is a clear contradiction as we can cancel out the binomial coefficients since they are odd. This is the desired contradiction, so $a_1$ and $a_2$ must in fact have the same parity, which we already resolved. This completes the proof. $\blacksquare$

The induction is now fairly clear. We see that \[\binom{2^{n+1}-1}{2a} = \frac{\prod_{i=0}^{a-1}(2^{n+1}-1-2i)}{\prod_{i=0}^{a-1}(1+2i)}\binom{2^n-1}{a}\equiv (-1)^a\binom{2^n}{a}\pmod{2^{n+1}},\]so \[\binom{2^{n+1}-1}{2a}\]generates all odd residues mod $2^{n+1}$, which completes the induction, and thus the proof.
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bora_olmez
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#15
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Very cool problem with an incredibly slick idea that only now reminds me of other problem which I only recalled during this write up. This solution is almost the same as #15 and in my opinion, is cleaner than the 'inductive/recursive' solutions above.

The fact that each of the binomial coefficients is odd is obvious because $\binom{2^n-1}{k} = \prod_{i=1}^{k} \frac{2^n-i}{i}$ where $i \leq 2^{n-1}-1$ hence $\nu_2(i) \leq n-1$ meaning that $\nu_2(2^n-i) = \nu_2(i)$ which means that $$\nu_2(
\frac{2^n-i}{i}) = \nu_2(2^n-i)-\nu_2(i) = 0$$and therefore $\binom{2^n-1}{k}$ is odd for all $k \in \{0, \cdots, 2^{n-1}-1\}$.

We will now prove that each of these binomial coefficients is distinct in $\pmod{2^n}$, assume for the sake of contradiction otherwise.

Then there exists $a>b$ among $\{0, \cdots 2^{n-1}-1\}$ such that $\binom{2^n-1}{a} \equiv \binom{2^n-1}{b} \pmod{2^n}$
Therefore, $$\prod_{i=b+1}^{a} \binom{2^n-i}{i} \equiv 1 \pmod{2^n}$$as odds have inverses.

Now notice the following result which should, in my opinion, be more popular. I actually had to come up with it myself again while solving the problem even though I have seen this is result in use for problem mentioned before.

$\textbf{Lemma 1:}$
For any set of consecutive positive integers, there exists an element with maximal $\nu_2()$.
$\textbf{Proof)}$
The proof is almost trivial, assume for the sake of contradiction that this is not the case, that is the maximal power of two is achieved twice, say at $u$ and $v$, then $u = 2^M \cdot k, v = 2^M \cdot t$ for odd positive integers $k,t$ such that without loss of generality $k < t$, then as $k \equiv t \equiv 1 \pmod{2}$, $k+1 < t$ as well and hence $s = 2^M(k+1)$ is among the integers achieving a higher $\nu_2()$, as desired. $\blacksquare$

Main Idea

Now, we can use $\textbf{Lemma 1}$ and consider the positive integer with maximal $\nu_2(t) = m$ among $b+1, \cdots, a$ for which $m \leq n-2$ because we are only considering $\binom{2^n-1}{0}, \binom{2^n-1}{1}, \cdots, \binom{2^n-1}{2^{n-1}-1}$. Then all but one of the fractions are of the form $\frac{2^T-s}{s}$ for odd $s$ and $T > n-m$ meaning that $$(-1)^{a-b-1} \cdot \frac{2^{n-m}-s}{s} \equiv \prod_{i=b+1}^{a} \binom{2^n-i}{i} \equiv 1 \pmod{2^{n-m+1}}$$for some $s$ odd which therefore means that $$(-1)^{a-b-1} \cdot s+s \equiv 2^{n-m} \pmod{2^{n-m+1}}$$as $2^{n-m} \not\equiv 0 \pmod{2^{n-m+1}}$ we must have that $$2s \equiv 2^{n-m} \pmod{2^{n-m+1}}$$which because $s$ is odd implies that $n-m = 1$ contradicting the fact that $m \leq n-2$, as desired. $\blacksquare$
This post has been edited 5 times. Last edited by bora_olmez, Feb 6, 2022, 12:31 AM
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HamstPan38825
8860 posts
#16
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Fairly sure this works.

The proof proceeds in two parts.
Part I: Parity. I claim that ${2^n-1 \choose i}$ is odd for all $0 \leq i \leq 2^n-1$. This follows by Lucas' Theorem because $2^n-1 = \overline{111\cdots 1}_2$, and thus subtracting any integer from it does not induce any carrying in base two.

Part II: Uniqueness. It remains to show that $${2^n-1 \choose i} \not \equiv {2^n-1 \choose j} \pmod {2^n}$$for $0 \leq i \neq j \leq 2^{n-1}-1$. Divide both sides of the congruence by ${2^n-1 \choose i}$, which is valid as the number is odd: $$\frac{{2^n-1 \choose j}}{{2^n - 1 \choose i}} \equiv \frac{(2^n-i-1)(2^n-i-2)\cdots(2^n-j)}{(i+1)(i+2)\cdots (j-1)j} \pmod {2^n}.$$We induct on $n$, with the stronger hypothesis that this expression cannot be congruent to 1 modulo $2^n$, and furthermore that it cannot be congruent to $-1$ modulo $2^n$ unless $|i-j| = 1$. The base case $n=2$ can be checked manually, and furthermore, notice that any two consecutive terms $$\frac{2^n - i}i \equiv -1 \pmod {2^n}$$if $i$ is odd, and we may reduce to the $2^{n-1}$ case if $n$ is even, so the $|i-j| = 1$ base case holds.

Next, suppose for the sake of contradiction that there exist $i < j$ such that the expression is congruent to $-1$ or 1 modulo $2^n$. Because $\nu_2(2^n-i-1) = \nu_2(i+1)$ for all $i$, we may remove terms on the numerator and denominator that are both odd and cancel out a factor of 2 from the remaining terms: $$(-1)^a \equiv \frac{(2^n-i-1)(2^n-i-2)\cdots(2^n-j)}{(i+1)(i+2)\cdots (j-1)j} \equiv (-1)^b \prod_{k=i'}^{j'} \frac{2^{n-1} - k}k \pmod {2^n}$$for some positive integers $i', j', a, b$. Notice that $j' \leq \frac j2 \leq 2^{n-2} - 1$ and obviously $i' \geq 0$, so we may apply the inductive hypothesis to obtain $$\prod_{k=i'}^{j'} \frac{2^{n-1} - k}k \not \equiv (-1)^{r_{n-1}} \pmod {2^{n-1}} \implies \prod_{k=i'}^{j'} \frac{2^{n-1} - k}k \not \equiv (-1)^{r_{n-1}} \pmod {2^n}.$$If there is only one term in the reduced expansion that is congruent to $-1 \pmod {2^{n-1}}$, notice that $$\frac{2^{n-1} - i}i \not \equiv -1 \pmod {2^n},$$so the quotient cannot be congruent to $-1$ either. This yields a contradiction, so the inductive step is complete.

As a result, all the binomial coefficients are congruent to distinct odd residues mod $2^n$. It follows they must be congruent to $1, 3, 5, \cdots, 2^n - 1$ in some order, as required.
This post has been edited 4 times. Last edited by HamstPan38825, Jul 20, 2022, 6:44 PM
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signifance
140 posts
#17
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Can someone check this? For some reason I'm terrible at algebra and this took me two hours.
The fact that they are all odd is evident by $\nu_2(i)<n-1\Rightarrow\nu_2(2^n-i)=\nu_2(i)$, so $\tbinom{2^n - 1}{2k}(1 + \tfrac{2^n - (2k+1)}{2k+1}) = \tbinom{2^n - 1}{2k}(\tfrac{2^n}{2k+1}) \equiv 2^n \pmod{2^{n+1}}$. Now by induction we prove that $\tbinom{2^n - 1}{4k+3} \equiv \tbinom{2^{n-1}-1}{2k+1} \pmod{2^n}$, base case evident, with inductive step following $$(\tfrac{2^n - (4k+7)}{4k+7})(\tfrac{2^n - (4k+6)}{4k+6})(\tfrac{2^n - (4k+5)}{4k+5})(\tfrac{2^n - (4k+4)}{4k+4}) \equiv (\tfrac{2^{n-1} - (2k+3)}{2k+3})(\tfrac{2^{n-1} - (2k+2)}{2k+2}) \equiv \tfrac{\binom{2^{n-1}-1}{2k+3}}{\binom{2^{n-1} - 1}{2k+1}} \pmod{2^n};$$similarly we deduce $\binom{2^n-1}{4k+1}\equiv\binom{2^{n-1}-1}{2k+1},\binom{2^n - 1}{4k} \equiv \binom{2^{n-1} - 1}{2k}\equiv\binom{2^n-1}{4k+2}\pmod{2^n}$. The problem is finished by inductive hypothesis of only odd residues appearing in tandem with noting that $\binom{2^{n+1}-1}{4k+1}\not\equiv\binom{2^{n+1}-1}{4k+3}\pmod{2^{n+1}}$, which follows from basic algebra.
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Mr.Sharkman
500 posts
#18
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Obviously, these are all odd by Lucas's Theorem. Now, we claim that they must all be distinct. Assume that, for some $i < j,$ we have
$${2^{n}-1 \choose i} \equiv {2^{n}-1 \choose j} \pmod{2^{n}}.$$Then,
\begin{align*}
    {2^{n}-1 \choose j} - {2^{n}-1 \choose i} 
    & = \frac{(2^{n}-1)! ((2^{n}-1-i)!i! - (2^{n}-1-j)!j!)}{i! j! (2^{n}-1-i)! (2^{n}-1-j)!}\\ 
    & = {2^{n}-1 \choose i} \cdot \frac{i!(2^{n}-1-j)! [(2^{n}-i-1) \cdots (2^{n}-j)-j \cdots (i+1)]}{j! (2^{n}-1-j)!} \\
    & = {2^{n}-1 \choose i} \cdot  \frac{i!}{j!} \cdot \left(\prod_{k = i+1}^{j} (2^{n}-k) - \prod_{k = i+1}^{j} k \right).
\end{align*}Notice that $j-i$ is even, so we can rewrite the first product as $\prod_{k = i+1}^{j}(k-2^{n}).$ So, when considering the number of factors of $2,$ the expression is just
$$\frac{i!}{j!} \cdot \sum_{k=1}^{j-i}2^{nk} S_{(j-i)-k}(i+1, \cdots, j).$$
Claim: $\frac{i!}{j!}S_{j-i-1}(i+1, \cdots, j)$ has a negative $2$-adic.

Proof: Notice that $$\frac{i!}{j!} S_{j-i-1}(i+1, \cdots, j) = \sum_{k=i+1}^{j}\frac{1}{k}.$$Consider the largest $k$ such that there exists an $\ell$ with $i < \ell \le j,$ and $\nu_{2}(\ell) = k.$ We claim that only such one $\ell$ can exist. Considering the smallest such $\ell,$ if there are more possible values, then $\ell+2^{k+1}$ must work as well, as it clearly has $2$-adic equal to $k.$ But, then, if both of these numbers are in the range $[i+1,j],$ then so is $\ell+2^{k},$ which has $2$-adic at least $k+1,$ a contradiction. So,
$$\sum_{k=i+1}^{j} \frac{1}{k} = \nu_{2}\left(\frac{1}{\ell}\right)=-k < 0,$$and thus we have proven the desired. $\blacksquare$

So, from the claim,
$$\frac{i!}{j!} \cdot \sum_{k=1}^{j-i}2^{nk} S_{(j-i)-k}(i+1, \cdots, j) $$$$= \nu_{2}\left(2^{n} \cdot \frac{i!}{j!} \cdot S_{j-i-1}(i+1, \cdots, j)+2^{2n}\left(\sum_{k=2}^{n} 2^{(k-2)n}S_{(j-i)-k}(i+1, \cdots, j)\right)\right) < n,$$and so
$2^{n} \nmid {2^{n}-1 \choose i} - {2^{n}-1 \choose j},$
as desired.
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Ilikeminecraft
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#19
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We show a stronger statement: Show that $\binom{2^n - 1}{2}, \binom{2^n - 1}{3}\dots\binom{2^n-1}{2^{n-1}-1}$ are $3, 5, \dots, 2^{n} - 3$ and $\binom{2^n-1}{1}\equiv-1, \binom{2^n-1}{0}\equiv1.$

We prove this with induction. Clearly, $n = 1$ is true. Now, assuming the statement for $n - 1$ is true, we will prove it for $n.$

Note that $\frac{\binom{2^n - 1}k}{\binom{2^n - 1}{k - 1}} = \frac{2^n- k}{k}.$ Note that $(\frac{2^n - k}k,2) = 1.$ Hence, if we define $a_i = \frac{2^n-k}{k}\pmod{2^n},$ it suffices to prove that there don't exist $i < j$ such that $\prod_{k = i}^j a_i\equiv1.$ Note that for even $i,$ this is equivalent to the previous sequence, and for odd $i,$ it is $-1.$ By induction hypothesis, we are done.
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scannose
1015 posts
#20
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choosing
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megarnie
5606 posts
#23
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Let $ n$ be a positive integer. Show that the numbers
\[ \binom{2^n - 1}{0},\; \binom{2^n - 1}{1},\; \binom{2^n - 1}{2},\; \ldots,\; \binom{2^n - 1}{2^{n - 1} - 1}\]are congruent modulo $ 2^n$ to $ 1$, $ 3$, $ 5$, $ \ldots$, $ 2^n - 1$ in some order.

Suppose otherwise and we had $a \ne b$ with $\binom{2^n -1}{a} \equiv \binom{2^n - 1}{b} \pmod{2^n}$ and $0 \le a < b < 2^{n-1}$. Note that $n = 1$ obviously works, so $n > 1$. Firstly, we see that for each integer $i$ in $[0,2^n - 1]$, $\binom{2^n - 1}{i}$ is odd by directly taking $\nu_2$: by Legendre's it becomes $s(i) + s(2^n - 1 - i) - s(2^n - 1)$ and then since you can't have any carries, this result is $0$.

We have \begin{align*}
\binom{2^n - 1}{a} - \binom{2^n - 1}{b} = (2^n - 1)! \left( \frac{1}{a! (2^n - 1 - a)!} - \frac{1}{b! (2^n - 1 - b)!} \right) \\
= \frac{(2^n - 1) ! (b! (2^n - 1 - b)!) - a! (2^n - 1 - a)!) }{a! b! (2^n - 1 - a)! (2^n - 1 - b)! } \\
= \frac{(2^n - 1) ! ((a+1) (a+2) \cdots b - (2^n - (a+1)) (2^n - (a+2)) \cdots (2^n - b)) }{b! (2^n - 1 - a)! } \\
= \frac{((a+1) (a+2) \cdots b - (2^n - (a+1)) (2^n - (a+2)) \cdots (2^n - b))}{(a+1)(a+2) \cdots b} \\
\end{align*}
Now, we will determine the $\nu_2$ of \[X =  (a+1) (a+2) \cdots b - (2^n - (a+1))(2^n - (a+2)) \cdots (2^n - b)\]
It must be at least $\nu_2((a+1)(a+2) \cdots b ) + n$ as $\binom{2^n - 1}{a} - \binom{2^n - 1}{b}$ has a $\nu_2$ of $\nu_2(X) - \nu_2((a+1)(a+2) \cdots b)$.

If a and b are opposite parities click hereNote that $X$ can be rewritten as \[ (a+1) (a+2) \cdots b - ((a+1) - 2^n)((a+2) - 2^n) \cdots (b - 2^n) \]
Again view the terms of in the expansion of $X$ when treated as a polynomial in $2^n$. Also let $k$ be the integer in $[a+1, b]$ with maximal $\nu_2$ (exists because if there were two or more, take the closest two and then average them).

Claim: The unique term in this expansion of $X$ that has a minimal $\nu_2$ is equal to $2^n \cdot \frac{(a+1)(a+2) \cdots b }{k}$.
Proof: Firstly, if a term has multiple copies of $2^n$, just replace the $2^n$ with some $a+i$ that is not included in the term, and we decrease the $\nu_2$, so it does not have a minimal $\nu_2$. Therefore, the term must have exactly one copy of $2^n$, after which we find that the term is equal to $2^n \cdot \frac{(a+1)(a+2) \cdots b}{i}$ for some integer $i \in [a+1,b]$, so we need to maximize $\nu_2(i)$ and this holds uniquely at $i = k$. $\square$

This implies that \[\nu_2(X) = \nu_2\left (2^n \cdot \frac{(a+1)(a+2) \cdots b}{k} \right) > \nu_2( 2^n \cdot (a+1)(a+2) \cdots b) = n + \nu_2((a+1)(a+2) \cdots b) ,\]absurd (where the inequality follows from the fact that $a < b$ are of the same parity, so there has to be an even number in $[a+1, b]$).
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