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jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
Circles tangent at orthocenter
Achillys   62
N 10 minutes ago by Rayvhs
Source: APMO 2018 P1
Let $H$ be the orthocenter of the triangle $ABC$. Let $M$ and $N$ be the midpoints of the sides $AB$ and $AC$, respectively. Assume that $H$ lies inside the quadrilateral $BMNC$ and that the circumcircles of triangles $BMH$ and $CNH$ are tangent to each other. The line through $H$ parallel to $BC$ intersects the circumcircles of the triangles $BMH$ and $CNH$ in the points $K$ and $L$, respectively. Let $F$ be the intersection point of $MK$ and $NL$ and let $J$ be the incenter of triangle $MHN$. Prove that $F J = F A$.
62 replies
Achillys
Jun 24, 2018
Rayvhs
10 minutes ago
Unsymmetric FE
Lahmacuncu   1
N 23 minutes ago by ja.
Source: Own
Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ that satisfies $f(x^2+xy+y)+f(x^2y)+f(xy^2)=2f(xy)+f(x)+f(y)$ for all real $(x,y)$
1 reply
Lahmacuncu
2 hours ago
ja.
23 minutes ago
find angle
TBazar   3
N 33 minutes ago by TBazar
Given $ABC$ triangle with $AC>BC$. We take $M$, $N$ point on AC, AB respectively such that $AM=BC$, $CM=BN$. $BM$, $AN$ lines intersect at point $K$. If $2\angle AKM=\angle ACB$, find $\angle ACB$
3 replies
TBazar
5 hours ago
TBazar
33 minutes ago
Polynomial and Irrational Coefficients
Iamsohappy   1
N an hour ago by tom-nowy
Find all natural numbers $n$ such that there exists a monic polynomial of degree $n$ and with irrational coefficients ( excepts its leading term ) and such that it has $n$ distinct irrational roots.
1 reply
Iamsohappy
Jul 5, 2019
tom-nowy
an hour ago
Summation
Saucepan_man02   5
N Today at 1:17 AM by Saucepan_man02
If $P = \sum_{r=1}^{50} \sum_{k=1}^{r} (-1)^{r-1} \frac{\binom{50}{r}}{k}$, then find the value of $P$.

Ans
5 replies
Saucepan_man02
May 3, 2025
Saucepan_man02
Today at 1:17 AM
Evan Chen Multivariable Calculus Book
Existing_Human1   7
N Yesterday at 7:16 PM by Blossom_tree_17
What do you guys think of Evan Chen's multivariable calculus book (or whatever you want to call it), through MIT? Is it useful for learning multivariable calculus? Here is a link: book
7 replies
Existing_Human1
Yesterday at 1:21 AM
Blossom_tree_17
Yesterday at 7:16 PM
Prove the statement
Butterfly   0
Yesterday at 11:13 AM
Given an infinite sequence $\{x_n\} \subseteq  [0,1]$, there exists some constant $C$, for any $r>0$, among the sequence $x_n$ and $x_m$ could be chosen to satisfy $|n-m|\ge r $ and $|x_n-x_m|<\frac{C}{|n-m|}$.
0 replies
Butterfly
Yesterday at 11:13 AM
0 replies
Morphism in a ring makes it a field
RobertRogo   1
N Yesterday at 10:16 AM by ysharifi
Source: Daniel Jinga, Ionel Popescu, RNMO SHL, 2003
Let $A$ be a ring with unity in which $1+1 \neq 0$ and there is a morphism $f$ from the group $(A,+)$ to the monoid $(A,\cdot)$ such that for all $a\in A\setminus \{0\}$, there is a $b \in A$ such that $f(b)=a^2$. Prove that $A$ is a field.
1 reply
RobertRogo
Tuesday at 4:02 PM
ysharifi
Yesterday at 10:16 AM
Putnam 2002 B3
ahaanomegas   2
N Yesterday at 7:36 AM by Rohit-2006
Show that for all integers $n>1$, \[ \dfrac {1}{2ne} < \dfrac {1}{e} - \left( 1 - \dfrac {1}{n} \right)^n < \dfrac {1}{ne}. \]
2 replies
ahaanomegas
Mar 12, 2012
Rohit-2006
Yesterday at 7:36 AM
Putnam 2013 B1
Kent Merryfield   13
N Yesterday at 5:38 AM by Rohit-2006
For positive integers $n,$ let the numbers $c(n)$ be determined by the rules $c(1)=1,c(2n)=c(n),$ and $c(2n+1)=(-1)^nc(n).$ Find the value of \[\sum_{n=1}^{2013}c(n)c(n+2).\]
13 replies
Kent Merryfield
Dec 9, 2013
Rohit-2006
Yesterday at 5:38 AM
Can cos(√2 t) be expressed as a polynomial in cost?
tom-nowy   2
N Yesterday at 4:54 AM by wh0nix
Source: Question arising while viewing https://artofproblemsolving.com/community/c51293h3562250
Can $\cos ( \sqrt{2}\,  t )$ be expressed as a polynomial in $\cos t$ with real coefficients?
2 replies
tom-nowy
May 6, 2025
wh0nix
Yesterday at 4:54 AM
Rolles theorem
sasu1ke   7
N Tuesday at 7:27 PM by GentlePanda24

Let \( f: \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that
\[
f(0) = 2, \quad f'(0) = -2, \quad \text{and} \quad f(1) = 1.
\]Prove that there exists a point \( \xi \in (0, 1) \) such that
\[
f(\xi) \cdot f'(\xi) + f''(\xi) = 0.
\]

7 replies
sasu1ke
May 3, 2025
GentlePanda24
Tuesday at 7:27 PM
Convergent series with weight becomes divergent
P_Fazioli   7
N Tuesday at 3:07 PM by P_Fazioli
Initially, my problem was : is it true that if we fix $(b_n)$ positive such that $b_n\underset{{n}\longrightarrow{+\infty}}\longrightarrow +\infty$, then there exists $(a_n)$ positive such that $\displaystyle\sum_{n\geq 0}a_n$ converges and $\displaystyle\sum_{n\geq 0}a_nb_n$ diverges ?

Thinking about the continuous case : if $g:\mathbb{R}_+\longrightarrow\mathbb{R}$ is continuous, positive with $g(x)\underset{{x}\longrightarrow{+\infty}}\longrightarrow +\infty$, does $f$ continuous and positive exist on $\mathbb{R}_+$ such that $\displaystyle\int_0^{+\infty}f(x)\text{d}x$ converges and $\displaystyle\int_0^{+\infty}f(x)g(x)\text{d}x$ diverges ?

To the last question, the answer seems to be yes if $g$ is in the $\mathcal{C}^1$ class, increasing : I chose $f=\dfrac{g'}{g^2}$. With this idea, I had the idea to define $a_n=\dfrac{b_{n+1}-b_n}{b_n^2}$ but it is not clear that it is ok, even if $(b_n)$ is increasing.

Now I have some questions !

1) The main problem : is it true that if we fix $(b_n)$ positive such that $b_n\underset{{n}\longrightarrow{+\infty}}\longrightarrow +\infty$, then there exists $(a_n)$ positive such that $\displaystyle\sum_{n\geq 0}a_n$ converges and $\displaystyle\sum_{n\geq 0}a_nb_n$ diverges ? And if $(b_n)$ is increasing ?
2) is it true that if we fix $(b_n)$ positive increasing such that $b_n\underset{{n}\longrightarrow{+\infty}}\longrightarrow +\infty$
and $\frac{b_{n+1}}{b_n}\underset{{n}\longrightarrow{+\infty}}\longrightarrow 1$, then $\displaystyle\sum_{n\geq 0}\left(\frac{b_{n+1}}{b_n}-1\right)$ diverges ?
3) is it true that if we fix $(b_n)$ positive increasing such that $b_n\underset{{n}\longrightarrow{+\infty}}\longrightarrow +\infty$
and $\frac{b_{n+1}}{b_n}\underset{{n}\longrightarrow{+\infty}}\longrightarrow 1$, then $\displaystyle\sum_{n\geq 0}\frac{b_{n+1}-b_n}{b_n^2}$ converges ?
4) if $(b_n)$ is positive increasing and such that $b_n\underset{{n}\longrightarrow{+\infty}}\longrightarrow +\infty$
and $\frac{b_{n+1}}{b_n}$ does not converge to $1$, can $\displaystyle\sum_{n\geq 0}\frac{b_{n+1}-b_n}{b_n^2}$ diverge ?
5) for the continuous case, is it true if we suppose $g$ only to be continuous ?

7 replies
P_Fazioli
May 5, 2025
P_Fazioli
Tuesday at 3:07 PM
Cool Integral, Cooler Solution
Existing_Human1   2
N May 6, 2025 by ysharifi
Source: https://youtu.be/YO38MCdj-GM?si=DCn6DaQTeX8RXhl0
$$\int_{0}^{\infty} \! e^{-x^2}\cos(5x) \,dx$$
Bonus points if you can do it without Feynman
2 replies
Existing_Human1
May 6, 2025
ysharifi
May 6, 2025
Prove that expression is always even.
shivangjindal   20
N Apr 18, 2025 by EVKV
Source: INMO 2014- Problem 2
Let $n$ be a natural number. Prove that,
\[ \left\lfloor \frac{n}{1} \right\rfloor+ \left\lfloor \frac{n}{2} \right\rfloor + \cdots + \left\lfloor \frac{n}{n} \right\rfloor + \left\lfloor \sqrt{n} \right\rfloor \]
is even.
20 replies
shivangjindal
Feb 2, 2014
EVKV
Apr 18, 2025
Prove that expression is always even.
G H J
Source: INMO 2014- Problem 2
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shivangjindal
676 posts
#1 • 2 Y
Y by Adventure10, cubres
Let $n$ be a natural number. Prove that,
\[ \left\lfloor \frac{n}{1} \right\rfloor+ \left\lfloor \frac{n}{2} \right\rfloor + \cdots + \left\lfloor \frac{n}{n} \right\rfloor + \left\lfloor \sqrt{n} \right\rfloor \]
is even.
This post has been edited 1 time. Last edited by ahaanomegas, Feb 2, 2014, 2:31 PM
Reason: Fixed LaTeX
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Altricono
42 posts
#2 • 5 Y
Y by Adventure10, Mango247, cubres, Japnoor2411, and 1 other user
Just induct. The idea is common.
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Rust
5049 posts
#3 • 7 Y
Y by makar, ashrith9sagar_1, Adventure10, Mango247, cubres, aidan0626, Japnoor2411
$\sum_{k=1}^n [\frac{n}{k}]=\sum_{k=1}^n\tau(k)$. $\tau(k)$ - is number of divisors of k.
If $k\not =m^2$ divisors $d,\frac{k}{d}$ different and $\tau(k)$ is even.
$\tau(k)$ is odd only if $k=m^2$. Therefore
$\sum_{k=1}^n[\frac{n}{k}]+[\sqrt n]=2[\sqrt n]\mod 2 =0\mod 2$.
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mathuz
1524 posts
#4 • 6 Y
Y by DanDumitrescu, Adventure10, Mango247, Math_DM, cubres, Japnoor2411
[Moderator: Please do not double-post and/or quote the post immediately before yours.]

$ \sum_{k=1}^n \left \lfloor \frac{ n }{ k }\right \rfloor=\sum_{k=1}^n \tau(k) $.
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Shravu
131 posts
#5 • 6 Y
Y by Pluto1708, ashrith9sagar_1, Adventure10, Mango247, cubres, Japnoor2411
Yes, Induction....

gif(n+1/k) > gif(n/k) if and only if k|n+1 gif(n+1/k) = gif(n/k)+1 so there is pairing

and if n+1 is a square gif(root(n+1)) > gif(root(n))
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Askr
181 posts
#6 • 8 Y
Y by Unknown-6174, Uagu, SHREYAS333, stevoo, Adventure10, Mango247, cubres, Japnoor2411
We prove the statement by Induction.

For n = 1, [1/1] + [1^(1/2)] = 1+1 = 2 (Base Case)
Lets assume that for n = n, [n/1] + [n/2] + [n/3] + .............. + [n/n] + [n^(1/2)] = even
Now, for n=n+1, we take two cases :

(i) n+1 is not a perfect square :
In this case n+1 has k factors (excluding 1 and n+1) where k is even.
If b is not a factor of n+1, then [(n+1)/b] = [n/b].
If b is a factor of n+1, then [(n+1)/b] = [n/b] +1. There will be k such cases.
Also, [(n+1)^(1/2)] = [n^(1/2)] since n+1 is not a perfect square.
therefore, [(n+1)/1] + [(n+1)/2] + [(n+1)/3] + ............... + [(n+1)/n] + [(n+1)/(n+1)] + [(n+1)^(1/2)]
= [n/1] + [n/2] + [n/3] + ............. + [n/n] + [n^(1/2)] + k + 2
= even (since k is even)

(ii) n+1 is a perfect square :
In this case n+1 has k factors (excluding 1 and n+1) where k is odd.
If b is not a factor of n+1, then [(n+1)/b] = [n/b].
If b is a factor of n+1, then [(n+1)/b] = [n/b] +1. There will be k such cases.
Also, [(n+1)^(1/2)] = [n^(1/2)] + 1 since n+1 is a perfect square.
therefore, [(n+1)/1] + [(n+1)/2] + [(n+1)/3] + ............... + [(n+1)/n] + [(n+1)/(n+1)] + [(n+1)^(1/2)]
= [n/1] + [n/2] + [n/3] + ............. + [n/n] + [n^(1/2)] + k + 3
= even (since k is odd)

We have proved both the cases.
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makar
1581 posts
#7 • 11 Y
Y by isomorphicdevikram, mathbuzz, Pluto1708, AlastorMoody, ashrith9sagar_1, Supercali, Adventure10, Mango247, cubres, TestX01, Japnoor2411
shivangjindal wrote:
Let $n$ be a natural number. Prove that,
\[ \left\lfloor \frac{n}{1} \right\rfloor+ \left\lfloor \frac{n}{2} \right\rfloor + \cdots + \left\lfloor \frac{n}{n} \right\rfloor + \left\lfloor \sqrt{n} \right\rfloor \]
is even.

By counting number of Lattice point in first quadrant and below or on the curve $xy=n$ in two different ways, we get:

$\left\lfloor \frac{n}{1} \right\rfloor+ \left\lfloor \frac{n}{2} \right\rfloor + \cdots + \left\lfloor \frac{n}{n} \right\rfloor =2\sum _{k=1}^{\left\lfloor \sqrt{n} \right\rfloor }\left\lfloor \frac{n}{k} \right\rfloor -\left\lfloor \sqrt{n} \right\rfloor^2 $ :!:

Now

$\left\lfloor \frac{n}{1} \right\rfloor+ \left\lfloor \frac{n}{2} \right\rfloor + \cdots + \left\lfloor \frac{n}{n} \right\rfloor + \left\lfloor \sqrt{n} \right\rfloor=\underbrace{2\sum _{k=1}^{\left\lfloor \sqrt{n} \right\rfloor }\left\lfloor \frac{n}{k} \right\rfloor}_\text{Even} -\underbrace{ \left\lfloor \sqrt{n} \right\rfloor\left(\left\lfloor \sqrt{n} \right\rfloor-1\right)}_\text{Even}=\text{Even} $ :)
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mathbuzz
803 posts
#8 • 8 Y
Y by AHKR, div5252, maths_arka, Adventure10, Mango247, Math_DM, cubres, Japnoor2411
Non-inductive solution --
Lemma -- $[\frac{n}{1}]+......+[\frac{n}{n}]=d(1)+d(2)+....d(n)$
Proof -- the number of elements in S={1,2,...,n} ,which are divisible by $q$ is $[\frac{n}{q}]$.
Now , the rest is trivial
end of lemma
Back to the main problem --
Now , note that ,
$\sum[\frac{n}{k}]+[\sqrt{n}] =d(1)+d(2)+.....+d(n)+[\sqrt{n}]=f(n)$ (say)
Now , note that , $d(k)=1(mod2)$ iff $k$ is a perfect square ..
There are only $[\sqrt{n}]$ perfect squares in {1,2,...,n}.
hence , $f(n)=[\sqrt{n}]+[\sqrt{n}]=0(mod2)$. :D
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BSJL
641 posts
#9 • 4 Y
Y by Adventure10, Mango247, cubres, Japnoor2411
We just need to notice that the following two points:

1. For $ n=1 $, It is true.

2.$ ([\frac{n+1}{1}]+......+[\frac{n+1}{n+1}]+[\sqrt{n+1}])-([\frac{n}{1}]+......+[\frac{n}{n}]+[\sqrt{n}])=2d(n+1) $
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AlastorMoody
2125 posts
#12 • 4 Y
Y by Arhaan, Adventure10, cubres, Japnoor2411
INMO 2014 P2 wrote:
Let $n$ be a natural number. Prove that,
\[ \left\lfloor \frac{n}{1} \right\rfloor+ \left\lfloor \frac{n}{2} \right\rfloor + \cdots + \left\lfloor \frac{n}{n} \right\rfloor + \left\lfloor \sqrt{n} \right\rfloor \]is even.
Solution: Notice, for $n,k \in \mathbb{N}$, $\left \lfloor \frac{n}{k} \right \rfloor$ denotes the number of positive multiples of $k$ below $\left \lfloor \frac{n}{k} \right \rfloor$. Also, any multiple in the form of $k \times \ell \le n$ will be counted twice, once in $\left \lfloor \frac{n}{k} \right \rfloor$ and in $ \left \lfloor \frac{n}{\ell} \right \rfloor$ where $k, \ell$ are distinct. Only Numbers that can be represented in the form of squares (eg: $n \times n$) are not counted twice, but adding a $\lfloor {\sqrt{n}} \rfloor$ to the expression makes all the counts twice and hence the resultant is even $\qquad \blacksquare$
This post has been edited 1 time. Last edited by AlastorMoody, Sep 5, 2019, 4:52 PM
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Mogmog8
1080 posts
#13 • 3 Y
Y by centslordm, cubres, Japnoor2411
It is well known that $\sum_{i=1}^n\left\lfloor\frac{n}{i}\right\rfloor=\sum_{i=1}^n d(i),$ where $d(i)$ denotes the number of factors of $i.$ Notice $d(i)\equiv 1\pmod{2}$ if and only if $i$ is a perfect square. We know there are $\left\lfloor\sqrt{n}\right\rfloor$ perfect squares from $1$ to $n$ so $$\left(\sum_{i=1}^n d(i)\right)+\left\lfloor\sqrt{n}\right\rfloor\equiv 2\left\lfloor\sqrt{n}\right\rfloor\equiv 0\pmod{2}.$$$\square$
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minusonetwelth
225 posts
#14 • 3 Y
Y by panche, cubres, Japnoor2411
We will first prove the following well-known result:
Claim:
If $d(n)$ is the number of divisors of a positive integer $n$, then
$$d(1)+d(2)+\ldots+d(n)=\lfloor\frac{n}{1}\rfloor+\lfloor\frac{n}{2}\rfloor+\ldots+\lfloor\frac{n}{n}\rfloor$$Proof of the claim:
\begin{align*}
			d(1)+d(2)+\ldots+d(n)&=\sum_{k=1}^{n}{\sum_{i|k}{1}}\\
			&=\sum_{i=1}^{n}{\sum_{\substack{1\le j\le n\\ i|j}}{1}}\\
			&=\sum_{i=1}^{n}{\lfloor\frac{n}{i}\rfloor}=\lfloor\frac{n}{1}\rfloor+\lfloor\frac{n}{2}\rfloor+\ldots+\lfloor\frac{n}{n}\rfloor
		\end{align*}as desired.

Now we proceed with induction. The base case $n=1$ is clear. Assume that $2| d(1)+\ldots+d(k)+\lfloor\sqrt{k}\rfloor$. Then for $k+1$
\begin{align*}
		2&|d(1)+d(2)+\ldots+d(k)+d(k+1)+\lfloor\sqrt{k+1}\rfloor\Longleftrightarrow\\
		2&|d(1)+d(2)+\ldots+d(k)+d(k+1)+\lfloor\sqrt{k+1}\rfloor+\lfloor\sqrt{k}\rfloor-\lfloor\sqrt{k}\rfloor\Longleftrightarrow\\
		2&| d(k+1)+\lfloor\sqrt{k+1}\rfloor-\lfloor\sqrt{k}\rfloor
	\end{align*}If $k+1$ is a square, then $d(k+1)$ is odd, while $\lfloor\sqrt{k+1}\rfloor-\lfloor\sqrt{k}\rfloor=1$. If $k+1$ is not a square, $d(k+1)$ is even, and $\lfloor\sqrt{k+1}\rfloor-\lfloor\sqrt{k}\rfloor=0$, finishing the proof.
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lifeismathematics
1188 posts
#15 • 2 Y
Y by Math_DM, Japnoor2411
first of all we have $\left \lfloor \frac{n}{1} \right \rfloor +\left \lfloor \frac{n}{2}  \right \rfloor +\cdots+\left \lfloor \frac{n}{n}  \right \rfloor=\tau(1)+\tau(2)+\cdots+\tau(n)$

clearly, $\tau(j) \equiv 0 \pmod 2$ if $ j$ is not a perfect square, and $\equiv 1 \pmod 2$ if it is a perfect square. We notice in set $\{1,2 , \cdots , n\}$ we have $\lfloor \sqrt{n} \rfloor$ many perfect squares, so we have \[ \left\lfloor \frac{n}{1} \right\rfloor+ \left\lfloor \frac{n}{2} \right\rfloor + \cdots + \left\lfloor \frac{n}{n} \right\rfloor + \left\lfloor \sqrt{n} \right\rfloor \]$\equiv 2 \lfloor \sqrt{n} \rfloor \equiv 0 \pmod 2$ as there are just $\lfloor \sqrt{n} \rfloor$ many perfect squares in $[n]$ so done. $\blacksquare$
This post has been edited 1 time. Last edited by lifeismathematics, Aug 11, 2023, 12:52 PM
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AshAuktober
1005 posts
#16 • 1 Y
Y by Japnoor2411
Basic idea:
d(1) + d(2) + \cdots + d(n) + [\sqrt{n}
and parity changes only when n is a square
where parity of [sqrt(n)] changes too
so parity of expression is invariant
now we see at n = 1 its even
so its always even
qed
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Siddharthmaybe
106 posts
#17 • 1 Y
Y by Japnoor2411
[n/1] + [n/2] +.... + [n/n] = d(1) + d(2) +.... d(n) (lemma)
no. of divisors of a square is odd
Case 1->
there are an even number of naturals <= [rootn] (the squares till n are even in no. )
<=> [rootn] is even and the sum of all d(i) from 1 to [rootn] is obv even and we are done
similar logic for the other case works :)
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cursed_tangent1434
623 posts
#18 • 1 Y
Y by Japnoor2411
This problem really made me appreciate how theory dependent Olympiad number theory really was. Knowing that one trick really made solving the problem much easier. The following lemma is essential.

Lemma : For all positive integers $n$ and $k$,
\[\left\lfloor \frac{n}{k} \right\rfloor - \left\lfloor \frac{n-1}{k} \right\rfloor = \begin{cases}1 & k \mid n \\ 0 & k \nmid n\end{cases}\]


Proof : The proof is clear. $[\left\lfloor \frac{n}{k} \right\rfloor  > \left\lfloor \frac{n-1}{k} \right\rfloor$ if and only if we jump to a new multiple of $k$.

Now, we proceed via induction. Note how the problem is clear for $n=1$. Now, assuming the problem when $n=k$ we show the result when $n=k+1$. Note,
\begin{align*}
\left(\sum_{i=1}^{k+1} \left \lfloor \frac{k+1}{i} \right \rfloor + \sqrt{k+1} \right) - \left(\sum_{i=1}^{k} \left \lfloor \frac{k}{i} \right \rfloor + \sqrt{k} \right ) &= d(k+1) + \sqrt{k+1}- \sqrt{k}
\end{align*}
Now, if $k+1$ is not a perfect square, $d(k+1)$ is even, and $\sqrt{k+1}=\sqrt{k}$. Thus, the result holds when $n=k+1$ as well. Further if $k+1$ is a perfect square, $\sqrt{k+1}-\sqrt{k}=1$ but $d(k+1)$ is odd, and thus again the result holds for $n=k+1$ and we are done.
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eg4334
637 posts
#19 • 1 Y
Y by Japnoor2411
It rewrites into $d(1)+d(2) + \dots + d(n) + \lfloor \sqrt{n} \rfloor$. All the $d(i)$ are even except the squares but the latter term makes up for it by adding one for each square.
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LeYohan
43 posts
#21 • 1 Y
Y by Japnoor2411
It's known that $\left\lfloor \frac{n}{1} \right\rfloor+ \left\lfloor \frac{n}{2} \right\rfloor + \cdots + \left\lfloor \frac{n}{n} \right\rfloor = d(1) + d(2) + \cdots + d(n)$ where $d(x)$ is the number of divisors of $x$, so the problem is now equivalent to showing that $d(1) + d(2) + \cdots + d(n) + \left\lfloor \sqrt{n} \right\rfloor$ is even.

Let's express $n$ as $k^2 + l$ where $0 \leq l \leq 2k$, and let's consider two cases:

Case 1: $k$ is odd.

In this case, there is an odd number of squares bewteen $1$ and $n$ inclusive, and notice that it does not matter where $n$ lies between $k^2$ and $(k+1)^2$ since we remember that $d(x)$ is odd if and only if $x$ is a square. This means that $d(1) + d(2) + \cdots + d(n)$ is odd, and $\left\lfloor \sqrt{n} \right\rfloor = k$ which is odd, so the total sum is even. $\square$

Case 2: $k$ is even.

Similarly, we don't care where $n$ lies between $k^2$ and $(k+1)^2$ and notice that there is an even number of squares between $1$ and $n$ inclusive. Therefore, $d(1) + d(2) + \cdots + d(n)$ is even and $\left\lfloor \sqrt{n} \right\rfloor = k$, which is also even, so the total sum is even. $\square$
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jasperE3
11305 posts
#22 • 1 Y
Y by Japnoor2411
shivangjindal wrote:
Let $n$ be a natural number. Prove that,
\[ \left\lfloor \frac{n}{1} \right\rfloor+ \left\lfloor \frac{n}{2} \right\rfloor + \cdots + \left\lfloor \frac{n}{n} \right\rfloor + \left\lfloor \sqrt{n} \right\rfloor \]is even.

Claim: $\tau(n+1)+\left\lfloor\sqrt{n+1}\right\rfloor-\left\lfloor\sqrt n\right\rfloor$ is even for all $n\ge0$

Let $\left\lfloor\sqrt{n+1}\right\rfloor=k$.

Case 1: $n+1$ is a square
We claim that $\left\lfloor\sqrt{n+1}\right\rfloor-\left\lfloor\sqrt n\right\rfloor=1$.

Write $n+1=k^2$. We have, since $k\ge0$:
$$k^2-1\ge(k-1)^2\Rightarrow n\ge(k-1)^2$$as well as $n<n+1=k^2$. Then $k-1\le\sqrt n<k$ so we must have $\left\lfloor\sqrt n\right\rfloor=k-1$, so:
$$\left\lfloor\sqrt{n+1}\right\rfloor-\left\lfloor\sqrt n\right\rfloor=1.$$
Then since $\tau(n+1)$ is odd, our claim is proven for this case.

Case 2: $n+1$ isn't a square
We claim that $\left\lfloor\sqrt{n+1}\right\rfloor-\left\lfloor\sqrt n\right\rfloor=0$.

By the definition of $k$, we have $k^2\le n+1<(k+1)^2$, but since $n+1$ isn't a square, this inequality is strict on both sides. Then $k^2\le n<n+1<(k+1)^2$ so $\left\lfloor\sqrt n\right\rfloor=k$, hence $\left\lfloor\sqrt{n+1}\right\rfloor-\left\lfloor\sqrt n\right\rfloor=0$.

Then since $\tau(n+1)$ is odd, our claim is proven.



We finish the problem by summing this from $0$ to $n-1$, giving that:
$$\sum_{i=1}^n\tau(i)+\left\lfloor\sqrt n\right\rfloor$$is even, and we're done since $\sum_{i=1}^n\tau(i)=\sum_{i=1}^n\left\lfloor\frac ni\right\rfloor$.
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alexanderhamilton124
391 posts
#23 • 3 Y
Y by Nobitasolvesproblems1979, Japnoor2411, L13832
Breh
Let us induct. We are clearly done for \( n = 1 \), so suppose we are done for \( n \in \{1, \dots, k\} \). Let \( f(n) \) be our expression. Note that

\[
\left\lfloor \frac{k + 1}{x} \right\rfloor - \left\lfloor \frac{k}{x} \right\rfloor = 1 \text{ if } x \mid (k + 1), \text{ and } 0 \text{ otherwise}.
\]
Hence:

\[
f(k + 1) - f(k) = d(k + 1) + \left\lfloor \sqrt{k + 1} \right\rfloor - \left\lfloor \sqrt{k} \right\rfloor
\]
If \( k + 1 \) is not a perfect square, \( d(k + 1) \) is even and

\[
\left\lfloor \sqrt{k + 1} \right\rfloor = \left\lfloor \sqrt{k} \right\rfloor
\]
which gives \( f(k + 1) - f(k) \) even, so \( f(k + 1) \) even as well, and we're done.

If \( k + 1 \) is a perfect square, \( d(k + 1) \) is odd and

\[
\left\lfloor \sqrt{k + 1} \right\rfloor = \left\lfloor \sqrt{k} \right\rfloor + 1,
\]
so \( f(k + 1) - f(k) \) is even again and we're done.
This post has been edited 1 time. Last edited by alexanderhamilton124, Mar 17, 2025, 10:19 PM
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EVKV
71 posts
#24
Y by
Trivial nt in INMO what a suprise :joy:

Just induction and use the fact that d(n)= odd iff n is a perfect square
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